WEBVTT 00:00:02.800 --> 00:00:08.760 What's the fuel consumption of a car? Say, on the highway moving at highway speeds? We 00:00:08.760 --> 00:00:13.100 can look it up, but can we understand those numbers? Can we predict them from our knowledge 00:00:13.100 --> 00:00:18.619 of science and engineering? That's what we're gonna do in this video. 00:00:18.619 --> 00:00:21.859 This video is part of the problem solving video series. 00:00:21.859 --> 00:00:27.460 Problem-solving skills, in combination with an understanding of the natural and human-made 00:00:27.460 --> 00:00:32.659 world, are critical to the design and optimization of systems and processes. 00:00:32.659 --> 00:00:38.399 Hi, my name is Sanjoy Mahajan, and I'm a professor of Applied Science and Engineering at Olin 00:00:38.399 --> 00:00:43.899 College. And I excel at streetfighting mathematics. Before watching this video, you should be 00:00:43.899 --> 00:00:49.359 familiar with free body diagrams and dimensional analysis. 00:00:49.359 --> 00:00:54.370 After watching this video, you will be able to: Model drag to predict terminal velocities; 00:00:54.370 --> 00:01:01.370 and Determine the fuel efficiency of a car. 00:01:02.329 --> 00:01:07.330 Fuel is consumed in fighting drag, on the highway at least. What's the force of drag 00:01:07.330 --> 00:01:12.770 on a car? That's fluid mechanics. We could try the hard way—solving the Navier-Stokes, 00:01:12.770 --> 00:01:15.430 the equations of fluid mechanics. 00:01:15.430 --> 00:01:21.690 Now after 10 years of learning mathematics, you'll discover this is a really hard problem. 00:01:21.690 --> 00:01:26.240 You can only solve it analytically in certain special cases like a sphere moving at really 00:01:26.240 --> 00:01:31.590 slow speeds and a car is not one of those special cases; certainly not a car moving 00:01:31.590 --> 00:01:36.430 at any reasonable speed. So we need another way. 00:01:36.430 --> 00:01:40.990 By applying approximations artfully, we're gonna find the drag force and the fuel consumption 00:01:40.990 --> 00:01:47.910 of a car with a simple experiment and some scientific and engineering reasoning. 00:01:47.910 --> 00:01:54.800 First, we need to model drag. To do this, we're gonna figure out how the drag force 00:01:54.800 --> 00:01:59.650 depends on the quantities that control it. Drag force F depends on the density of the 00:01:59.650 --> 00:02:04.240 air, to some power which we don't know yet, times the speed of the car moving through 00:02:04.240 --> 00:02:09.250 the air, to some power we don't know yet. It depends on how big the car is, which we'll 00:02:09.250 --> 00:02:14.220 represent by the area; traditionally, the cross- sectional area, to some power. And 00:02:14.220 --> 00:02:19.190 it also can depend on the viscosity here, the kinematic viscosity of the air to some 00:02:19.190 --> 00:02:25.560 power. And what those powers are, we don't yet know. To figure them out, let's do an 00:02:25.560 --> 00:02:26.880 experiment. 00:02:26.880 --> 00:02:32.000 For the experiment, we make two cones. First, by making a large circle and cutting out a 00:02:32.000 --> 00:02:37.480 quarter of it. The large circle has a radius of seven centimeters. And the small three-quarter 00:02:37.480 --> 00:02:42.800 circle has half the radius. Cut them out and then tape this edge to that edge and this 00:02:42.800 --> 00:02:46.950 edge to that edge in order to make the 2 cones. 00:02:46.950 --> 00:02:53.850 Next we're gonna race them, the big cone vs. the small cone. But first, make a prediction. 00:02:53.850 --> 00:02:58.680 The question is: what is the ratio of fall speeds of the big cone and the small cone 00:02:58.680 --> 00:03:03.959 approximately? Is the big cone roughly twice as fast, is the small cone twice as fast, 00:03:03.959 --> 00:03:10.200 or do they fall at roughly the same speed? Pause the video and make your best guess. 00:03:10.200 --> 00:03:17.200 You may want to try the experiment yourself. 00:03:19.990 --> 00:03:25.300 The experimental result is that they both fall at roughly the same speed. Now what does 00:03:25.300 --> 00:03:31.870 that mean for the exponents in the drag force? To decide, use a free body diagram. Here is 00:03:31.870 --> 00:03:37.430 a free body diagram of a cone as it falls. What are the forces on the cone? Pause and 00:03:37.430 --> 00:03:44.430 fill in the diagram. 00:03:45.209 --> 00:03:50.020 There is drag and gravity. Because the cones are falling at their terminal speed, in other 00:03:50.020 --> 00:03:54.709 words at the constant velocity that they reach pretty quickly, the drag must equal the force 00:03:54.709 --> 00:03:59.160 due to gravity. So the drag force can easily be measured just 00:03:59.160 --> 00:04:03.910 by sticking the cones on a scale. We don't actually have to stick the cones on a scale 00:04:03.910 --> 00:04:07.709 because we know what their relative weights are and that's what we're gonna use to figure 00:04:07.709 --> 00:04:09.819 out one of the exponents. 