1 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 2 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 3 00:00:13,100 --> 00:00:18,619 of science and engineering? That's what we're gonna do in this video. 4 00:00:18,619 --> 00:00:21,859 This video is part of the problem solving video series. 5 00:00:21,859 --> 00:00:27,460 Problem-solving skills, in combination with an understanding of the natural and human-made 6 00:00:27,460 --> 00:00:32,659 world, are critical to the design and optimization of systems and processes. 7 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 8 00:00:38,399 --> 00:00:43,899 College. And I excel at streetfighting mathematics. Before watching this video, you should be 9 00:00:43,899 --> 00:00:49,359 familiar with free body diagrams and dimensional analysis. 10 00:00:49,359 --> 00:00:54,370 After watching this video, you will be able to: Model drag to predict terminal velocities; 11 00:00:54,370 --> 00:01:01,370 and Determine the fuel efficiency of a car. 12 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 13 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, 14 00:01:12,770 --> 00:01:15,430 the equations of fluid mechanics. 15 00:01:15,430 --> 00:01:21,690 Now after 10 years of learning mathematics, you'll discover this is a really hard problem. 16 00:01:21,690 --> 00:01:26,240 You can only solve it analytically in certain special cases like a sphere moving at really 17 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 18 00:01:31,590 --> 00:01:36,430 at any reasonable speed. So we need another way. 19 00:01:36,430 --> 00:01:40,990 By applying approximations artfully, we're gonna find the drag force and the fuel consumption 20 00:01:40,990 --> 00:01:47,910 of a car with a simple experiment and some scientific and engineering reasoning. 21 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 22 00:01:54,800 --> 00:01:59,650 depends on the quantities that control it. Drag force F depends on the density of the 23 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 24 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 25 00:02:09,250 --> 00:02:14,220 represent by the area; traditionally, the cross- sectional area, to some power. And 26 00:02:14,220 --> 00:02:19,190 it also can depend on the viscosity here, the kinematic viscosity of the air to some 27 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 28 00:02:25,560 --> 00:02:26,880 experiment. 29 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 30 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 31 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 32 00:02:42,800 --> 00:02:46,950 edge to that edge in order to make the 2 cones. 33 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. 34 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 35 00:02:58,680 --> 00:03:03,959 approximately? Is the big cone roughly twice as fast, is the small cone twice as fast, 36 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. 37 00:03:10,200 --> 00:03:17,200 You may want to try the experiment yourself. 38 00:03:19,990 --> 00:03:25,300 The experimental result is that they both fall at roughly the same speed. Now what does 39 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 40 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 41 00:03:37,430 --> 00:03:44,430 fill in the diagram. 42 00:03:45,209 --> 00:03:50,020 There is drag and gravity. Because the cones are falling at their terminal speed, in other 43 00:03:50,020 --> 00:03:54,709 words at the constant velocity that they reach pretty quickly, the drag must equal the force 44 00:03:54,709 --> 00:03:59,160 due to gravity. So the drag force can easily be measured just 45 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 46 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 47 00:04:07,709 --> 00:04:09,819 out one of the exponents. 48 00:04:09,819 --> 00:04:15,090 Knowing that the drag equals the force due to gravity, let's compare the drag forces, 49 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 50 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, 51 00:04:25,319 --> 00:04:30,960 four times the mass, and four times the force due to gravity. Therefore, four times the 52 00:04:30,960 --> 00:04:32,960 drag force. 53 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 54 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 55 00:04:43,340 --> 00:04:49,110 the exponent. Now, what about the speed? The speeds were the same; that was the result 56 00:04:49,110 --> 00:04:53,979 of our experiment, so no matter what the exponent here, the speed contributes to the effect 57 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 58 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 59 00:05:07,330 --> 00:05:14,330 is times four and raised to the unknown exponent. And here, the kinematic viscosity, that's 60 00:05:14,330 --> 00:05:20,919 the kinematic viscosity of air and both cones feel the same kinematic viscosity. Therefore, 61 00:05:20,919 --> 00:05:25,520 that contributes a factor of one just like the density does. 62 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 63 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. 64 00:05:39,960 --> 00:05:44,169 Now that we have found one exponent, we can find the remaining three using dimensional 65 00:05:44,169 --> 00:05:46,308 analysis. 66 00:05:46,308 --> 00:05:51,539 We make sure that the exponents chosen on the right side produce dimensions of force, 67 00:05:51,539 --> 00:05:56,729 which are the dimensions on the left side. A force has dimensions of mass, length per 68 00:05:56,729 --> 00:06:01,740 time squared, that's mass times acceleration. 