Welcome to lecture number 33 on measure and integration. From today onwards, we will be looking at some special topics in measure and integration. We will start with looking at how to define integral for complex valued functions defined on measure spaces. So, topic for today's discussion is going to be integrating complex valued functions. So, for the coming one or two lectures, we will be fixing a measure space X as mu, which is sigma finite and complete. So, all the discussion will be on a fixed sigma finite complete measure space X as mu. We will denote by this letter c with a line in between that is called script c to be the field of complex numbers. For a complex valued function f defined on x taking values in c, we say that we define its real part and imaginary part. So, for any point x f of x is going to be a complex number because its values are in the complex field. So, that complex number f of x has got a real part and imaginary part. So, we define real f at the point x to be the real part of the value f of x. Similarly, imaginary f at x to be the imaginary part of the value f of x. .. So, let us observe that x going to real f x and x going to imaginary f of x are real valued functions. The first one is called the real part of the function f and imf is called the imaginary part of the function f. So, every function f which is complex valued has got a real part function and the imaginary part function and both are real valued functions. So, let us define what is called the measurability of a complex valued function. So, a complex valued function f from x to c is said to be measurable if both the real part of f and the imaginary part of f are measurable functions. So, note that f is equal to the real part f plus imaginary part of f. So, if both real part of f and imaginary part of f which are real valued functions are measurable on x with respect to the sigma algebra s, then we say that the function f itself is a measurable function. We will say f is integrable with respect to the measure mu if both the real part f and imaginary part f are integrable functions. So, real part f is a real valued function and imaginary f is also a real valued function. So, if both of them are integrable as real valued functions, then we say that the function f which is complex valued is integrable and we define the integral of f to be denoted by the symbol integral f d mu to be equal to integral of the real part of f plus i times the integral of the imaginary part of f, where this i is the square root of minus 1 which is normally used to write complex numbers. So, for a complex valued function f, we define it to be measurable if both real part and imaginary part are measurable and similarly we define f to be we say f to be integrable if both the real part and the imaginary part of f are integrable and that case we define the integral of f to be equal to integral of the real part plus i times the integral of the imaginary part of the function f. So, that is the definition of the integral of a complex valued function. .. We will study properties of this integral. Let us also denote the set of all complex valued integrable functions on x by l 1 x s mu. That is the symbol we are used to denote the integrable functions real valued integrable functions. We will denote the space of complex valued integrable functions on x s mu by the same symbol. In case we are referring to specifically the real valued functions, we will put a suffix r l lower 1 upper r x s mu. So, that will denote the space of all real valued integrable functions on x s mu. So, whenever need be and we want to specify that we are in the space of real valued functions integrable we will use this symbol. Otherwise the space of all complex valued integrable functions will be denoted by the symbol l 1 x s mu. So, next we will like to study properties of integral namely we will like to study the space l 1 of x s mu the space of integrable functions. So, let us start with the very basic property. So, let us take a function f which is a measurable function f is a measure complex valued measurable function on s. Then the first property as for the case of real valued functions is that f is integrable if and only if mod f which is a real valued function is integrable with respect to is integrable real valued integrable function. And further in that case we want to claim that the integral of mod f modulus of the integral is less than or equal to integral of the absolute value. So, let us look at how does one prove these properties. So, we have got a function f which is l 1 of x s mu. So, that means what is the definition of this? That says that the real part of f and the imaginary part of f are both integrable functions l 1 of x. Now, let us consider. So, look at the function absolute value of f.  So, that is the definition of because it is a complex function complex valued function. So, absolute value of f can be written this way. And from here it is easy to see that this is always less than or equal to 2 times square root 2 times the absolute value of the real part of f plus absolute value of imaginary part of f. This is very easy in equality about complex numbers. So, it is basically saying that if you take a complex number then it is always less than or equal to square root. The absolute value of the complex number is always less than or equal to square root of 2 times the real part plus the imaginary part. One way of looking at this would be if you look at this definition. So, this is real part of f square plus imaginary part of f square. So, this will be less than or equal to 2 times the maximum value of real part or the imaginary part. And that is less than or equal to the absolute value of the real part plus the absolute value of the imaginary part. So, this term in the under the square root you can easily see it is less than or equal to 2 times the absolute value of the real part of f plus imaginary part of f. So, square root is less than or equal to this quantity. So, this is very easy in equality to prove for complex numbers. So, let me just write it as an exercise that you verify it yourself.  So, these are both integrable functions. So, mod f is a non-negative real valued function less than or equal to 2 times an integrable function. So, this will imply that mod f belongs to is a real valued integrable function. So, this inequality implies that this is integrable with respect to mod f is. So, if f is integrable, so that implies mod f is a real valued integrable function. Let us look at the converse part. So, let us suppose that so conversely let mod f be integrable. So, is l 1 r of x. Then the real part of f absolute value is less than or equal to absolute value of f. That is a very simple state forward in equality that for any complex number the real part is less than or equal to absolute value of the. So, and similarly the imaginary part of f is also less than or equal to absolute value of f. So, these are real valued functions and they are less than or equal to a function f mod f which is integrable.  So, we have proved. So, this proves so hence f belonging to l 1 of x if and only if mod f belongs to l 1 r of x. So, that proves the first part of the statement. So, f is integrable if and only if mod f is integrable. Let us look at the second part. So, we want to show now that mod of integral f d mu is less than or equal to integral of mod f d mu. So, this is what we want to show. So, let us write. So, let us denote by alpha the number on the left hand side.  Note that f d mu is a complex number and its absolute value is denoted by alpha. So, we can write.  So, you can write as alpha times a raise to the power i theta for some theta between 0 and 2 pi. So, every complex number can be written as its absolute value at times e raise to the power i theta for some theta between.  And this we will just show in the next property that integral of f a scalar multiple is same as I can write this as e raise to the power minus i theta times f d mu. We can write this as this. We will just in a minute we will prove this property that for any scalar complex scalar times the integral of f is same as integral of scalar times f whenever f is integrable. So, using that property we will have. So, we will just put it that we are going to prove this property soon. So, this is equal to this. So, let us write e raise to the power minus i theta f. It is a complex value function. So, it will have the real part.  So, let this be equal to this. Then alpha which is equal to e raise to the power minus i theta f d mu its integral I can write it as integral can be written as integral of f 1 d mu plus i times integral of f 2 d mu. Because this function e raise to the power i theta f has got real part f 1 imaginary part f 2. So, its integral must be equal to integral of the real part plus i times integral of the imaginary part. But note alpha is real because what was alpha? Alpha was nothing but the absolute value of the integral of f d mu. So, it is a non-negative real number.  