00:04:09.819 --> 00:04:15.090 Knowing that the drag equals the force due to gravity, let's compare the drag forces, 00:04:15.090 --> 00:04:20.459 which are the factors on the right side, of the small and large cones. The large cone 00:04:20.459 --> 00:04:25.319 was made out of a paper circle with twice the radius so it has four times the area, 00:04:25.319 --> 00:04:30.960 four times the mass, and four times the force due to gravity. Therefore, four times the 00:04:30.960 --> 00:04:32.960 drag force. 00:04:32.960 --> 00:04:38.379 Now let's look at the causes one by one. This density here is the density of the air and 00:04:38.379 --> 00:04:43.340 both cones feel the same air so there's no difference here, so that's times one no matter 00:04:43.340 --> 00:04:49.110 the exponent. Now, what about the speed? The speeds were the same; that was the result 00:04:49.110 --> 00:04:53.979 of our experiment, so no matter what the exponent here, the speed contributes to the effect 00:04:53.979 --> 00:05:00.960 by a factor of one; in other words, nothing. Now the area. This is traditionally the cross-sectional 00:05:00.960 --> 00:05:07.330 area. The big cone has four times the cross-sectional area of the small cone so this area factor 00:05:07.330 --> 00:05:14.330 is times four and raised to the unknown exponent. And here, the kinematic viscosity, that's 00:05:14.330 --> 00:05:20.919 the kinematic viscosity of air and both cones feel the same kinematic viscosity. Therefore, 00:05:20.919 --> 00:05:25.520 that contributes a factor of one just like the density does. 00:05:25.520 --> 00:05:30.270 So now we know what the question mark here has to be. The question mark here must be 00:05:30.270 --> 00:05:37.270 one so that this four to the question mark is equal to this four. And so does this one. 00:05:39.960 --> 00:05:44.169 Now that we have found one exponent, we can find the remaining three using dimensional 00:05:44.169 --> 00:05:46.308 analysis. 00:05:46.308 --> 00:05:51.539 We make sure that the exponents chosen on the right side produce dimensions of force, 00:05:51.539 --> 00:05:56.729 which are the dimensions on the left side. A force has dimensions of mass, length per 00:05:56.729 --> 00:06:01.740 time squared, that's mass times acceleration. 00:06:01.740 --> 00:06:07.559 Density has dimensions of mass per length cubed, raised to this unknown exponent. Velocity 00:06:07.559 --> 00:06:13.689 has dimensions of length per time, raised to this unknown exponent. Area has dimensions 00:06:13.689 --> 00:06:19.409 of length squared, raised to the power 1, we've already figured that out. And kinematic 00:06:19.409 --> 00:06:24.529 viscosity; now, that's the regular viscosity divided by the density. It has dimensions 00:06:24.529 --> 00:06:31.439 of length squared per time raised to this unknown exponent 00:06:31.439 --> 00:06:36.689 Now let's solve for these unknown exponents one at a time. Looking first at the mass, 00:06:36.689 --> 00:06:42.039 there is mass here and mass here on the left side, but there is no other mass. So the only 00:06:42.039 --> 00:06:46.529 way to get the mass to work out is to make this exponent one. 00:06:46.529 --> 00:06:52.719 Now, what do we have left? We have to make sure the lengths and times work out and we 00:06:52.719 --> 00:06:57.719 have two unknowns, this exponent and this exponent. 00:06:57.719 --> 00:07:02.300 Let's see how many more lengths we need, given what we already have. We have mass worked 00:07:02.300 --> 00:07:08.129 out, we have two lengths here over three lengths so we have mass per length. So we have length 00:07:08.129 --> 00:07:11.619 to the minus one so far. 00:07:11.619 --> 00:07:16.949 And what about time? Well, we have no times yet except in these question marks so we need 00:07:16.949 --> 00:07:22.279 to get two more times on the bottom two more lengths on the top. So we need to multiply 00:07:22.279 --> 00:07:27.930 by length squared over time squared, and the only way to do that is to make this exponent 00:07:27.930 --> 00:07:32.520 a two... and this exponent here a zero. 00:07:32.520 --> 00:07:36.819 You can pause the video and go ahead and try other combinations of exponents. You'll find 00:07:36.819 --> 00:07:43.819 that none of them will get both length and time correct. 00:07:46.149 --> 00:07:52.279 So now we have all the exponents, we can actually enter that into our force formula. Drag force 00:07:52.279 --> 00:07:58.659 is proportional to density to the first power, speed squared, area to the first power we 00:07:58.659 --> 00:08:04.399 got from our home experiment and viscosity here to the zeroth power. 00:08:04.399 --> 00:08:09.649 Now let's test our formula with a second demonstration and see if it predicts this new situation. 00:08:09.649 --> 00:08:15.279 We're gonna race four small cones stacked on top of each other. One, two, three, four. 00:08:15.279 --> 00:08:19.689 So we stack them to make one small cone that is four times the mass of the other small 00:08:19.689 --> 00:08:21.830 cone. Next we're going to race them. 00:08:21.830 --> 00:08:28.779 But first, what's the drag force ratio between the small cone and 4 small cones stacked together. 