Welcome back friends. In the previous lecture, we have seen the gating system, the meaning of the gating system. We have seen that gating system refers to all the sections through which the molten metal passes while entering into the mould cavity. We have also seen that these are the elements of the gating system. One is the pouring cup, sprue, next the sprue well, next the runner, runner extension, in gates. These are also known as the gates and finally the razor. So we have seen the significance of different elements of the gating system. Now let us see further, there are different types of the gates are there. One is the top gate. Second one is the parting line gate, third one bottom gate and fourth one the side gate. Now let us see what these gates are. Let us first see the top gate. Now this is the top gate. Now you can see here this is the mould cavity and this is the pouring basin and this is the sprue. Now here we pour the molten metal through this way. Finally the molten metal fills the pouring basin and it passes through the sprue and this is the choke and this is also the gate then it enters into the mould cavity. Now how the molten metal is entering into the mould cavity? From the top surface of the mould cavity that is why it is known as the top gate. Now in the top gate the molten metal is poured at the top of the mould. Now here we can see AM means mould cross sectional area. This is the mould cross sectional area. Usually we express this in square centimeters. Next one AS is the gate cross sectional area. This also is expressed in square centimeters. Next one HM is the height of the mould and it is usually expressed in centimeters. Next one HT this is the HT and this is the filling height or the pouring height and it is usually expressed in centimeters. Now velocity of liquid metal at ingate. What is the velocity of the liquid metal at this point? What is that? Vg is equal to root of 2g HT where HT is the pouring height. This is the pouring height. Usually it is the 2g HT. Now pouring time what is the pouring time? Pouring time is equal to volume of the mould divided by gate cross sectional area multiplied by velocity of melt at gate. That is equal to AM into HM divided by AS into Vg and this is expressed in seconds. So this is the formula for pouring time in the case of the top gate. Now what are the advantages of the top gating system? Simple design. Then what are the disadvantages? Turbulence and erosion is caused in case of the large castings. If the casting is a small casting or if the mould cavity is a small one then we pour the molten metal from the top it does not matter. But if it is a large casting and the mould cavity is a large one then we pour the large quantity of the molten metal from the top gate then what will happen? From the choke the molten metal falls at a very high speed and it heats at the bottom surface of the mould cavity because of that erosion takes place and because of that turbulence takes place. So this is not useful for the large castings, this will be useful only for the small castings and this can be used where moulds are erosion resistance where we think that the erosion cannot take place in such a case we can use the top gating system. Next one the parting line gate. Now in the case of the top gating system the molten metal enters into the mould cavity from the top of the mould cavity whereas in the case of the parting line gate the molten metal enters into the cavity along the parting line. Now you can see here this is the drag box, this whole thing is the mould, this is the drag and this is the cope and this is the parting line, this is the parting line. Now this is the pouring basin and this is the sprue and this is the choke. Now the molten metal is poured at the pouring basin and it enters into the what is a sprue then it enters into the runner choke finally this is the mould cavity and here we can see the razor. So here the molten metal enters into the mould cavity along the parting line, the gate is located along the parting line that is why it is known as the parting line gate. The third one is the bottom gate, first one is the top gate where the molten metal enters into the mould cavity from the top surface of the mould cavity and the second one the parting line the moulding what say the molten metal enters along the parting line. Now in the case of the bottom gate the molten metal flows into the mould through the bottom of the mould cavity in the drag box you see. So this is the mould cavity and this is the cope and this is the drag, this is the sprue. The molten metal is poured along the sprue, now the molten metal comes this way. Now there is a kind of horn sprue means a curved sprue is there. Now the molten metal comes like this and it passes through the curved path finally it enters into the mould cavity and it fills the mould cavity. Now here because of this the turbulence and erosion are minimum. Now what are the disadvantages of the bottom gate? Metal at the top of the cavity will be at a lower temperature is it not? The metal that is filled initially will be going up and the new metal will be occupying at the bottom. Now what happens? At the top there will be a lower temperature metal and at the bottom the temperature of the metal will be higher. Now the razor is here you can see here the razor is here. Now what is the principle