69 00:06:01,740 --> 00:06:07,559 Density has dimensions of mass per length cubed, raised to this unknown exponent. Velocity 70 00:06:07,559 --> 00:06:13,689 has dimensions of length per time, raised to this unknown exponent. Area has dimensions 71 00:06:13,689 --> 00:06:19,409 of length squared, raised to the power 1, we've already figured that out. And kinematic 72 00:06:19,409 --> 00:06:24,529 viscosity; now, that's the regular viscosity divided by the density. It has dimensions 73 00:06:24,529 --> 00:06:31,439 of length squared per time raised to this unknown exponent 74 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, 75 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 76 00:06:42,039 --> 00:06:46,529 way to get the mass to work out is to make this exponent one. 77 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 78 00:06:52,719 --> 00:06:57,719 have two unknowns, this exponent and this exponent. 79 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 80 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 81 00:07:08,129 --> 00:07:11,619 to the minus one so far. 82 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 83 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 84 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 85 00:07:27,930 --> 00:07:32,520 a two... and this exponent here a zero. 86 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 87 00:07:36,819 --> 00:07:43,819 that none of them will get both length and time correct. 88 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 89 00:07:52,279 --> 00:07:58,659 is proportional to density to the first power, speed squared, area to the first power we 90 00:07:58,659 --> 00:08:04,399 got from our home experiment and viscosity here to the zeroth power. 91 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. 92 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. 93 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 94 00:08:19,689 --> 00:08:21,830 cone. Next we're going to race them. 95 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. 96 00:08:28,779 --> 00:08:35,169 Pause the video and use the formula to make a prediction. 97 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 98 00:08:40,419 --> 00:08:46,560 side. What about the right side? Again, both contestants feel the same air density. The 99 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 100 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 101 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 102 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 103 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 104 00:09:12,810 --> 00:09:14,779 answer should be two to one. 105 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. 106 00:09:21,980 --> 00:09:28,100 It looks like our prediction was correct! Now we feel pretty confident that we can use 107 00:09:28,100 --> 00:09:35,100 this formula to try to figure out the fuel consumption of a car. 108 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 109 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 110 00:09:44,899 --> 00:09:49,769 gonna use to estimate the fuel consumption of a car on the highway. 111 00:09:49,769 --> 00:09:53,920 Fuel efficiency is measured as the distance that that the car travels on one gallon or 112 00:09:53,920 --> 00:09:57,000 liter of gasoline. 113 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 114 00:10:01,139 --> 00:10:06,449 drag force, which is the formula we've just found. 115 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 116 00:10:13,389 --> 00:10:20,329 4 liters. And gasoline is roughly like water so, there are 1,000g/liter. 117 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 118 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 119 00:10:30,639 --> 00:10:33,509 come from 11 grams of butter. 120 00:10:33,509 --> 00:10:38,170 But now we want to convert kilocalories to some reasonable metric unit. So that's four 121 00:10:38,170 --> 00:10:45,170 joules/calorie or four kilojoules per kilocalorie. And that all divided by the drag force so 122 00:10:46,910 --> 00:10:50,470 let's put in those terms one by one. 123 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 124 00:10:56,660 --> 00:11:03,060 for highway speeds. Say 100 kilometers an hour, 65 miles an hour. That's roughly 30 125 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 126 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 127 00:11:16,550 --> 00:11:23,550 you do that, pause the video to check that the dimensions all cancel appropriately. 128 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? 129 00:11:32,550 --> 00:11:37,959 A quick numerical approximation will tell you that you get 50 kilometers, which is 30 130 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; 131 00:11:45,129 --> 00:11:52,129 or 50 kilometers per gallon. In typical European Union units, that's eight liters per hundred 132 00:11:52,319 --> 00:11:59,319 kilometers.To Review With our approximation methods, we've come to pretty reasonable numbers 133 00:12:01,189 --> 00:12:05,519 for the fuel efficiency of a car, without having to solve any of the complicated equations 134 00:12:05,519 --> 00:12:10,910 of fluid mechanics. Instead, we used one simple home experiment, dropping the cones, and the 135 00:12:10,910 --> 00:12:17,699 powerful principle of dimensional analysis. From that, we were able to find all the exponents 136 00:12:17,699 --> 00:12:23,579 here, in the relationship between drag and density, speed, and cross-sectional area. 137 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 138 00:12:28,560 --> 00:12:32,949 as dimensional analysis, they give you an understanding of how systems behave without 139 00:12:32,949 --> 00:12:38,449 having to solve every single last bit of the detailed mathematics. This understanding allows 140 00:12:38,449 --> 00:12:44,529 us to redesign the world.