00:08:28.779 --> 00:08:35.169 Pause the video and use the formula to make a prediction. 00:08:35.169 --> 00:08:40.419 The four cones weigh four times as much as one cone so that's times four on the left 00:08:40.419 --> 00:08:46.560 side. What about the right side? Again, both contestants feel the same air density. The 00:08:46.560 --> 00:08:50.250 velocity ratio, that's what we're trying to figure out, and let's see what all the other 00:08:50.250 --> 00:08:55.519 pieces are. The area; well here in this case, the stack of four has the same cross- sectional 00:08:55.519 --> 00:09:01.399 area as the one cone so that's just one. And the viscosity doesn't matter. So, this factor 00:09:01.399 --> 00:09:07.100 of four on the left side has to be produced by the v squared; in other words, the v goes 00:09:07.100 --> 00:09:12.810 up by a factor of two so that when it's squared, you get a factor of four. That means the correct 00:09:12.810 --> 00:09:14.779 answer should be two to one. 00:09:14.779 --> 00:09:21.779 Now, let's test our prediction with a demonstration to check whether nature behaves as we predict. 00:09:21.980 --> 00:09:28.100 It looks like our prediction was correct! Now we feel pretty confident that we can use 00:09:28.100 --> 00:09:35.100 this formula to try to figure out the fuel consumption of a car. 00:09:36.500 --> 00:09:40.180 Let's clean up our formula here just a bit and erase the viscosity since it comes in 00:09:40.180 --> 00:09:44.899 with a zero exponent so it doesn't matter. And this is our drag formula, which we're 00:09:44.899 --> 00:09:49.769 gonna use to estimate the fuel consumption of a car on the highway. 00:09:49.769 --> 00:09:53.920 Fuel efficiency is measured as the distance that that the car travels on one gallon or 00:09:53.920 --> 00:09:57.000 liter of gasoline. 00:09:57.000 --> 00:10:01.139 So it depends on how much energy you can get out of a gallon of gasoline divided by the 00:10:01.139 --> 00:10:06.449 drag force, which is the formula we've just found. 00:10:06.449 --> 00:10:13.389 So let's put in some numbers. What's the energy from a gallon of gasoline? A gallon is roughly 00:10:13.389 --> 00:10:20.329 4 liters. And gasoline is roughly like water so, there are 1,000g/liter. 00:10:20.329 --> 00:10:25.220 How much energy do you get out of each gram of gasoline? Gasoline is like fat and from 00:10:25.220 --> 00:10:30.639 the side of a butter packet, you can read that 100 kilocalories, 10 to the 2 kilocalories 00:10:30.639 --> 00:10:33.509 come from 11 grams of butter. 00:10:33.509 --> 00:10:38.170 But now we want to convert kilocalories to some reasonable metric unit. So that's four 00:10:38.170 --> 00:10:45.170 joules/calorie or four kilojoules per kilocalorie. And that all divided by the drag force so 00:10:46.910 --> 00:10:50.470 let's put in those terms one by one. 00:10:50.470 --> 00:10:56.660 First the density here, one kilogram per meter cubed is the density of air roughly. Now v 00:10:56.660 --> 00:11:03.060 for highway speeds. Say 100 kilometers an hour, 65 miles an hour. That's roughly 30 00:11:03.060 --> 00:11:08.329 meters per second. Don't forget to square it. And then the area of a car. Well, it's 00:11:08.329 --> 00:11:15.329 about... two meters wide by 1.5 meters high. Now we have to work all the numbers, but before 00:11:16.550 --> 00:11:23.550 you do that, pause the video to check that the dimensions all cancel appropriately. 00:11:27.290 --> 00:11:32.550 You should have found that the result has units of meters, which is good! But how many? 00:11:32.550 --> 00:11:37.959 A quick numerical approximation will tell you that you get 50 kilometers, which is 30 00:11:37.959 --> 00:11:44.959 miles roughly. The fuel efficiency of a car on the highway is about 30 miles per gallon; 00:11:45.129 --> 00:11:52.129 or 50 kilometers per gallon. In typical European Union units, that's eight liters per hundred 00:11:52.319 --> 00:11:59.319 kilometers.To Review With our approximation methods, we've come to pretty reasonable numbers 00:12:01.189 --> 00:12:05.519 for the fuel efficiency of a car, without having to solve any of the complicated equations 00:12:05.519 --> 00:12:10.910 of fluid mechanics. Instead, we used one simple home experiment, dropping the cones, and the 00:12:10.910 --> 00:12:17.699 powerful principle of dimensional analysis. From that, we were able to find all the exponents 00:12:17.699 --> 00:12:23.579 here, in the relationship between drag and density, speed, and cross-sectional area. 00:12:23.579 --> 00:12:28.560 So the moral of this is that there is an art to approximations, and reasoning tools such 00:12:28.560 --> 00:12:32.949 as dimensional analysis, they give you an understanding of how systems behave without 00:12:32.949 --> 00:12:38.449 having to solve every single last bit of the detailed mathematics. This understanding allows 00:12:38.449 --> 00:12:44.529 us to redesign the world.