in the solidification? The razor should be the one which has to solidify at the end then only it can supply the molten metal to the casting which is undergoing the shrinkage. Now in this case what will happen? The upper portion of the liquid metal will be at a lower temperature naturally even the liquid metal in the razor will be at a lower temperature whereas the liquid metal at the bottoms of the mould cavity will be at a higher temperature then what will happen? The upper portion will solidify earlier the lower portion of the casting will solidify later. That be the case razor may solidify before the casting the razor may not supply the liquid metal to the casting during freezing. So this is the drawback of the bottom gate. Next one the side gate now you can see here this is the mould cavity and here the molten metal enters into the mould cavity along the sides on one side of the mould cavity we can see several ingates are there. Now this is the pouring cup and this is the sprue now the molten metal is poured here and the molten metal enters this way it enters this way enters this way enters this way finally the whole cavity is filled with the molten metal. Now this solves the problems raised in the bottom gate. Now there is another classification of the gating systems. Now the first classification is based on how the molten metal enters into the mould cavity whether it is through the top or through the sides or through the bottom. So that is the criteria for the earlier classification. Now there is another type that is the it is based on the pressure of the gating system. So here we can see another classification now one type is pressurized gating system and the second type is unpressurized gating system. In the pressurized gating system the total cross sectional area gradually decreases from choke to ingates. Now when we are pouring the molten metal right so the molten metal enters into the sprue and it enters into the cavity at the choke sorry not at the cavity at the end of the sprue it passes through the choke. Now at this choke the pressure will be more as it enters into the runner as it enters into the mould cavity the pressure will be decreasing whereas in the case of the unpressurized gating system the total cross sectional area gradually increases from choke to ingates. So naturally the pressure of the molten metal will be what will happen will be decreasing as it passes from the sprue to the runner and ingates the pressure will be decreasing that is why it is the unpressurized gating system whereas the first one is the pressurized gating system. Now let us see what are the differences of these two systems and what are the advantages and limitations of these two systems. Now these are the typical gating ratios. Again different gating ratios have been recommended for different casting alloys. Now for grey cast iron this is the gating system under the pressurized gating system. Now here we can see under pressurized gating system for grey cast iron A C A R A G is equal to 1 is to 1.3 is to 1.1. What does it mean? This 1 corresponds to choke area, 1.3 corresponds to runner cross sectional area and 1.1 corresponds to the total cross sectional area of the all the ingates. For aluminum alloys this is the gating ratio 1 is to 2 is to 1 and for steel the gating ratio A C is to A R is to A G is equal to 1 is to 2 is to 1.5. So these are the gating ratios for important cast alloys under the pressurized gating system. Now let us see the unpressurized gating system under this again for the same important cast alloys the gating ratios have been recommended after extensive experimentation. Now for grey cast iron A C is to A R is to A G is equal to 1 is to 4 is to 4. Similarly for aluminum alloys A C is to A R is to A G is equal to 1 is to 3 is to 3 and for steel A C is to A R is to A G is equal to 1 is to 3 is to 3. So these are the typical gating ratios under pressurized gating system and unpressurized gating system for important cast alloys. Now comparison of the gating systems. So here we see the comparison and the differences of these two systems. Now here let us see the what is the merits and demerits of the pressurized gating system and here we see the merits and demerits of the unpressurized gating system. Now in the case of the pressurized gating system the total cross sectional area decreases towards the mould cavity that is why the pressure will be increasing as the molten metal is entering into the mould cavity whereas in the case of the unpressurized gating system the total cross sectional area increases towards the mould cavity that is why it is the unpressurized gating system. Next one what happens in the case of the pressurized gating system as the cross sectional area is decreasing from choke to the ingates then what will happen the pressure will be intensifying for the liquid metal because of that more turbulence and chances of mould erosion naturally. But here because the what is a cross sectional area is increasing the pressure will be decreasing that is why there will be less turbulence. Now in the case of the pressurized gating system casting yield is more. What is casting yield? Suppose if we are making a casting of 10 kgs do we think that we pour only 10 kgs of the metal no if we are making a casting of 10 kgs we may pour about 13 kgs. 13 kgs means say we say we have