Dot Physics

Maybe you have noticed how much material there was (for me at least) in last week’s MythBusters. One of the myths they looked at was the bus jumping over a gap in the road from the movie Speed. I am not looking at that myth, it has been discussed many times in many places. Rather, I am going to talk about scaling the motion. As typical with the MythBusters, they like to make a scaled down version of the event. It’s cheaper that way. In this case, they made a 1/12th scale model of the bus and the road. The question was: how fast should the model go?

The first question to ask is: what do you mean by scale? I will interpret scale to mean that the trajectory of the model bus will have the same shape as the full size bus’s trajectory. I will also assume the dimensions of the model bus trajectory are scaled the same as the other stuff (in this case, 1/12th). The MythBusters can make smaller models. They can make them go at different speeds. However, they can not change the gravitational field (well, at least not very easily). Also, they can not scale time. So, then here is the real question:

How should the speed change such that the trajectory of the model bus is also 1/12th the scale of the real trajectory?

Range of a projectile

To make this problem slightly easier, I am going to first look at the range of an object in projectile motion. Let me assume the following:

  • The bus is launched at a speed v and at an angle theta
  • The bus starts and lands at the same vertical position (which is not exactly true for the scene from Speed – but close)
  • Air resistance is negligible
  • The bus can be treated as a point particle (I will ignore rotational effects)

So, then this is now the projectile motion you will see in a textbook. I have gone over this before, but let me quickly solve for the range the bus will go if it is launched in the above manner.

The key to projectile motion is to realize that the horizontal and vertical motions are independent of each other (except for the time each motion takes). This essentially makes a 2-d problem, 2 1-dimensional problems. Here is a diagram of the forces on an object after it left the ground.

i-183c563df1f29c29aa87fbd2ff31575e-2009-12-15_untitled_1.jpg

There is only one force on the object while it is in the air. That is the gravitational force from the interaction with the Earth. I also put an arrow to indicate the direction of the velocity, just because. Since the only force is the gravitational force in the y-direction (vertical), then there is only an acceleration in the y-direction. There is no acceleration in the x-direction (horizontal). I can write the following two kinematic equations for these two directions (assuming the acceleration in the y-direction is -g):

i-aa439331975707bde918e218a4dbb7d5-2009-12-15_la_te_xi_t_1.jpg

For both cases, I need the initial velocity components in that direction.

i-a475d09e47e4ef8c1a1466943767791f-2009-12-15_la_te_xi_t_1_1.jpg

Where v0 is the magnitude of the launch velocity. Ok, some simplifications. If the object is launched and landing at the same y value, then the y-motion equation is: (which I can solve for time)

i-63333e9b76c4556d63d1e3000fdfa9a4-2009-12-15_la_te_xi_t_1_2.jpg

Quick check, does that have the correct units for time? Yes. Now on to the x-direction. For simplicity, let me say that it starts at x0 = 0

i-69bbc5ee16eebf7802b412ae9625bba6-2009-12-15_la_te_xi_t_1_3.jpg

And now I can use the time from the y-motion. This gives:

i-f3a9a85eccb07dd70922a1d7207f1484-2009-12-15_la_te_xi_t_1_4.jpg

So there – I have a relationship between range and initial velocity. One thing I should note – the angle in the scale model should be the same as in the full version.

Scaling

Ok, suppose that I want to scale the stuff by a factor s such that my new range will be:

i-c189a004b62fa4f1d2e81fb993d51d66-2009-12-15_la_te_xi_t_1_5.jpg

For this particular case, the MythBusters used the scaling factor of s = 1/12. However, left it this way in case you want to biggie size your motion. So, the question is: by what factor should I multiply the initial velocity? First, let me solve the range equation for the initial velocity:

i-46580a1c2bf2408bba1a61b55655b85c-2009-12-15_la_te_xi_t_1_6.jpg

Now, what if I let x = x’/s?

i-d377df0be8e08b720c536dd87e2e5444-2009-12-15_la_te_xi_t_1_7.jpg

If I let:

i-ad216c24f62fdfabecc772974f913b63-2009-12-15_la_te_xi_t_1_8.jpg

Then I can write:

i-08845b306d7e77a2bd837d665b9f2966-2009-12-15_la_te_xi_t_1_9.jpg

In short, if you want to grok this, think of it in the following way. You scale x by a factor s. You can not scale time or g. The range depends on v2, so your scaled velocity is going to scale a little differently.

Back to the MythBusters

In the bus jump episode, they used s = 1/12. The want the real bus to have a launch velocity of 70 mph (just like in the movie). This gives a scaled velocity of:

i-aba1dc886fabc9c814c23ac5f12bb7ce-2009-12-15_la_te_xi_t_1_10.jpg

Which is exactly what Grant (the MythBuster) calculated on the side of the bus for his model. Of course, he derived it a little differently:

i-fa73680c684988bef93bd7d51aa316b6-2009-12-15_i_photo_1.jpg

Or…maybe it is the same. I really can’t tell.

Oh wait! Grant fell victim to one of the classic blunders – The most famous of which is “never get involved in a land war in Asia” – but only slightly less well-known is this: the initial velocity is the velocity RIGHT AFTER it leaves the ground. Check out this equation Grant wrote:

i-3af14a538ab6a5136911867dfb7390ad-2009-12-15_i_photo_2.jpg

To me, this looks like he is saying the initial y-velocity is zero. Which is true before it hits the ramp. However, for projectile motion to be solvable, you have to look at it after the object is launched. I don’t know how he got the correct answer. Maybe he googled it.

Comments

  1. #1 Stretch
    December 15, 2009

    Sharp eyesight on your part. Now, is there a mathematical representation to calculate the size of a crashing vehicle to the amount of fun caused by watching it?

  2. #2 Fran
    December 15, 2009

    My usual classic blunder is to drop a negative sign somewhere. I had a terrible time today trying to relate the acceleration of the center of mass of an object rolling down a ramp to the angle of the ramp, and I kept getting it wrong whether the positive direction was down the ramp or up the ramp… Really annoying when it happens in front of the class I am trying to teach…

  3. #3 Markk
    December 16, 2009

    I thought when he did this there was no ramp – it was straight out. They did the ramp later when this scale model showed the bus would crash every time. So there would be no vertical velocity an instant after the launch.

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