So, imaginary part must be 0. So, that implies that integral f 2 d mu is equal to 0 and alpha must be equal to integral of f 1 d mu. Alpha is non-negative. So, that implies I can write this also equal to absolute value of integral f d mu because alpha is a non-negative real number. So, that is equal to f 1 d mu and that for real valued functions we know this is less than or equal to integral of mod f 1 d mu. So, here we are using the property that for a real valued integrable function integral of f 1 d mu absolute value is less than or equal to absolute value of the integral. So, that is this. As a next step, let us observe. So, f 1 was real part of e raise over i theta times f. That was a real part of this function. So, that means what? That means that absolute value of f 1 is less than or equal to absolute value of e raise power minus i theta of f. So, real part of any complex number is less than the absolute value of that complex number. But e raise power i theta is a complex number of absolute value 1. So, this is same as mod of f. So, absolute value of f 1 is less than or equal to absolute value of f. So, that implies that integral of mod f 1 d mu is less than or equal to integral of mod f d mu. So, now let us combine these two facts that alpha was equal to less than or equal to integral of f 1 d mu and integral f 1 d mu is less than or equal to integral of mod f d mu. So, these two facts together imply that integral of f d mu which was equal to alpha is less than or equal to integral of mod f d mu. So, that proves the fact that f is integrable if and only if mod f is integrable and integral of absolute value of the integral f d mu is less than or equal to integral of the absolute value of the function. So, this proves the first property about integrals that for a integrable function f is integrable a complex valued function f is integrable if and only if absolute value of this complex valued function which is a real valued function is integrable and integral of absolute value of the integral f d mu is less than or equal to integral of f d mu. So, that proves the first property. Next we want to prove the linearity property part of which we have used already in the previous proof namely that if f and g are integrable functions complex valued integrable functions and alpha and beta are complex numbers then alpha f plus beta g the linear combination is also integrable and integral of alpha f plus beta g d mu is equal to alpha times integral f plus beta times integral g. So, to prove this inequality we will split the proof into two parts.  Let us prove this part first because alpha times let us look at proof of this. So, let us write alpha as a plus i b it is a complex number.     So, they are real valued integrable functions.   So, the difference is integrable.   So, I can write now the integral of this function. So, integral of alpha f d mu by definition it is integral of the real part plus i times integral of the imaginary part.  So, that is integral of alpha f. So, let us now use the properties that integrals is a linear operation for real valued function. So, the first integral I can write it as a times integral of real part of f minus b times integral of imaginary part of f d mu d mu.   Just open up the bracket and that is same as this. That means this is equal to alpha times integral of f d mu. So, we have shown that integral of alpha f d mu is equal to alpha times integral of f d mu. So, that is the first part of the linearity property that if I take a function f which is l 1 and multiply it by scalar alpha, then that is also integrable and integral of alpha f is equal to alpha times integral of f. Now, let us look at the second part of a requirement.  So, that is easy to verify because mod of f plus g, we want to show that is integrable. So, let us look at note. So, note as in real case is less than or equal to absolute value of f plus absolute value of g.   So, that is finite. So, that implies that f plus g is integrable is an integrable function. And to compute the integral of f plus g, that is simple thing. So, let us observe to compute the integral.  That is easy to verify and the imaginary part of f plus g is equal to imaginary part of f plus imaginary part of g.  So, real part of f plus g will be integrable. Similarly, imaginary part of f plus g is also integrable and further we can write down the integrals. So, the integral of f plus g d mu by definition is equal to integral of the real part of f plus g plus integral of the imaginary part of f plus g by definition.  So, this is the first one is real part of f plus real part of g d mu plus integral imaginary part of f plus imaginary part of g d mu. So, that follows from the simple result that we have just now shown. Now, these are all real valued functions. So, integration is linear. So, that splits into four integrals. So, integral of real part of f d mu plus integral of real part of g d mu. So, there is i here because real and imaginary plus i times integral of the imaginary part of f plus i times integral of imaginary part of g d mu. Now, the first and the third term combined together will give you that is equal to integral of f d mu plus integral of g d mu. So, integral of f plus g is equal to integral of f plus integral of g. So, that proves property three completely namely integral of f plus g is equal to integral of f plus integral of g. Let us look at some more properties. So, let us look at the property third property namely if f is an integrable function and e is any measurable set then the indicator function of e times f is also a integrable function and we write this as integral of the indicator function of e times f d mu is written as integral of f over e. So, that is one property and we want to show something more namely that if e and f are disjoint measurable sets then integral of f over e union e 2 is same as integral over e 1 plus integral over e 2. So, let us approve this property. Let us take a function f which is l 1 and e is a set which is measurable. Now, we want to look at e times indicator function of e times f. So, that is same as indicator function of e times real part of f plus i times imaginary part of f which I can write as indicator function of e times real part of f plus i times indicator function of e multiplied with the imaginary part of f. So, what we are saying is that for the function indicator function of e times f the real part is indicator function of e times v f part of f and this imaginary part is indicator function of e times the imaginary part of f. integral of. Because, real part of f is integrable, imaginary part of f is integrable. So, multiplying with the indicator function of E also gives them integrable. So, both real part of the indicator function of E times f is integrable, and the imaginary part of indicator function of E times f which is indicator function of E times imaginary part of f, that are real valued integrable functions. And, hence we can write the indicator function of E f, which we are denoting by integral over E of f d mu to be equal to integral of the indicator function of E of the real part of f plus i times integral over E of the imaginary part of f. And that if you recall, we had denoted it by integral over E of the real part of f plus at i or sometime also called data integral over E imaginary part of f d mu. So, what we are saying is the following, namely for an integrable function for f integrable and E belonging to a measurable set, integral of over E of f d mu is well defined and is nothing but integral over E of the real part of f plus i times the integral of imaginary part of f over E d mu and also d mu. Now, let us come to the second part. Now, suppose E 1 and E 2 are two sets, which are measurable and they are disjoint. E 1 intersection E 2 is empty. In that case, by the above claim integral over E 1 union E 2 of f d mu will be equal to integral over E 1 union E 2 of real part of f d mu plus i times integral over E 1 union E 2 of imaginary part of f d mu. Now, let us recall that for real valued integrable functions, integral over E 1 union E 2 whenever E 1 and E 2 are disjoint is nothing but the integral over E 1 plus the integral over E 2. So, this right hand side, I can use that property and write it further as, so that is equal to, I can write the right hand side as equal to integral over E 1 of real part of f d mu plus integral over E 2 of real part of f d mu. So, that is the first one plus i times the second integral gives me integral over E 1 of imaginary part of f d mu plus i times integral over E 2 of the imaginary part of f d mu. So, that is using the properties that for real valued functions, integral over E 1 union E 2 whenever they are disjoint spreads into 2 into the sum of that.  So, I can write that as integral over E 1 of f d mu and the second combination will give me integral over f d mu over E 2. So, what we have shown is that integral over E 1 union E 2 whenever they are disjoint of a function f is integral over E 1 plus integral over E 2. So, that proves the required property 3 completely. We will extend this property a bit further. So, the next property is an extension of this property which says let f be a integrable function and let us take a sequence E n be a sequence of pairwise disjoint measurable sets and let us write E as the union of the sets E n. Then the claim is then the series look at the complex series summation 1 to infinity integral over E n of f d mu. This series is absolutely convergent and the sum of the series and implies as a consequence that integral of chi E times f is a integrable function and the integral of f over E is nothing but the sum of this series complex series sigma 1 to infinity integral of E n f d mu. So, to prove this property, we will be using the corresponding property for the real valued integrable function. So, let us write E is equal to union disjoint sets E i 1 to infinity where E i is belong to S and f is a integrable function. So, let us look at the series. So, the series we are concerned with is sigma i equal to 1 to infinity integral over E i of f d mu. So, we want to show that is absolutely convergent. So, that means what? That means we want to show that the series i equal to 1 to infinity absolute value of integral over E i of f d mu is convergent. So, this is what we want to show. But this note that just now we proved that the absolute value of the integral is less than or equal to integral of the absolute value. So, to prove it is convergent it is enough if we prove that the series. So, we will prove enough to show that the series i equal to 1 to infinity integral of absolute value of f d mu over E i is finite. But for that let us just observe that absolute value of f is a real valued integrable function and that we have already observed while discussing the integral of non-negative measurable functions that this is finite if f is mod f is integrable. So, this