to fill the what say sprue with the liquid metal we have to fill the riser with the liquid metal and the runner and the ingates these are not the part of the casting only the casting which solidifies in the mould cavity is the final casting. So the weight of the poured metal is always greater than the weight of the casting. Now what is this casting yield? Weight of the casting divided by weight of the poured metal multiplied by 100 usually this is between 70 percent to 80 percent in the industries. Now in the case of the pressurized gating system what happens as we go inside the gating system the cross sectional area is gradually decreasing. So the less metal is required for filling the whole mould cavity and also the gating system in what happens less metal is required then that be the case casting yield will be more. Then what happens in the case of the unpressurized gating system as we go from choke to inside the mould cross sectional area sorry the gating gate cross sectional area will be increasing then what happens it requires more metal for filling the gating system and the mould cavity then the total consumption of the liquid metal will be increasing then what happens finally the casting yield is less. So this increases the cost of the production for the foundry. Next one now what happens in the case of the pressurized gating system complex and thin sections can be successfully cast sometimes the casting may have a complex and very thin sections. Now if you chose the unpressurized gating system then what happens because of the lesser pressure of the liquid metal these thin sections may not be filled with the liquid metal. In such a case to fill such what say complex and thin what say sections we need the pressurized gating system only when we choose the pressurized gating system the molten metal will be forced to inject into those thin sections and we can successfully cast such a thin and complex sections whereas on the other hand if you choose the unpressurized gating system these thin and complex sections are very difficult to be cast. Now let us see the logical approach for a proper design of the gating system. Now this is the logical approach what is this first one estimation of the optimum pouring time second one calculation of sprue choke area next one selection of appropriate gating ratio. First one selection of gating type and ingate location next one calculation of runner or ingate size. So this is the sequence in the logical approach for a proper design of gating system. Now calculation of the pouring times for grey iron castings and if the weight of the casting is less than 450 kgs pouring time which is indicated by T is equal to K multiplied by 1.41 plus T by 14.59 multiplied by root of W seconds where K is the fluidity factor which depends upon the temperature and composition of the molten metal K that is equal to fluidity of the iron in inches divided by 40 what is this there will be a spiral will be there. So a spiral casting so the if we stretch it it will be equal to 40 inches. Now in this spiral mold we pour the liquid metal then what happens depending upon the fluidity of the melt some metals what say some alloys may flow up to may be 39 inches or some metals may go up to 35 like that different alloys will flow to different what say lens again it depends upon the temperature of the melt. Now this fluidity of right is equal to right K is equal to fluidity of iron in inches divided by 40. T is the average thickness of the casting in millimeters W is the mass of the casting in kgs kilograms. So this is the what say pouring of the what say pouring time for the grey iron castings when the weight of the casting is less than 450 kgs. Now let us see the second case now what happens if the weight of the casting is more than 450 kgs again from the excellence it has been recommended that pouring time T is equal to K multiplied by 1.236 plus T by 16.65 multiplied by Q through W seconds where K is the fluidity factor the same way we have to calculate as we have discussed before T is the average thickness of the casting in millimeters W is the mass of the casting in kilograms. Next one shell moulded ductile iron now how to calculate the pouring time pouring time T is equal to K 1 root of W seconds where K 1 is equal to 2.08 for castings of thinner sections or when the thickness is less than 10 millimeters that is equal to 2.67 for castings of medium sections when the thickness is between 10 to 25 mm that is equal to 2.97 for castings of heavier sections when the thickness is more than 25 millimeters and W is the mass of the casting in kilograms. Next one the fourth case steel castings how to calculate the pouring time for the steel castings pouring time T is equal to 2.435 minus 0.3953 log W multiplied by root W seconds where W is mass of the casting in kilograms. Now let us see some problems let us solve some problems. Now problem one calculate the optimum pouring time for a casting whose mass is 20 kilograms and having an average section thickness of 15 millimeters the material of the casting is grey cast iron take the fluidity of the material as 28 inches. Now for grey iron castings when the weight of the casting is less than 450 kilograms pouring time T is equal to K multiplied by 1.41 plus T by T by 14.59 multiplied by W where K is the fluidity factor that is the fluidity of