CHAPTER 07.06: DISCRETE DATA INTEGRATION: via Regression Method
In this segment, I'm going to show you another method of integrating discrete functions. We have already talked about how to integrate discrete functions by using trapezoidal rule with unequal segments, or by using the polynomial method, or by using the spline method. This is yet another method where regression is allowed for discrete data, to be able to find out what the integral under that discrete function is. So we're going to take a pragmatic example here to be able to do this part of the problem, or to illustrate this problem. Here is a case of a trunnion which needs to be shrink-fit into a hub, so this is the trunnion, that is the trunnion, so this is simply a hollow shaft which you have there, and this is the hub. Actually, this example has been taken from the assembly of the fulcrum of a bascule bridge, which you must have seen if you live close to the waterways. So the way they make the fulcrum is by taking this trunnion here, and they shrink-fit it into the hub so as to make the single assembly of the trunnion hub. So one of the ways to shrink-fit this trunnion into the hub is by taking this trunnion, cooling it down in some liquid medium, such as dry ice and alcohol, or liquid nitrogen, and what happens is that the diameter of the trunnion becomes smaller by cooling it into a . . . by cooling it in a cooling medium, and only then can it go into the hub, otherwise it won't fit into the hub. So let's go ahead and look at this example, how this is related to integrating discrete functions, and using regression to do that. So in this particular case, we're taking an actual real life problem, where the magnitude of contraction which was needed for the trunnion was 0.015 inches, so they want it to be put into a cooling medium so that you get at least that much contraction when you are . . . when you are taking it out of the liquid medium and trying to put it in the hub. So let's go ahead and see, from the data which is given to us, whether we get this much contraction or not. If we are not getting this much contraction, then the trunnion is not suitable to go through the hub for a shrink-fitting procedure. So the example which we have taken here is that, in order to be able to find out what is the change in the diameter of the trunnion, this is the change in the diameter of the trunnion, is given by the diameter of the trunnion multiplied by some integral of the thermal expansion coefficient. And the reason why we have to integrate the thermal expansion coefficient with respect to temperature is because, as you can see from this table right here, that the thermal expansion coefficient varies quite a bit when you go to cryogenic values. So this is the thermal expansion coefficient of cast steel at room temperature, which is 6.47 10 to the power -6, but if you're getting very close to dry ice and alcohol mixture, your temperature . . . your thermal expansion coefficient is somewhere between 4.72 and 5.43, and if you are going to liquid nitrogen, then you are very close to this number for the thermal expansion coefficient. So you cannot simply use your regular physics . . . you cannot use your regular physics formula D times alpha times delta T, and the reason why that is so is because alpha is not a constant. So one of the ways to do the problem is be regressing this data, this discrete data which is given to you, to a regression model for alpha, and now if we can do that, then we can simply integrate it by using our integral calculus knowledge to be able to do that. So in order to be able to appreciate that whether we have all the inputs to be able to calculate this delta D, which is . . . which we want to be at least 0.015 in magnitude. We are given that the room temperature is 80 degrees Fahrenheit, that goes right here, we're given that this is the dry ice alcohol mixture temperature, and this is given as the diameter of the trunnion. So we want to see that whether this information which we are given, so we're given Ta, we're given Tc, which is the temperature of the cooling medium, we are given alpha as a function of temperature at certain points, and we want to be able to be able to find out what delta D is. So let's go ahead and see how that is done. So the way it is done is that what we're going to do is we are going to take this data which was given to us, and we are going to regress it to a second-order polynomial, and the second-order polynomial which we are wanting to regress to is a0, plus a1 T, plus a2 T squared. So we are using a second-order polynomial to regress the eight data points which are given to us, and then we can use the second-order polynomial to integrate under the curve, because everybody knows how to integrate a second-order polynomial, to be able to get how much is the change the diameter. So this is what we obtained, we obtained alpha as a function of temperature. So this is what we obtained as the second-order polynomial regression model. How we obtained the second-order polynomial regression model is shown in a separate segment, and you can refer to that to be able to see that how do we get the coefficients of the second-order polynomial, which we've seen. So now it is pretty straightforward, because we now know what alpha is, so we should be able to substitute the value of the second-order polynomial into our integral formula to be able to find out how much is the contraction taking place. As you can see, the second-order polynomial regression is doing a good job of, visually at least, we can see that it does a good job of approximating the data which is given to us, there are other ways to figure out whether this particular second-order model is adequate or not. And so, recapping, we have all the information available to us. We know that the room temperature is 80 degrees Fahrenheit. We know that the dry ice alcohol temperature is -108 degrees Fahrenheit. We also know that the diameter of the trunnion is 12. . . . the outside diameter of the trunnion is 12.363 inches. We also calculated the value of alpha as a second-order polynomial by regressing the eight data points which are given to us to a second-order polynomial. So all we would need to do is to substitute the different values which are given to us, this is the diameter, this is the room temperature, this is the dry ice and alcohol temperature, this is the second-order polynomial which just derived, and then we are integrating with respect to temperature. Now, keep this in mind, that this one right here, we are multiplying by 10 to the power -6 because the alpha which we calculated right here is given in micro-inch per inch . . . is given in micro-inch per inch per degree Fahrenheit, so that's why we are multiply by 10 to the power -6 so as to be able to get the answer in inches, and this is the magnitude of contraction which we are getting by integrating the alpha expression with respect to temperature, and multiplying it by the diameter here. But as we see that the diametric contraction need was how much? 0.015 inches or more, but that's not what we are getting. So we have to be wary about putting this particular trunnion into a dry ice and alcohol mixture, and we are finding out that we're getting 0.0137 inches as the diametric contraction, when actually we need 0.015 inches. So as an exercise, what I would like you to do is go ahead and . . . go ahead and change this -108 degrees Fahrenheit, which is dry ice and alcohol mixture temperature, to -321 degrees Fahrenheit, which is actually the boiling temperature of liquid nitrogen. So if you're going to, instead of putting it in dry ice and alcohol mixture, which we find out is not enough, you're getting only this much contraction, go ahead and put it in the liquid nitrogen mixture of -321 degrees Fahrenheit, and see whether you are getting enough contraction which is more than 0.015 inches. And that's the end of this segment. |