property is true by the property of real valued integrable functions because f in L 1 that is same as saying that mod f is a real valued integrable function and E i is being disjoint and we know this series is convergent. So, as a result we will have that the series summation over i integral over i E i's of f d mu is an absolutely convergent series. And now we only want to prove the claim namely, so claim is that integral over E of f d mu is equal to summation i equal to 1 to infinity of integrals over E i of f d mu. So, this is what we want to check. So, let us look at the partial sums of this series. So, let us look at integral over E of f d mu minus summation i equal to 1 to n integral over E i of f d mu. So, let us look at this absolute value of this. So, this is same as saying that this is this is integral of absolute value of the integral minus this sums. So, this we can write it as equal to absolute value of summation i equal to 1 to n of f over integral of f over E d mu minus integral over E i of f d mu. So, that is equal to this. Now, using the triangle inequality for the absolute value, I can write this as we can write this as less than or equal to summation i equal to 1 to n absolute value of integral over E f d mu minus integral over E i of f d mu. So, but that using the fact that absolute value of the integral is less than or equal to integral of the absolute value, I can write this is less than or equal to summation i equal to 1 to n and this is nothing but integral of chi E minus chi of summation. I think we may have to slightly modify the proof, but let us see this may also work out.  This I think will not work out. So, let us modify the proof a bit, because we want to say that as n goes to infinity, this goes to 0. This may not exactly happen in this case. So, let us modify the proof slightly. Let me go back to this step. So, this is what we want to analyze. So, to analyze this, we proceed as follows. Let us first write this is equal to, this is the finite sum i equal to 1 to n integral over E i and that we have already observed is equal to integral over f d mu minus. So, this I can write as integral over union of E i's i equal to 1 to n of f d mu. I can write it that.  Let me write the integral over the indicator function of union E i i equal to 1 to n of f d mu. This I can write as the indicator function of E. So, this is less than or equal to integral over chi E of f minus chi union of E i i equal to 1 to n f d mu by using the fact that absolute value of the integral is less than or equal to. So, this is minus the union and the union I can write as the indicator function of E times f and now absolute value of the integral is less than or equal to integral of the absolute value. So, using that I come up to here and now let us observe that this where does this is a sequence of functions and where does it converge. So, let us observe this point that indicator function of union E i i equal to 1 to n f converges to the function indicator function of E of f. So, this converges to this function point wise. So, if I look at the difference. So, this difference. So, the chi E of f minus chi union of E i i equal to 1 to n f d mu f this goes to 0 as n goes to infinity. So, that is one observation. So, the integrant goes to 0 as n goes to infinity and this is less than or equal to. So, and this is absolute value of chi E f minus indicator function of union E i is 1 to n f d mu f is less than or equal to absolute value of the first one plus absolute value of the second function and both are less than or equal to f. So, 2 times mod f and mod f is a. So, this this function goes to 0 and this function is dominated by 2 times mod f which is a integrable function and mod f is a integrable real valued integrable function. So, by the dominated convergence theorem we will have that this goes to 0 as n goes to infinity. So, what we are saying is an application of dominated convergence theorem will tell me that. So, by dominated convergence theorem integral over E of f d mu minus summation i equal to 1 to n which we are trying to analyze absolute value of that. So, integral over E i of f d mu absolute value of that. So, this is what we wanted to show goes to 0 and this we got is less than or equal to. So, we got this is less than or equal to this integral and the integrant goes to 0 and is. So, integrant goes to 0 we showed that and is dominated by an integrable function. So, Lebesgue's dominated convergence theorem says this goes to 0 as n goes to infinity. So, that proves. So, hence integral over E of f d mu is going to be equal to summation i equal to 1 to infinity integral over E i of f d mu. So, this complex series converges and this is the sum. So, that proves the result for complex valued functions namely that if E is a disjoint union of sets measurable sets and f is integrable then integral of f over E is equal to summation of integrals over E n. So, these are the basic properties of complex valued integrable functions and one can actually also prove dominated convergence theorem for complex valued functions. Recall that for real valued functions we had three important theorems. One was the monotone convergence theorem and another one was for 2's lemma and the third one was dominated convergence theorem. The monotone convergence theorem was true for non-negative measurable functions and for 2's lemma also is valid for non-negative real valued measurable functions, but for complex valued functions no way of saying that a function is non-negative or something. So, you cannot expect monotone convergence theorem and for 2's lemma to hold for complex valued functions, but it is fortunately enough the dominated convergence theorem which is true for real valued integrable functions is also true for complex valued functions and let us look at a proof of that. So, the theorem says that if f is a sequence of actually measurable functions is enough complex valued measurable functions such that f converges to the limit f n converge to a limit f point wise almost everywhere and there is a function g which is integrable real valued integrable function such that mod of absolute values of f n's are dominated by g then the f is the limit function is integrable and its integral is equal to limit of the integrable. So, basically this is a straightforward application of. So, let us look at the proof of this.  So, that automatically implies is a non-negative function.  So, we will imply that mod f is less than or equal to g almost everywhere and that also will imply that f is also an integrable function. Now, we want to prove that the integrals will converge. So, let us look at the integral of.   Similarly, the imaginary part. So, integral of the imaginary part of f n d mu converges to integral of imaginary part of f d mu and we already shown that for complex scalar multiplication the corresponding result told. So, i times that will converge to i times that and adding this to implies that integral of f n d mu converges to some of this limits. So, that is equal to integral of f d mu. So, what we have shown is that for complex valued functions monotone convergence theorem also holds. So, what we have done today is we have extended the notion of measureability and integrability from real valued functions to complex valued functions and we have shown that the integral for complex valued functions has same properties as that of real valued functions and monotone convergence theorem also gets extended. So, we will be using these results to in the next lecture to define what are called the p th power integrable functions and their properties. Thank you.


Hello and welcome back to the channel. This video is going to be a version of a keynote lecture that I delivered last month at the Brazilian Congress of Biomechanics on biomechanical considerations in flywheel resistance training. If you want to see some of the great work being done out in Brazil then check out the Neuromech TV channel. They've got some great presentations in English as well as in Portuguese. For my presentation I'll include open access links to the studies mentioned in the description below. Hope you enjoy the video. Before I start, I need to acknowledge the contribution of two people in particular. Firstly, when I first joined the University of Suffolk a few years ago, this was an area of research that my colleague Marco Biato was already specialising in from a strength and conditioning perspective and then as time has progressed we began to collaborate more and more and I began to tackle some of the same flywheel paradigm issues but from more of a biomechanical perspective. Secondly, Kevin DeKaiser at the time when this work started was an undergraduate student with his dissertation supervised by myself and Marco. Kevin has now progressed through to doing his masters and is now being a PhD student again with myself and Marco supervising. So a big thank you to both Marco and Kevin. Almost everything that I will present has been contributed to in one way or another by either or both of these two people. What am I going to present then? There's a brief overview. I'll start with what is flywheel resistance exercise. I will then focus mainly on a comparison of the flywheel squat with more traditional gravitational resistance exercise such as a barbell back squat or the Smith machine squat and I'll cover that from both a kinematic and a kinetic perspective. Finally, I'll do a little bit of application to acute performance enhancement and as I do this I'll briefly touch upon a few theoretical concepts which include the force-velocity relationship, a comparison of discrete and continuous biomechanical analysis methods, looking at some coordination and control, the difference between post-activation potentiation or post-activation performance enhancement and finally touching on dynamic correspondence. So to begin with what is flywheel resistance exercise? The story goes that the flywheel ergometer