the iron in inches divided by 40 T is the average thickness of the section in millimeters W is the mass of the casting in kilograms. Now just let us substitute in the formula pouring time is this is the formula for the grey iron castings for less than 450 grams K is that is equal to 28 by 40 T is the 15 mm you can see here this is the thickness is 15 millimeters. Now W mass of the casting is 20 cases 20 cases just we will substitute in this formula now pouring time T is equal to when we substitute we get T is equal to 7.632 seconds. So this is the way to calculate the pouring time for grey iron castings when the what is a pouring what is a weight of the casting is less than 450 kilograms. Now problem two solve the previous problem if the material is steel for steel castings pouring time T is equal to 2.435 minus 0.3953 log W multiplied by root of W and here W is the weight of T casting. Now just let us substitute the W in this formula then we will get pouring time is equal to 8.5825 seconds. So this is the way to calculate the pouring time when the material is steel. Now let us take another problem design the gating system for a casting made up of cast iron whose dimensions are 500 into 250 into 50 millimeters. Density of the solid cast iron is equal to 7.86 grams per cubic centimeters. Density of liquid cast iron is equal to 6.9 grams per cubic centimeters. Fluidity length is equal to 22 inches. Height of the cope is equal to 100 millimeters. Now here the dimensions of a casting are given simple dimensions the length is 500 mm width is 250 mm and the thickness is 50 mm. How to what is a design the gating system means initially we have to design the choke what is the diameter of the sprue at the choke we have to design. Next one what is the cross sectional area of the runner after that we have to design the ingates how many ingates should be there and what should be the cross sectional area of each ingate. So this is the meaning of the design of the gating system. Now solution initially we need to find out the volume of the casting that is equal to 500 into 250 into 50 that is equal to 6.25 into 10 to the power of 3 cubic centimeters. Now mass of the casting is equal to volume into density it is a well known formula that is equal to now this is the density 7.86 into 6.25 into 10.3 that is equal to 49.125 kilograms. So this is the mass of the casting assuming casting yield as the 70 percent again we need to recollect what is the meaning of the casting yield. Now here what is the weight of the casting or the mass of the casting it is 49.125 kilograms if this is the mass of the casting what should be the mass of the poured metal. Poured metal do not think that the weight of the poured metal also will be 49.125 kilograms. So this much will be occupied in the mould cavity what about the sprue, what about the pouring cup, what about the runner, what about the ingates, what about the riser, here also liquid metal will be filled so this requires extra metal then the weight of the casting that is why the mass of the poured metal is always greater than the mass of the casting. So now what is the casting yield, casting yield means mass of the casting divided by mass of the poured metal multiplied by 100 usually we have seen that it varies from 70 to 80 percent. Now let us assume the casting yield as the 70 percent. Now weight of the poured metal it is not the casting you see weight of the poured metal is equal to mass of the casting divided by casting yield. Now casting yield is we assume 70 percent means here it is 0.7. Now it becomes weight of the poured metal is equal to 70.18 kilograms. Now pouring time T is equal to this is the formula K multiplied by 1.41 plus T by 14.59 multiplied by root of W that is equal to where K is 22 by 40. Now here we are substituting for T that is the thickness of the casting that is the 50 mm and here we can see this is W is the 70 we have rounded it to 70. Now pouring time is almost equal to 23 seconds. Now we need to calculate the choke area. Now in the previous lecture we have derived this expression for the choke area. Choke area Ac is equal to W divided by C rho T root of 2g hc where Ac is the choke area and in this case we have to find out W is weight of the poured metal that is in this case it is equal to 70 kilograms C is the efficiency factor what is that because of the friction involved in the flow of the liquid metal. So this efficiency factor comes into picture whereas for an ideal fluid this will be 1 but here generally it varies between 0.7 to 0.9. And here we are taking 80 as as 0.8 rho is the density of the liquid metal that is equal to 6.9 grams per cubic centimeters that is also equal to 6.9 into 10 to the power of minus 6 kilograms per cubic millimeters. T is equal to pouring time just now we have calculated that is the 19 seconds. Ac is the acceleration due to gravity that is 9800 millimeters per second square hc is the effective height of the metal head usually it is height of the mould box that is 100 mm. Now choke area is given by Ac is equal to 70 that is the weight of the poured metal 6.9 into 10 to the power of minus 6 into 19 this is the pouring time and here we have taken say 0.826 whole multiplied by 2 into 9800 into 100 choke area is equal to 477 square millimeters this is the what say choke area generally the choke area will be a circle that