 In this video I want to talk about disc clutches. So disc clutches are basically a mechanism for causing two things to rotate at the same speed where they initially are not. And in its simplest form basically what it means is we take two rotating discs, rotating at different speeds, and we press them together until they're rotating at the same speed. And that's kind of a simplification of what we're talking about. So in order to analyze this situation we're going to go ahead and say, you know, assume that we have a rotation occurring and we're going to apply a force to press these two discs together. Now we have a contact material in here because it isn't, you know, terribly efficient to press two solid circles together to make this happen. But we have a little bit of a friction lining here which provides that contact interface. So if I was to take this drawing and look at it from a side view, I'd have a contact surface that looks something like this where it has an inner radius r i and an outer radius r o. And we want to understand the forces and the torques necessary for this all to come into play. You know, what sort of torque would we expect to see transferred between these two things. So to do that we're going to take a little differential section of the contact surface here. So this differential section is a circle with a width dr. So as when we do, you know, differential units of things, we're going to call this dr. So it's a differential piece of radius, which means small, more or less. And some assumptions we're going to make or an assumption we're going to assume a uniform wear rate. So we're going to assume that the entire range of radius from r i to r o wears at the same rate, which means that we can expect a pressure differential as we move out. Because what this is saying is that pressure times radius is a constant due to, you know, the different rotational speed and such from the inside to the outside. So if pressure times radius is constant, that means that the max pressure occurs at the inner radius. And this is useful to us because we can then use that as a as a known point when we look at our differentially sized piece of material here that I'm kind of trying to shade on the picture. So taking that, then we can say that there is a force applied over that differential area. And it's going to be equal to pressure times area. So pressure is just p max. We can go ahead and say that. And we need to think about what the area is. Well, the area is going to be two pi r i, which is circumference. And I'm picking our eye here, because it's where we know a value of the pressure that it's at the maximum. So two pi r i is circumference times width, which is dr. So that gives us an area circumference times radius, or circumference times width of that circular section. And area times pressure then is equal to force. And it's a differential force because it only applies to this small little differentially sized unit that we have going around here. So we've got this little differential equation. We can integrate this and get force. We're integrating over the radius r i to r naught, or r o, excuse me, two pi p max r i dr. So I'm integrating over r. You can see there's no r in my equation. r i is just a constant value. So that's not our differential variable. So we can just integrate over this. And it's going to be equal to all this stuff in the middle times r evaluated from r i to r o, right? If we remember how to carry out our our integrations when we know the limits. So p max r i r o minus r i. Great. We can do a similar thing now in terms of torque. So torque, again talking about this differential piece of torque, we can say it's equal to the differential force, where this is the the normal force applied. We have a friction coefficient, and we have r. So we have a force applied multiplied by a friction coefficient to get a tangential force, which is where we get our torque from, and multiplied by the torque arm, which is going to be radius. And carrying this out in a similar way, or rewriting this and substituting in what we have, we have two pi r i d r p max mu and r. So now I've pulled down this d f quantity up here that I had already written down into this equation and substituted it in and get this over here. So now I can integrate this piece and I get t equals the integral from r i to r o. So two pi r i p max mu r dr. So not much has really changed, but now I have an r inside my integral and I'm integrating over r. So I just taken that into account. So integral of r is one half r squared. So I get, I've dropped the two now because of the one half pi r i p max mu for friction and my r values, the difference between them squared. All right. So now I want to go ahead and do something with this. And one thing that I don't probably have by default is p max. So we want to go ahead and see if we can substitute that with something else. So I'm actually going to take this equation and rearrange it to solve for p max. And it's going to be equal to f over two pi r i r o minus r i. And I can take this and substitute this in to this equation here into this this t equation. Okay. So what we get then is t is equal to f mu over two times r o squared minus r i squared divided by r o minus r i. All right. So we're almost to where we want to be. One thing we need to make note of or that we can make as a simplification, r o squared minus r i squared is equal to r o minus r i times r o plus r i. So this is a nice little algebraic thing that we can try to remember. And what you're going to notice here then is that this r o minus r i would then cancel with this one down here and drop that out of the equation. So what we end up with then is t is equal to f mu over two r o plus r i as an equation that gives us the torque transmitted between two shafts for an applied force and friction and unknown values of the radius. Now one thing I'm going to add in here I'm going to put an n and this is to account for the number of friction interfaces that we have in our clutch system. So I've derived this equation by just saying we're taking two plates and pressing them together. But of course a clutch typically doesn't consist of just two plates being pressed together. There's usually multiple stacks of plates which are being pressed together. So however many friction interfaces we have which is number of contacts between the two sides of the rotating members would tell us how many times we can multiply this torque by. And that gives us that equation. Now we could actually go back and evaluate our equation and dig in a little deeper. But one thing we find is that generally the maximum torque is found when r i is about equal to 0.58 r o. So this is one of those the reasons that we would have this this shape right where we have this inner radius and this outer radius where the contact is happening because there's actually a point of diminishing returns where we get our best outcome when r i is equal to 0.58 r o. Alright so I will stop there.