was developed by NASA for astronauts during space travel. There are well known issues with muscle atrophy during space travel which present the importance of resistance training. However, the limitations of gravitational resistance in a low gravity environment are hopefully quite self-explanatory. So the solution was the flywheel ergometer. The way that it works is essentially that during the concentric phase of the movement the user rotationally accelerates the flywheel with maximal effort. The resistance is due to the flywheel moment of inertia. This results in flywheel kinetic energy and inertial torque that then imparts a high linear resistance during the subsequent eccentric phase of the movement. To simplify that slightly, I like to think of this being like a toy car that I used to have when I was a child. So think of those cars where you would pull the car back and it would essentially wind the car up. Once you let go, the car would then race off. The flywheel works in a similar way. So on the way up the user essentially winds up the wheel. Once they reach the top of this movement, a squat in this case, the flywheel will then essentially pull them back down. And so the resistance is not due to gravity, which is why it was able to be used during space travel. It's typically seen either in this kind of device pictured here or similar to a cable machine that you might see in the gym that then enables the flywheel ergometer to be adapted for pretty much whatever movement it is that the user or practitioner has in mind. To see it in action now, the user typically wears a harness that's attached to the flywheel ergometer, and then as I said before, concentric phase will rotationally accelerate the flywheel, and then on the way down, as they are pulled back down, the user will resist eccentrically, essentially acting as a brake. You're likely familiar with Huxley's sliding filament theory and with the associated force velocity relationship. According to the theory, the decrease in force that we observe at increasing concentric velocities of muscle contraction is caused by one or both of either an increasing likelihood of acting and myosin pairs passing each other without forming a crossbridge, or the increasing proportion of active myosin links that will not disassociate in time, and therefore generate a force in the opposite direction. Important for flywheel resistance training is the eccentric phase of the movement, and the enhanced force production eccentrically is believed to be due to one or both of either an increasing number of crossbridges attached to acting or an increasing force per crossbridge. For more information on the muscle mechanics specifically, I strongly recommend this video lecture on YouTube that Walter Herzog did for me as part of the Sports Biomechanics lecture series last year. To put the force length and force velocity relationships into some example data, this is just an isochronetic dynamometer example from my PhD thesis, but the point I wanted to highlight here is that regardless of joint angle, the force or the torque in this instance is always greater in eccentric contractions or negative joint angular velocities than it is during concentric or positive joint angular velocities. Practically, this means that the eccentric phase of most resistance training exercises is almost always submaximal. If we imagine for a moment that I was capable of lifting 100 kilograms above my head, I would probably be capable of lowering in a controlled manner something like 140 kilograms or in any instance more than 100. And so this means I would typically perform the eccentric phase of the movement with the weight that I was capable of lifting concentrically, and so that eccentric phase would be submaximal. However, during the flywheel resistance exercise, we are able to exert eccentric resistance that can sometimes be greater than the concentric force that is generated. And so this is one of the most frequently cited advantages of flywheel resistance training is the potential for greater intensity during the eccentric phase of the movement as the flywheel pulls you back down. And we would have, if we compared it to traditional forms of resistance training. And in one survey from Kevin DeKaiser's PhD thesis, he asked a number of practitioners that use flywheel resistance training with athletes. And the majority of practitioners did believe that this potential for an eccentric overload was an advantage of flywheel resistance training. The second question looked at the most frequently prescribed exercises, and the flywheel squat came out on top. And so for the next part of this presentation, I will focus on the most frequent exercise, which is the flywheel squat. And I will compare that with gravitational resistance exercises from both a kinematic and a kinetic perspective. In the first study that I'm going to talk about, we recruited 11 recreationally active males who each performed three sets of six repetitions of each of flywheel squats and barbell back squats with the resistance matched for peak concentric power. So each participant performed the flywheel squats with the same flywheel moment of inertia. We recorded their peak concentric power with the inbuilt transducer. And we then tested their barbell back squat using an external resistance that matched their peak concentric power recorded during the flywheel squat. All of the squats were recorded using an eight-camera Qalysis system. And from that 3D motion capture data, we extracted ankle, dorsi, and plantaflexion time series data. We looked at knee joint angles for flexion extension, abduction, adduction, and internal and external rotation. At the hip, we looked at flexion extension angles and the ab and adduction time series. And then finally, we also looked at the trunk segment angle in the sensual plane. However, at this point, there are a number of different ways in which we could analyze the data. Most biomechanical curves such as the joint angle time histories that we're interested in are typically sampled at frequencies in the hundreds or thousands per second. In our example, it was 300 hertz for the motion capture data, but it might be much higher for ground reaction force data, for example. However, typically in biomechanics, at least historically, we've tested our hypotheses using a relatively small number of those data points. For example, we might look at the minimum knee joint angle or the peak vertical ground reaction force. And so we might be throwing away 299 of those 300 data points per second. And in subsequent Ptaki et al studies in 2013 and 2016, they showed that if our hypothesis specifically relates to kinetic or kinematic time series, then it's generally biased to test that hypothesis using only discrete values. For more on that, then again, I recommend a video lecture this time by Todd Ptaki himself as part of the same lecture series. An alternative is statistical parametric mapping or SPM. There's one key paper highlighted on screen. SPM is a method that uses random field theory to investigate the entire time series and not just single discrete data points. Luckily for us, it's open source. So Todd Ptaki has an SPM1D website at SPM1D.org where you can download these scripts in even MATLAB or Python. And there are numerous examples there. For anybody without coding experience, I recorded this year a series of short videos just showing how you can take your data from something like Excel, copy and paste it into MATLAB, run the SPM analysis in a few lines of code, and then get it back out again. What do the results look like? Here for our ankle, dorsi and plant deflection angles, I've plotted the mean and standard deviation joint angle time history for both the barbell back squat, which will always be in black in these examples, and the flywheel squat, which will always be in red. I've time normalized each squat from minus 100% to zero being the eccentric phase and zero to plus 100% being the concentric phase. So all squats are aligned where zero is the bottom of the squat or the transition from eccentric to concentric. We also have our SPM paired samples t-test results, and here we have a dotted red line representing the critical t-value threshold, and the black line shows the t-value at all times during that time series. So to simplify it, if the black curve goes beyond either of the red dashed lines, it's a significant difference for that part of the time series, which here are shaded gray. Say I, the example code from Todd at the website has shaded gray, and if it fails to cross that red line, then it's not a significant difference. This will be the same for all of the graphs that I'm going to show, but in this example for ankle, plantar and dorsiflection, we have a significant shaded difference in the early eccentric phase from minus 88 to minus 68%, and if we look at the same time period in the graph on the left, we can see that the barbell back squat has a more plantar flexed angle at the start or early in the eccentric phase. Then again at the end of the concentric phase, so from 90 to 100% of the concentric phase, we again see a more plantar flexed ankle in the barbell compared to the flywheel squat. This could simply be a case of participants anticipating the flywheel pulling them back down, and so they're not locking out or fully extending their joints at the end of the concentric phase in the flywheel squat in preparation to be pulled back down, whereas in the barbell squats they will fully extend because it's up to them when they then start the subsequent eccentric phase. Despite the fact that multiple studies have shown relationships between ankle joint range of motion and depth or other metrics during a squat, it's important to remember that the results I just showed had significant differences at the start and end of movement, so not at the extremes of the range of motion towards the bottom of the squat. So we're not necessarily