be the case what should be its diameter. Now if we simplify this the diameter of the choke will be 24.60 millimeters or that is almost equal to 25 millimeters. Now what is the gating ratio for grey cast iron for grey cast iron the gating ratio is Ac is to Ar is to Ag is equal to 1 is to 4 is to 4 and we are taking here unpreserved gating system why because the casting is a simple one you see a simple what is a rectangular cross sectional one no complex shapes are there no intricate shapes are there that is why we are choosing the unpreserved gating system. So this is the unpreserved gating system ratio 1 is to 4 is to 4. Now we have already found out the Ac the choke area. Now next we need to find out the design the runner what does it mean what is the cross sectional area of the runner cross sectional area of the runner in the case of the unpreserved gating system Ar is equal to 4 into Ac that is equal to 4 into 477 that is equal to 1908 square millimeters. Now usually the runner will be of square cross section that be the case height will be equal to runner let us assume that is equal to A. Now what is the cross section of the runner what is a height multiplied by width that is equal to A square A square is equal to 1908 and if we simplify this A is equal to 43.6 that is almost equal to 44 millimeters. Now let us what is a specify the dimensions of the runner height of the runner is equal to 44 millimeters width of the runner is equal to 44 mm same then the question is what is the length length we have not specified length it what say always it depends upon the length of the casting. So, we need not specify. Now we need to design the ingates the total cross sectional area of ingates Ag is equal to 4 into Ac that is equal to 4 into 477 that is equal to 1908 square millimeters. Now in this case we are taking two ingates the cross sectional area of each ingate is equal to total cross sectional area divided by 2 that is equal to 954 square millimeters. Now usually the cross section of an ingate will be a rectangle and the width of the rectangle will be double then height of the rectangle. So, if the height of the ingate is A then the width of the ingate will be 2A. Now this is the cross section of the ingate 2A square is equal to 954 and if we simplify this A is equal to 21.8 that is almost equal to 22 millimeters. Now let us specify the dimensions of the cross section of the in each ingate height of each ingate is equal to 22 millimeters width of each ingate will be equal to 44 millimeters. Now let us take another problem. Design the gating system for a casting of size 400 into 200 into 40 millimeters made up of steel. The casting has two thin fins on the two sides. The dimension of each fin is 250 into 50 into 3 millimeters. Density of the solid steel is equal to 7.86 grams per cubic centimeter. Density of liquid steel is equal to 6.9 grams per cubic centimeter. Next one height of the cope box is equal to 150 millimeters. Now we need to design the gating system means what is the cross sectional area of the choke, what is the cross sectional area of the runner, what is the cross sectional area of the ingates. So, that is our task. Now initially we need to find out the volume of the casting that is equal to this much that is 3.2 into 10 to the power of 3 cubic centimeters. Next one volume of the two fins there are two fins are there on two sides. Volume of two fins is equal to each fin dimension is 250 into 50 into 3. So, 2 multiplied by 250 into 50 into 3 that is equal to 75 cc. Total volume of the casting is equal to 3.2 into 10 to the power of 3 plus 75. This 75 is the volume of the thin fins. So, total volume of the casting is equal to 3275 cc. Now mass of the casting is equal to volume into density. Now the density of the solid casting solid steel 7.86 7.86 into 3275 that is equal to 25.741 kilograms. Now again we will assume casting yield as 70 percent that be the case weight of the poured metal is equal to mass of casting divided by casting yield that is equal to 25.741 divided by 0.7 that is equal to 36.773 kilograms. Now we need to find out the pouring time. Now what is the material of the casting here? It is the steel you can see here it is steel. So, for steel the pouring time T is equal to 2.4335 minus 0.3953 log W multiplied by root of W and where W is the mass of the poured metal in kilograms and if we substitute the mass of the poured metal in this formula we will get the pouring time as 11 seconds. Now we are going to design the choke area choke area AC is equal to W divided by C rho T root of 2 G H C and AC is the choke area and we need to find out that W is the weight of the poured metal that is equal to 36.773 kilograms C is the efficiency factor that varies between 0.7 to 0.9 and here we have taken 0.8 and density of the liquid metal that is equal to 6.9 into 10 to the power of minus 6 kilograms per cubic millimeter pouring time is equal to 11 seconds G is the acceleration due to gravity 9800 millimeters per second square H C is the effective height of metal head that is equal to 150 millimeters. Now this is the choke area we are substituting different terms in the formula this is the weight of the poured metal and this is the what say see now this is the weight of the liquid density of the liquid metal and this is the pouring time. Now choke area AC is equal to 353.2 square millimeters. Now the choke area is usually a circle now from this if we simplify this the diameter of the choke is equal to 21.2 millimeters or that is almost equal to 21 millimeters. So, we have designed the choke area next one we need to design the runner before that we need to specify the gating ratio since the casting has got two fins on the two sides. Now if we want the molten metal to successfully fill those what say thin sections we must to what say adopt the pressurized gating system. So, in the pressurized gating system AC is to AR is to AG is equal to 1 is to 2 is to 1.5 when we are doing with the steel when the material of the casting is steel. Now we will design the runner design of the runner. Now the cross sectional area of runner AR is equal to 2 into AC that is equal to 2 into 353.2 that is equal to 706.4 square millimeters. Now for a runner the cross section is a square height of the runner will be equal to width of the runner let us assume that is equal to A then A square that is the cross sectional area of the runner is equal to 706.4 and if we simplify this A is equal to 26.57 or it is 27 mm. Now let us specify the dimensions of the cross section of the runner height of the runner is equal to 27 millimeters width of the runner is equal to 27 millimeters. Next we need to design the ingates. Now the total cross sectional area of ingates AG is equal to 1.5 AC that is equal to 1.5 into 353.2 that is equal to 529 529.8 square millimeters. Number of ingates here we are taking 2. Now the cross sectional area of each ingate is equal to 529.2 529.8 divided by 2 that is equal to 264.9 square millimeters and again the cross section of an ingate is a rectangle and the width of the cross section will be double than its height. Now the height of the ingate is equal to A then the width of the ingate will be 2A. Now this is the cross sectional area 2A square is equal to 264.9 then if we simplify this AG is equal to 11.5 and 2A is equal to 23 mm. Now let us specify the dimensions of the cross section of the ingate height of each ingate is equal to 12 mm this is almost 12 mm width of each ingate is equal to 24 mm. So we have designed the ingates also. Now let us consider some case studies. Now you see this one this is the casting. Now the here is the what say sprue means here the sprue is perpendicular to this spot. Now and this is the what say runner and here we can see the gate and the molten metal flows and here and it fills the cavity the molten metal stream it flows like this and it flows like this and it flows like this. This is the pore design this is a pore design then what should be the better one and what is the alternative you see here this is the sprue and this is the casting mould cavity this is the mould cavity. Now the runner is like this the runner is surrounding the mould cavity you see this whole thing this is the what says sprue and this is the runner. The runner starts from here and it goes like this and it goes like this and up to here this is the runner again the runner is this side also it is like this and here you can see this side there are three ingates here is one ingate second ingate and fourth ingate and this side also one ingate second ingate and third ingates. So the runner is surrounding the mould cavity. So this is the better runner compared to the previous one. So this is a case study this is better. Next one let us take another case let us consider a hitch housing the hitch housing acts as an alignment and locking fixture for connecting the trailer tow hook to the truck tow bar. So this is the shape of the casting and how should be the what say pattern oriented inside the mould box how this runner should be designed. So this is a again a case study. Let us analyse now how should be the pattern orientation now option A now the thing is the pattern has got some complex and intricate features you see these are all the complex and intricate features. Now in this option A first option these complex features are facing upwards and the flat portion is facing downwards. So that is the first option let us take second option in the second option the what say complex and intricate shapes are facing bottom and the flat portion is facing upwards. So this is the second option third option the pattern is like this. Now we have seen three options which is better for this casting. Now you see this option A the intricate shapes are facing upwards that be the case the molten metal may not exactly fill these shapes these intricate shapes that is why this is not preferred. Next one let us option three it is what say almost perpendicular to the moulding board this is also not good what happens again the problem comes with the complex shapes again the these complex shapes may not retain the molten metal but you see the option B the complex shapes are facing downwards. So definitely the molten metal will go inside these shapes and they retain the molten metal. So the option B is preferred for this casting. Now let us analyze further how to design the runner for this casting gating and runner option. Now you see this is the what say pattern now this is the runner you can see here and this is the sprue the molten metal goes like this and it fills and here is one in gate and here is one in gate and it fills like this again this is the again part of the runner the molten metal goes