saying that there's a difference in range of motion, it's just that early and late in the squat, there's a difference in how they are moving at that joint. If I move on to the knee, we see that the curve on the right does not cross the red dashed line either positively or negatively, so there are no significant differences in knee flexion or extension angle between the two squats. However, we can see that it approaches the red line both in the middle of the movement where we almost have a more flexed knee in the barbell squat, and then in the opposite direction at the end of the movement where we almost have a more extended knee in the barbell squat, so again almost seeing this result again of more extension at the end of the movement in a barbell squat, perhaps with participants anticipating being pulled back down and not fully extending. Whilst I hate the term trending towards significance or anything like that, the reason I chose to mention that things are almost significant in inverted commas is because this is an underpowered exploratory study, so the purpose here was to form hypotheses to then test specifically in subsequent studies, and especially with a recent publication in Journal of Biomechanics where Robinson et al. investigated statistical power during both 0d i.e. discrete values and one-dimensional i.e. time series or statistical parametric mapping style of analyses, and one of the main takeaways from this paper was that the 1d or continuous sample sizes required were always larger than for discrete values, ranging from just one participant more required up to needing an extra 20, depending on the details of the study and the biomechanical time series. Moving on to the other axes at the knee then, there was no significant difference in abduction or adduction angle, and there was also no significant difference in internal or external rotation, which is important especially if we consider that there may be injury risk factors associated with some of these non-sagittal plane rotations. At the hip, we do see a significant difference from minus 22 to plus 34%, so around that transition between eccentric and concentric phase. Late eccentric and early concentric, we see a more flexed hip angle in the barbell squat than the flywheel squat. Again, no significant difference for abduction or adduction at the hip, just like at the knee. Finally, we looked at sagittal plane trunk angle, and we would hope to see in our participants a more upright trunk rather than a more anterior or a more flexed forward trunk, particularly as studies have shown that lumbar shear forces increase with greater forward lean, and we also know that less skillful lifters exhibit greater forward lean, albeit in alternative types of squat, where the resistance is positioned slightly differently in a flywheel squat. But in our data, we saw no significant difference for sagittal plane trunk angle, and so it seems that all of the other lower extremity joint angles have balanced out to maintain a similar trunk angle between the two types of squat. To summarize all of those results then, for the flywheel compared to the barbell back squat, we saw a more dorsiflexed ankle early in the eccentric phase and late in the concentric phase. No differences at all at the knee and a less flexed hip at the bottom of movement, with no difference in ab or adduction, and no difference in sagittal plane trunk angle. So interestingly, we can also say that for the joint angles we looked at and for our statistical power, there were no significant differences in any joint angle outside of the sagittal plane. Something else we can look at with the same data is the coordination and control of the squats in both the flywheel and the barbell back squat. Coordination and control then. I've put some definitions on screen, but for example, intra-individual coordination refers to the coupling relationship between body segments or joints. Measures of control include amplitude, velocity, acceleration or force, and both coordination and control lead to the performance outcome. To highlight some examples of coordination patterns that I will refer to in the coming minutes, in-phase coordination means that both segments or both joints rotate in the same direction. Contrary to that, anti-phase coordination is where both segments or joints rotate in opposite directions. Proximal dominance means that the proximal segment or joint is the dominant contributor to relative movement, whereas distal dominance means that the distal segment or joint is the dominant contributor. At this point, I owe another thank you. This time to Rob Needham from Staffordshire University. A lot of the data visualizations and figures that are coming up in the next few slides, I say a lot, basically all of them, I owe thanks to Rob, especially as part of our collaboration on coordination and control of counter-movement jumps that I presented at ISB. Subject to permission, by the time you see this, that might now be publicly available on YouTube, but it depends whether I'm allowed to do so. Anyway, it's possible to qualitatively assess coordination using an angle-angle diagram. In this case, we've got a distal and proximal segment angle plotted against each other, but it could just as easily be joint angles, which is what it will be when I come to assess the flywheel and barbell squat in a moment. In our example, we have an increase in distal angle and a greater decrease in proximal angle, which suggests anti-phase coordination with proximal dominance. Vector coding builds upon the angle-angle plot to provide a quantitative assessment of coordination. It refers to the vector orientation between adjacent data points on our angle-angle diagram, measured relative to the right horizontal, and the outcome is referred to as a coupling angle, in this case 327 degrees. We can then map that coupling angle onto a coordination pattern classification. So, sticking with the same example, that 327 degrees relative to the right horizontal would be anti-phase with proximal dominance again, but with a dominance of 65% that we're able to quantify. Why 65%? Well, each quadrant of a unit circle is 100 gradient, so it's therefore relatively easy to convert that to a percentage dominance, where a coupling angle of 45 degrees would be equal 50-50 dominance between the proximal and distal segment or joint. If we take that coordination pattern classification at each normalized time point during a movement, we can use color mapping to show how the coordination changes throughout the movement, or to compare the coordination between different movements, as I will progress to with our squats. For more information on the data visualization techniques, please check out the Needem et al paper that's currently highlighted on screen. In this example, which happens to be the couple between hip joint angle and knee joint angle in a barbell squat, the bar color represents coordination pattern classification again, as it has done in the previous slides, but bar height represents the level of segmental joint dominance, ranging from 50 to 100%. What vector coding alone cannot do, however, is provide a measure of control, such as range of motion or angular velocity, and control can often differ, even when there are commonalities in coordination between individuals or between conditions, so it's important that we include this when we're assessing the technique. So in this data visualization and in the ones to follow on subsequent slides, the dotted line represents the inter-datapoint range of motion of the dominant segment or the dominant joint. To get into some results then, here I have the group mean coordination and control for both the barbell and flywheel conditions. This is from an angle-angle plot of proximal being the hip joint angle, and distal being the knee joint angle. So we're looking at the couple between two joint angles rather than between two segments here. The first thing that stand out to me are the similarities. For example, both conditions throughout the majority of the eccentric phase have distally dominant in-phase coordination, with that level of dominance decreasing as the phase progresses. This means that both joints, the knee and the hip, are flexing at the same time, as we might expect. During the second half of the concentric phase, again, there are similarities. So both conditions are in-phase, this time with both joints extending together, and again, it's distally dominant, with that level of dominance increasing as the phase progresses. The third similarity is in control, so the inter-datapoint range of motion. This looks very similar between the two conditions, at least at the group level. The differences, however, occur in coordination at around the transition point and the early concentric phase. In the barbell condition at the top of the screen, we see a change from dark green to grey. This means that we've gone from in-phase to anti-phase, where the proximal joint has reversed its direction from flexion to extension before the distal joint does, which is what you would expect from a proximal to distal sequence. Likewise, during the concentric phase, it begins light red and then becomes darker red. This means whilst it's in-phase, it changes from proximal dominance to distal dominance. So to sum up that top coordination pattern, it's exactly what we would expect from a proximal to distal sequence. The flywheel looks slightly different though. Here, during the in-phase eccentric part of the movement, there is a transition from dark green distal dominance to light green, where proximal dominance takes over. However, both joints then change direction on average at the same time as we go straight from green to red with no grey or blue in the middle. The proximal joint, i.e. the hip, never really takes over the dominance during the concentric phase. However, the onset of the distal dominance is rather delayed, happening at later than 70% of the overall movement compared to the barbell, where it's happening at about 60%. This seems to suggest a more even 50-50 approximate dominance between proximal and distal joints for a longer period of the movement rather than any progression from proximal to distal. We can also use either bar height or bar colour to show the inter-individual coordination variability. Again, there are similarities. Both conditions show the greatest amount of variability at roughly the same period of the movement, which again is the transition and the early concentric phase. However, the flywheel has substantially more variability at that key time. There are a number of limitations with this approach. One typical limitation is that we often can't tell whether this variability is spatial or temporal in nature. However, because it's happening at around the transition, when we've aligned all of the squats with 50% being the bottom of the squat or the transition between phases, hopefully we can be quite confident that it's more of a spatial variability than temporal. However, there's also a lot of artifact when the range of motion is small. We typically see a roughly inverse relationship between inter-datapoint range of motion and coordination variability. That's because of artifact in the coupling angle when the range of motion is very small. A number of alternatives have been proposed to address this, such as the use of ellipses or confidence intervals on the angle-angle plot to measure variability or uncertainty by stock, Muller-New and others. But for now, my purpose here isn't to necessarily investigate the variability itself, but rather is just to highlight it and then look at some of these strategies that lead to that variability, which we could do by plotting each of our 11 participants at the same time. Here, it's in order from the participant with the greatest peak concentric power during the squat through to the participant with the lowest peak concentric power during their flywheel squat. I won't go into too much detail here, but again, the things that stand out initially at least are the similarities, so that dark green in phase eccentric and dark red in phase concentric. And again, we can see that the majority of the differences are happening around that transition and early concentric phase, which is exactly what the coordination variability chart or figure showed on the last slide. I'll pull out a few individual participants to compare their barbell and flywheel squats against each other now. And again, I'll just pull out kind of one group level difference and then one difference from each of the three participants. The first one that seems to be common among the three people is this difference almost a bimodal shape in control during the eccentric phase. And I imagine this is because of the common instruction during a flywheel squat that the participants should not resist the eccentric for the first roughly third of the movement, i.e. they have to allow the flywheel to pull them back down slightly initially, and then they begin to resist and essentially break during the rest of the eccentric phase. And we can see that here with the control or the range of motion, where it begins to increase quickly, but then that range of motion decreases as the participant begins to decelerate, which isn't the case as much in the barbell. To look at the individual participants, the difference that stands out for the top participant is that during the concentric phase, they have three separate phases of dominancy. So they go from distal to proximal and then back to distal dominancy, whereas in the barbell they simply go from proximal to distal. For the second participant, the thing that stands out or the one that I want to highlight is the decrease and then increase in dominancy during the eccentric phase. So we see the level of distal dominancy decreases initially, and it's not due to the range of motion. So it is due to that range of motion that we can also see decreasing rather than being due to any increase in the opposite segment. So it's not that the hip is increasing its contribution. We can see from the control that it's simply a decrease in the contribution from this dominant knee joint at that time. Finally, for the bottom participant, the thing I wanted to highlight that stands out is during the barbell squat this time. We can see quite the big period of proximal dominancy during that eccentric phase, which isn't there for the group on average and isn't there for the same participant doing a flywheel squat. So the purpose there was simply to highlight some of the things that can be identified by looking into coordination and control of individual participants that would perhaps be lost if we only focus on comparing those two sets of group mean results. Just to pull out, I think, that top participant in a bit more detail from the last slide. For the barbell squat, they follow pretty much the exact same pattern as the group average, where the proximal joint reverses its direction first. And then we see a pattern of proximal followed by distal dominancy concentrically. During the flywheel, we see that the distal segment actually reverses its direction first. And then during the concentric phase, as highlighted a moment ago, we have both distal and back to proximal and then distal again, dominancy. And so we could argue that these changes in dominancy and the lack of a clear proximal to distal sequence represents inefficient coordination by this individual. However, we would need to link it to performance outcome measures in order to define what is effective and what is ineffective for this individual. To conclude that section of vector coding related results, we saw many commonalities at the group level in both coordination and control with the main differences occurring at around the transition between phases. In the barbell squat, we saw more of an apparent proximal to distal sequencing on average, and in two of the three individual participants that we look at. Whereas in the flywheel squat, there was a more even dominancy around that transition and early concentric phase with a delayed onset of knee-dominancy concentrically. A greater inter-individual variation during the flywheel squat and more of a multi-phase eccentric or multi-phase concentric in some individuals, where dominancy changed from proximal to distal multiple times or when the level of control or range of motion decreased and increased again. The next steps in this analysis would be to look at segment couples, for example, thigh shank around the knee joint or trunk thigh around the hip joint. We could look at other measures of control such as angular velocity, for example, and we can also look to link both coordination and control to performance outcomes. Finally, on this bit, I'd conclude that the group level analysis of coordination and control masks important inter-individual variation in movement strategies. Those results link quite nicely to a recent paper that assessed EMG muscle activity during flywheel squats. They found that at lower flywheel moments of inertia, they did get a proximal to distal sequence of muscle activation. However, as they increased the flywheel inertia, at higher loads, they found a reorganization of the coordination and the sequence of muscle activation. We have higher inertia requiring a specific and stable muscle coordination pattern that was not proximal to distal, which quite nicely reflects what we saw in coordination from a kinematic perspective looking at those joint couples. It also suggests that our results could perhaps be specific to the flywheel moment of inertia used, and we may find that proximal to distal sequence if we used a lower inertia, or it could suggest that the inter-individual differences may reflect the strength capabilities of that particular individual. Relatively, how difficult was the intensity and were they able to use a proximal to distal sequence or did they need to reorganize? Speaking of the effect of flywheel inertia, our recently published study in Journal of Sports Sciences investigated the effect of inertia on velocity and power parameters during both the concentric and eccentric phases of the squat. We recruited 15 recreationally active males, each of whom performed a set of flywheel squats after familiarization at four different flywheel moments of inertia. During each squat, we measured peak power from an inbuilt transducer, and we also assessed peak and mean velocity using 3D motion capture from markers on both greater tricentres. Before we look at the results, it's worth noting that previous studies, including our own, were using peak power to prescribe intensity or load during flywheel squats, mainly due to the ready availability of peak power data from inbuilt transducers within the ergometers. For mean velocity, we saw what you may expect that as the difficulty increased, so as flywheel inertia increased, both concentric and eccentric mean velocity decreased. There was no significant effect of inertia on the ratio of eccentric to concentric mean velocity, although it's worth noting that on average, at all inertias, the eccentric velocity was faster than concentric. That dashed line will be on a few of these graphs, and it represents an even ratio of eccentric and concentric values. For peak velocity, we again had similar results, where as inertia increased, there was a decrease in concentric and eccentric peak velocity. Again, no significant effect of inertia on the ratio of eccentric to concentric peak velocity, although this time the concentric part of the movement was faster on average at the lowest inertia, whereas at the three higher inertias, the eccentric part was faster. But peak power then, which was the parameter that had been most used to prescribe intensity beforehand, as you perhaps may imagine if you put some thought into it, there was no overall relationship of flywheel inertia on peak