inside like this and it goes like this then finally the molten metal fills the cavity. So this is the option one let us say take another option now this is the sprue here is the sprue now here is the sprue the runner is surrounding the casting like this again this side also it is surrounding the casting. Now here is one in gate and here is one in gate and here is one in gate and here is one in gate and here is one in gate and here is one in gate likewise there are several ingates and the runner is surrounding the casting. Now, we have studied two cases, two options for the runner design, which one is better? This is not preferred, because the molten metal may be it may not what say feel uniformly, the molten metal initially it feels then slowly it progresses or it flows to the extreme what say side, in which case it may not be good. By the time it travels from this portion to that portion, there will be a temperature drop will be there. On the other hand, this is preferred, why? Because the runner is surrounding the casting, from all directions the molten metal is parallely entering into the mould cavity, then the filling of the mould cavity will be uniform. So, the second option is preferred in this case. Friends, in these two lectures, we have seen that the elements of the gating system are pouring cup, sprue, sprue well, runner, runner extension, ingates and the riser. And we have seen the significances of different elements of this gating system. Now, the primary question is when to design the riser? First question is whether to design all these elements first and finally to design the riser or to design the riser first? That is the most important question. First one has to design the riser, though in this list riser appears in the last, but riser should be designed at first, then only one should design the choke area, the runner cross sectional area and the ingates. So, this is very important. Again I am telling the riser must be designed first, then only all other elements must be designed. So, with this we are completing the design of the gating system. In the next class, we will see the casting defects. Thank you.


Everybody forced here. We've been a lot of five packs 2022 Penany father's day packs is five. I'm here for David a thanks. David. Good luck. Here we go I'm gonna travel Lawrence and a Crystal Lave rookie A lot of the rookie receivers are really awesome Starbury Jaden Hardy too bad Berks got hurt Forget Dotson's out for a couple weeks to now the hamstring Brett and a cunning ham rookie. Oh Berks is a turf toe Yeah, turf token could be like Week to week it could be like a month Luca Donchance there got a red Nice is Patrick Mahomes guy looked like money last night 19 of 99 Red Raiders color match, too. That's awesome. You tore up a good bucks defense last night Redded them. They really want a back Mike Trout and a rookie Gary Wilson as well. Good five packs there for you, David. Thank you again We'll get the cards out to you shortly


You're right. So I just wanted to give some feedback. I know this was from Mr. Diligent who Who posted this as a CPR and it's not quite a CPR. So the key to any trading strategy is the fact that you want Traders to be convinced of the level because traders, you know, there's many different trading strategies, but they all revolve typically around Support and resistance. So support and resistance pretty much I'm drawing it on a price chart.  That's basically how it works But it has to be obvious that that is a rejection and that is a rejection Right, and then prices come back and they might trade off that level and then, you know Then it might come down and it might trade off that level. Yeah, so That's pretty much what we're looking at now This is slightly different because what you've done is you just put the wick there and The wick there now, although and this is an hourly chart as well Now, although I can see what you're saying, but it's not obvious, right? So if you were to go up to another time frame, let's say a four hour You would not see this as a Level, you know, one of the key things as well another key thing is that, you know For the best levels, it has to be obvious on really kind of like lower and higher time frames I know that if you're trading sometimes like a 15 minute or a five minute time frame It's not gonna levels aren't always going to be obvious on a daily, for example, or weekly Because they're quite extreme But typically it wants to, you know, that level wants to be First of all, obviously, you know rejected, right?  Whereas this is not really You know any kind of rejection that wick And so if I was drawing support and resistance, I would have to draw it like that That would be where You know the resistance because that's obvious to everybody if you went down to a 15 minute or if you went up to, you know, some sort of four hour or daily You would see that there was a rejection there and you can see it also as well here where you get that large rejection and so, um, I would never draw the The resistance or support from that area there just based off of one hourly candle wicks It's all about the price rejection So what you would want to really see is then prices do something like that because this is now becomes obvious, right? This becomes obvious This because because you look to the left and you can see that that is resistance