power, whether during the concentric or eccentric phase, which on an individual level, is perhaps what we would have expected, because although the force velocity relationship dictates that an increase in force, which you'd expect with an increase in inertia, would result in a decreasing movement velocity, power is force multiplied by velocity. So as you increase the inertia, if force goes up and velocity goes down, you might expect that power remains somewhat roughly similar. It would also be true, we would expect that different participants would have their peak power at different loads of different inertia, and so on the group level is perhaps not surprising that we concluded there was no significant relationship between flywheel inertia and peak concentric or eccentric power during the flywheel squad. No significant relationship again with the ratio of the two, although it's noteworthy that we only saw an eccentric overload, i.e. a greater eccentric value than concentric in the two higher inertias, whereas for the two lower inertias, there was no eccentric overload on average. We also fit both linear and nonlinear logarithmic curves to the inertia velocity relationships for each individual participant, and then averaged the r squared to see which had the better fit. And the goodness of fit r squared ranged from 0.62 for a nonlinear relationship with mean eccentric velocity through to 0.97 for nonlinear peak concentric velocity. So for both linear and nonlinear, we found better participant-specific relationships between inertia and velocity when peak concentric velocity was used, and there was not very much difference in fit between linear and nonlinear. And so we recommended that people, practitioners interested in velocity based training, could perhaps use peak concentric velocity to prescribe the intensity of a movement and that they could use linear relationships, therefore drawing on the vast amount of literature available using linear relationships between load and velocity to prescribe exercise intensity during gravitational resistance exercise. If you're interested in the concept of velocity-based training, then this practical review narrative review article by Weakly et al is a really good read and gives a lot of practical scenarios and advice. Moving on then, I was very fortunate earlier this year that Carlos Galliano, a PhD student from Seville in Spain, came to stay at our university for a few months and being this year, I stay obviously mean virtually, but I was lucky to supervise one of his PhD studies from a distance alongside his other PhD supervisors in Spain. This study used or assessed ground reaction force under each foot during both a flywheel squat and a Smith machine squat. For each of 30 recreationally active males, the resistance was matched this time for mean concentric velocity to be the same between both conditions. And from the ground reaction force, we calculated vertical velocity, displacement, power and force as well as the difference in force between the two feet. Again, we've used statistical parametric mapping as explained earlier. So hopefully you're familiar with the graphs by now. To start with velocity, there are quite a few differences there. To start with, at the beginning of the eccentric phase, the flywheel had a significantly more negative velocity, so it was significantly faster downwards. Based on what we saw with the vector coding results, where I explained the initial lack of resistance by the participant during the first third of the eccentric phase, this is perhaps not surprising that it would generate a faster velocity during that part of the movement where it's not being resisted and where the flywheel is pulling the person downwards. Concentrically, though, considering that both conditions were matched to have the same mean concentric velocity, it's interesting that the flywheel has a faster early concentric velocity and the barbell has a faster late concentric velocity with those two matching out to result in an equal mean concentric velocity. In terms of displacement, perhaps not surprising due to the faster eccentric velocity, the flywheel also has a greater negative displacement towards the middle of the eccentric phase. At the end of the concentric phase, the barbell squat has a greater concentric displacement. If we think back to the joint kinematics mentioned right at the start, I explained that some participants perhaps don't fully extend their joints in the lower extremities at the top of the flywheel squat in preparation to be pulled back down. Those same results here may lead to a lower vertical displacement or height for the flywheel compared to the barbell at the end of the movement. If we look at vertical power, then the flywheel squat actually recorded greater power during a big portion of the eccentric phase negatively and also the concentric phase positively. So we do have an eccentric overload in terms of power, but the flywheel squat also leads to greater power both eccentrically and concentrically during certain portions of the movement, which may be useful for practitioners seeking to get that high power value, whether it be eccentrically or concentrically. When we look at force, however, there was no eccentric overload for the flywheel in force only in power. Linked back to the velocity and displacement results, it again might not be surprising that when there's little resistance from the participant, there was lower force at the start of the eccentric phase. However, the barbell squat also had greater force at the transition between phases and the flywheel squat actually had greater force at the end of the concentric phase or towards the end of the concentric phase. Finally, when looking at asymmetry, the flywheel squat had a greater difference in force between the legs normalized to body mass throughout the whole of the concentric phase of the movement. And so to repeat or summarize those significant differences only in a chronological manner, at the start of the eccentric phase, the participants in the flywheel squat generated less force to resist that motion and so therefore also had more negative velocities. We saw more negative power in the middle of the eccentric phase and also more negative displacement. At the transition between phases, there was again less force for the flywheel squat. Early in the concentric phase, the flywheel squat had greater vertical velocity. Towards the end of the concentric phase, there was higher force, more positive power and then less positive velocity and displacement. Now it's greater force asymmetry during the concentric phase for the flywheel compared to the Smith machine squat. In an attempt to summarize the results from both SPM analyses that I've presented, as well as the vector coding analysis, I thought that perhaps the results could be summed up when moving from a more traditional exercise to the flywheel as redistribution and variability. For redistribution to choose one result from each analysis, we saw differences in ankle joint kinematics and hip joint kinematics in the sagittal plane. However, there was no overall difference in the depth of the squat or in trunk kinematics in the sagittal plane. We saw less proximal to distal sequencing and a more even dominance around the transition and then we saw a faster early concentric but slower late concentric phase leading to similar average concentric velocities. For variability, we saw greater inter-individual coordination variability around the transition and early concentric phase, as well as greater asymmetry during the concentric phase for the flywheel squat. To summarize the discrete or continuous analysis discussion, SPM enables comparison of continuous biomechanical variables such as kinematics or kinetics at time at time points other than discrete local maxima or minima. Vector coding provides information regarding differences in proximal to distal joint coordination throughout a movement and these continuous methods can increase validity and intuitive application of biomechanical conclusions. However, the methodology selected for studies should always be made with reference to the specific research question and hypotheses being addressed and tested. I'll finish this presentation with a brief overview of some applications. We haven't yet done much chronic application work but there's a nice review here to sum it up. The chronic adaptations to flywheel training are generally greater than or similar to other forms of resistance with the specific results depending on details of the study, the intervention method and the outcome of interest. Where we've done a lot more work is in acute performance enhancement, so I'll go through a few of our studies very quickly to finish off with now. Both post-activation potentiation and post-activation performance enhancement involve the idea that a user can perform a preload activity such as a squat, rest for a short period of time and then following that rest period their muscular performance will be acutely improved compared to their baseline level before they perform the preload activity. Historically post-activation potentiation has been studied frequently and is often attributed to an increased phosphorylation of myosin regulatory light change. This is a phenomenon observed in muscle, so enhanced muscle force at submaximal concentrations of calcium, i.e. submaximal muscle activities. It's therefore been recently questioned whether this is actually the cause of sporting or whole body performance improvement that's been observed in the literature because these activities are often maximal, not submaximal in nature and the time course of their improvement doesn't match well with the acute enhancement in muscle activity due to post-activation potentiation. It's therefore been proposed that this enhancement in sporting or whole body performance should be termed post-activation performance enhancement and that it might be due