that becomes support traders go long and then Let me go back a bit And then what happens is let's say for example, you can start to see, you know prices now are, you know traders are buying here somewhere around here and if you get something like this We get the breakout traders Then you get Those traders being caught on the wrong side of the market then prices pulled back to that area And that's the capture That's the pain And that's the relief That's what you want to see but the level starts from down here Yeah, so look for the rejections Massive, you know the bigger the big price rejections not these Not these candle stick wicks. Yeah, so price has to really kind of move away From that level. Yeah price didn't really move away from that level. It touched the level But it didn't move away from it there price touched the level. Yeah, it moved away, but It's not really You know, I mean What I would consider a support support level, especially if you look to the left So again, just to reiterate you want levels that some that um, the best levels anyway are the ones that you know, you can have the rejections Yeah, so you can obviously see that that would be resistance But even better if you're looking at for example You know, um looking at historical Levels you want to see Resistance as well as support for you to then start to look to see what kind of setup So around here, whether that's going to be a potential for a stop hunt Yeah, and then potential for a cp up All right, take care and uh, hope that helps


But I found that when when annotating in a social setting where students can see one another engaging with ideas. It helps students very quickly recognize that they can do hard things or that you know their question is a question that was shared widely by everybody else in the class. Not only does this help to increase engagement, but it allows students to have some a sense of authority or self worth. And it comes to directly engaging with content, what really whether it's, you know, a two pack song or, you know, the chapter two of the doubt aging, or even, you know, an article from vice magazine so I find that the power of collective annotation is, magnifies engagement tenfold, but also really is a tool that helps with this piece about belonging.


There's Hatton and Magnesites. That was on the way out dope to hit the gym. When I saw on the computer that, uh, going through Yahoo News basically, that someone made a George Lucas cut of the trailer. So I said I gotta see this. I gotta react for the Magnesites before I hit the gym.  I just woke up about half hour ago, but I said let's have a good morning laugh or a good morning reaction. Let's get excited before hitting the gym and doing my thing. So without further ado, let's do this. Shall we? About this trailer and what he did to his first three movies. We all think, okay, because when he did those reboots or whatever you want to call them or those, you know, those remakes, those special editions, he added all the extra special effects. He's telling us this is what my original vision was. And masses of Star Wars fans was like, well, this was your original vision.  Your vision ain't 20, 20. Like people were angry at what he did. They did not like it. They hated what he did. This is my question. See, being in sales and studying psychology and stuff like that, I know that really do not like change. They actually hate change, especially when it's routine or it's something, it's a change in something that is beloved, something that is true and near and dear to their heart. So having said that, were those remakes, okay, those re edits, whatever you want to say of them, were they really that bad? In other words, if when the series first came out in Star Wars, you know, you know, episode three and voting Friday, when they first came out, would we have hated them? And I don't think so. I don't think that that is the case. I do not think that we would have hated them. I think that there would have still been hits. I still think they would still be classics and everything else. I just think it's just my own personal view. I just think that because he made those changes, that a lot of a lot of ways seemed to hookie to us.  You didn't have to do that. You know what I mean?  But he wanted to do stuff.  So he did what he could with what he had until the technology was available. You know, that kind of reminds me of really, what was it? Iron Man 2? Was it? I think it was two. When he, when his dad had kind of figured out how to have the arc reactor, the extra power or whatever it was, but he knew he couldn't do it in his time because, you know, maybe his mind wasn't as ready yet or the technology wasn't there. But he knew Tony would eventually figure it out. Kind of, kind of reminds me, kind of on that same plane. Kind of like he had a vision. He did the best that he could with what he did because his mind was so great at science fiction. He ended up creating a masterpiece. But when he went to change, re-edit, revise his masterpiece, people's like, no, no, no, no, no, no. You're not going to change what I love. So tell me what y'all think. And my question is this. If the re-edits, the remakes, the re-envisions that came out as the originals, do you think they would have been as big of a hit? I say yes. Let me know what y'all think. Or me and subscribers.