to other factors including muscle temperature, muscle pH, blood flow, water content, etc. We have shown that the post-activation performance enhancement following flywheel squats is similar in magnitude to that following traditional forms of the squat exercise. However, for practitioners to program such an intervention to acutely enhance performance in training or competition, it's important to know how to optimize or vary some of these parameters. So what rest period is ideal? What moment of inertia should be programmed? What volume in terms of repetitions and sets? And what effect does the exercise chosen have on the subsequent performance enhancement? So I'm going to very quickly go through a brief summary of some of our recent published work from more of a strength and conditioning perspective tackling some of these questions. To start with the idea of rest period, the reason this is important is that both the factors contributing to performance enhancement but also fatigue are both increased after the preload activity. However, the factors contributing to performance enhancement will decay at a different rate to the decay of fatigue. We're therefore looking for an optimal time window at which the fatigue has decreased to a greater degree than the performance enhancement that still remains. And if the rest period goes on too long, then both will have decreased and there'll be no performance enhancement left. Well, that's the idea anyway. So in this study before I joined the university and the research group, they investigated the performance enhancement of counter movement jumps and iso-kinetic dynamometry peak talks at a number of different time intervals. And they found that performance was enhanced following flywheel squats between three and nine minutes after completion of the final set of the exercise. Performance wasn't enhanced at less than three minutes likely due to fatigue. And there was no assessment of performance later than nine minutes. So at this point, we don't know whether it continues longer. We can say three to nine minutes plus. When investigating moments of inertia in the first study that I was involved in on this topic after I joined, we compared the effect of flywheel squats at the two lower inertias of the four that I mentioned earlier in our inertia velocity and inertia power studies. For both standing long jump and counter movement jump performance, we observed similar enhancements for both inertias. So at least for performance outcomes, it wasn't clear which inertia was ideal at this point. Looking at volume in terms of the number of sets, each was a set of six repetitions. So we looked at one set, two sets, or three sets. There was no enhancement at any time period following only one set of flywheel squats. There was no enhancement three minutes after any number of sets. However, six minutes after either two or three sets of flywheel squats, there was a performance enhancement in jumps. So we therefore concluded that in terms of volume, one set was not sufficient to stimulate this post activation performance enhancement. But two or three sets of six repetitions followed by a six minute rest was sufficient. There's a review article in International Journal of Sports, Physiology and Performance that covers these topics and also provided recommendations for future studies. One of which was looking at alternative exercises. So in a study in sports biomechanics, we compared the effect of flywheel squats and an alternative exercise, the flywheel deadlift. Both improved eccentric knee flexor strength on an isokinetic dynamometer, but neither improved concentric strength of the knee flexors or extensors or conventional or functional HQ ratio. And so one of the conclusions or suggestions in the discussion here was that perhaps there's some specificity from the eccentric overload in power perhaps during the exercises to then enhancing eccentric but not concentric muscle strength. That's definitely something for further study. The study that I'd like to finish on was one that went into a bit more detail and included some ground reaction force measures. We assessed counter movement jump, standing broad jump and a five meter change of direction, both in a control condition and six minutes after three sets of six reps, using both of the inertias from the previous study, which we had shown enhanced performance similarly. However, this time we weren't just looking at the performance outcome, such as jump height. We were looking at peak force, peak power and peak rate of force development during each exercise in the appropriate direction. The counter movement jump that was peak vertical force power and rate of force development for the standing broad jump. It was the resultant of vertical and horizontal. And for the change of direction, it was horizontal. We found that for the three exercises, there was no enhancement in any of the kinetic parameters for standing broad jump or change of direction, but all three were enhanced during the counter movement jump. And so one suggestion was that there may be some direction specificity between the vertical squat exercise and then enhancement in the vertical counter movement jump. Although in reality, it's probably more complex than simply the vertical nature and may involve the muscle length, muscle velocity or even the coordination and control of the flywheel squat exercise resembling that of the counter movement jump. In the post hoc analysis, when we looked at the two inertias separately, we found that peak force was enhanced at the higher inertia and peak power was enhanced at the lower inertia, which again suggested some specificity from the region of the force velocity curve occupied during the preload exercise of the flywheel squat to then the region of the force velocity curve that was enhanced during the subsequent whole body exercise. It should be noted that there was no overall effect of inertia on these parameters. So it should be very cautious when interpreting a difference in two post hoc results without directly comparing those two results. This reminded us of the concept of the force vector theory, which is based on chronic training studies, but was a suggestion that a vertical exercise such as a front squat generated greater enhancement in vertical exercises such as a jump, whereas horizontally orientated exercises relative to the individual such as a hip thrust would enhance horizontal performance such as a horizontal jump or sprinting. However, authors have stated that it's more complicated than that. So in this Fitzpatrick et al study, they have a really nice figure showing that during both a sprint and a vertical jump, although the ground reaction force vector is orientated differently, relative to the participant, it is the same. And this is actually the exact same image just rotated. So the point they're making here is that during what would be considered a vertical jump or a horizontal sprint, relative to the individual, the ground reaction force vector is in a similar direction. And they stated that rather than simply vertical or horizontal, it's likely a case of the dynamic correspondence between the training intervention and the performance that we want to improve. And this could include factors such as the region of peak force, so the muscle length, the velocity of the activity, the range of motion, motor unit recruitment. I think to link this back to some of our earlier work with the more biomechanical side, then with joint kinematics, we know the difference in joint kinematics between a flywheel squat and the barbell back squat. So we can perhaps consider which kinematics more closely resemble the subsequent performance activity. We've looked at the coordination and control so we can maybe consider, are we trying to enhance a proximal to distal sequence or is it something else and which coordination and control may transfer better. We also have the SPM results from the force plate analysis of a flywheel squat and a Smith machine squat. And so we can consider the timing of peak velocity, for example, where one had greater velocity in the early concentric phase and one had greater velocity in the late concentric phase. We could consider the eccentric overload in peak power or in power, eccentrically and concentrically, and consider some of these biomechanical results to start to think about the dynamic correspondence between our preload exercise and the performance we want to improve and start to make more informed recommendations. So I think that's a good area to finish with for future studies where hopefully now we will start to combine some of the more biomechanics related studies with the more applied strength and conditioning side, perhaps looking at some chronic training adaptations as well as just the acute performance enhancement. But to summarize this acute part, rest periods of around six minutes seem to be ideal. Possibly a lower inertia is better for enhancing power, but a higher inertia could be better for peak force. Two or three sets of six repetitions have been used successfully and we need to consider that dynamic correspondence between preload and subsequent activities. To summarize some of the topics that we've covered, we've gone through an overview of what is flywheel resistance exercise. I compared some kinematic or kinetic measures against both the barbell back squat and the Smith machine squat. We've briefly covered some studies applying that to acute performance, so some published work rather than our unpublished under review work on the biomechanical side of things. And as we've done so, I've touched upon briefly a few theoretical concepts, including force velocity relationship, some continuous analysis methods, coordination and control, difference between PAP and PAPE, as well as dynamic correspondence at the end. Thank you very much for watching and making it to the end. 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