Mythbusters Bus Jump Small Scale

  1. In the movie Speed a speeding bus is shown jumping a long gap in the elevated freeway. The Mythbusters test the physics at 1/12 scale. In the TV show Grant writes an equation to find the scaled down ramp exit velocity, where supposedly one must compensate the small scale speed downward because gravity does not scale down.

    Let the big bus have mass M. The little bus has mass m = M/(12 x 12 x 12).

    If there is zero atmosphere (air density rho equal zero) I figure this does not effect the scale down study. The mass drops out of the equation. However the distance travelled is 1/12 so the ramp exit velocity needed to cross the gap need not be as large.

    I'll be looking closely at the scale down logic eventually, but figured I'd pose the challenge here for comment.
     
  2. jcsd
  3. diazona

    diazona 2,156
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    Interesting question, I've actually often wondered about the accuracy of the Mythbusters' small-scale experiments. (Of course, they're usually good enough to get a sense of what's going to happen at full size, which makes them perfectly accurate for the show's purposes... I'm not faulting the Mythbusters for bad science!) My first thought is that in the absence of air resistance, the bus's track is a parabola, and since a parabola is not scale-invariant (in the sense of a power law), the situation wouldn't scale down perfectly. But that's just a first thought.

    I'll get back to this after I watch the rerun at midnight.
     
  4. diazona

    diazona 2,156
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    OK, saw the episode... sure, the velocity is less at small scale, but not as low as 1/12 of the full velocity. Even if you could scale gravity, it still wouldn't be 1/12 as much.
     
  5. Neglecting air resistance the solution for range and time is easily calculated:

    http://hyperphysics.phy-astr.gsu.edu/Hbase/traj.html#tra9

    Where in full scale there is a two foot drop in height, a very small launch ramp angle, a 50 foot total jump range x = R, and a known top speed for the bus.

    Cut the geometry down to 1/12 scale and solve the same formulas. Why try to scale down gravity if mass and weight do not enter the calculations?

    However with significant deceleration due to wind resistance, there is a compensation factor requiring one to boost the ramp velocity, since the small scale bus will decelerate in the x direction more rapidly than its full scale replica. Again, this does not require scaling for gravity as far as I can tell.
     
  6. Hi, I'm new to this forum and I've been trying to teach myself some basic physics for a while, so I'm not very advanced. I have a basic question - what is meant by "scaling" gravity. I thought the force exerted by gravity was given by the equation
    F=GmM/r squared (haven't figured out equation editor yet either).

    So if a 1/12 scale bus has 1/12 the mass of the original bus, doesn't the equation account for that?

    Thanks for any advice.
     
  7. Well, in the first place, there is no bus in the market which will actually have an axial so strong so as to so as to withstand that huge impulse in such a small amount of time.

    The axial needs an excuse to snap off...hit the breaks too hard, and the probability of it breaking is pretty high.
     
  8. diazona

    diazona 2,156
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    Equation editor? There's no equation editor here... (it's LaTeX, a way to type equations in text, which you've got plenty of time to learn ;-)

    Anyway, yes the gravitational force is F = GMm/r^2, and as you noticed, it is proportional to mass. But Newton's second law, which relates the force and acceleration, is F = ma... also proportional to mass. When you put those together to figure out the acceleration, the mass cancels out. This is why all objects fall at the same rate on Earth (or would, if it weren't for air resistance). So the small scale bus falls at the same rate as the real bus, not at 1/12 the rate.
     
  9. Hey guys I'm new to this forum too, and I made an account after reading this because I was so curious about the episode in question. Diazona, what I'm getting from what you're saying is that they wouldn't have to compensate for scaled down gravity, which is what I originally thought and why I was so confused as to why they attempted to adjust the speed to compensate for their inability to scale down gravity.

    Also on a side note, what type of mathematics/physics backgrounds do most of the experienced editors have here? I'm just finishing up Calc III but I'm still in high school and have no experience with "LaTeX." This isn't the first time I've come to this site for help and I'm looking to use it more.

    Thanks.
     
  10. DaveC426913

    DaveC426913 16,223
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    I have never understood how the bus in Speed could ever clear the landing ramp, under any circumstance.

    Highways are not designed with launch ramp angles. The launch ramp and landing ramp point virtually right at each other. That means that, no matter how fast it's going, the bus's trajectory when it leaves the launch ramp will at best, fall below the landing ramp. The only thing to resolve is how far below.
     
  11. Cursor over the equation below and click on it to get a look at the LaTex code. You can cut and paste to write similar equations or use Show LaTex reference button in the post editor for tools.

    Near-earth gravity is relatively uniform (within 1% error at 100,000 feet) at the standard gravity g, so the acceleration of freefall is derived as:

    [tex]a = \frac{W}{m} = \frac{mg}{m} = g[/tex]

    and this is constant for all bodies falling near earth in zero atmosphere.

    Range R in the scale down study is:

    [tex]R_{S} = \frac{R_{B}}{12} = \frac{50}{12} = 4.17[/tex] feet

    This gives some idea of using LaTex and some of the initial variables one needs to find the scale down ramp velocity assuming zero atmosphere (neglibible wind resistance) to clear the jump.
     
  12. Thank you Diazona. I guess my question then becomes, if a big bus and a small bus fall at the same rate, why do we have to be concerned about "scaling" gravity. In other words, how does simulating 1/12 gravity add to the realism of the scaled down bus jump?
     
  13. This thought just struck: it might generate an equal time of flight. Perhaps the notion of scaling down gravity is worth looking into after all ... just to gain some better insight.

    An estimate of full scale speed is simple using this link:

    http://hyperphysics.phy-astr.gsu.edu/Hbase/traj.html#tra13

    plug in R = 50 feet and angle theta = 2 degrees, neglect the estimated two foot drop from ramp to landing zone, and full scale launch velocity is vB = 152{ft/s}. Note the formula used to compute this velocity.

    Next visit google search field and enter: 152 ft/s in miles per hour

    152 (feet / second) = 103.636364 miles per hour

    So the bus at full scale needs a minimum speed of 104 miles per hour! Maybe just a little less if we account for the 2 foot ramp elevation over the landing zone.

    DON'T TRY THIS AT HOME!

    Edit: work the same calculation at small scale R = 50/12, convert ft/s to mph, and you'll get v0 = 29.9{mph}.
     
    Last edited: Dec 10, 2009
  14. Integral

    Integral 7,345
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    I saw that episode last night. Though I have not worked it out, his attempt at calculating the necessary speed was wrong. There is NO way 20mph was fast enough to jump that gap. They totally blew the physics so it is no surprise that the bus did not even come close to doing what they wanted. Their quandary about scaling gravity is nonsense.

    Frankly, all to frequently Mythbusters makes mistakes like this.
     
  15. So neglecting air resistance and the 2 foot elevation this formula applies:

    [tex]v_{0} = \sqrt{ \frac{Rg}{sin 2 \theta}}[/tex]

    Then if small R and g are both divided by 12 then it appears the launch velocity may be scaled down by 1/12, but I haven't run actual numbers yet.
     
  16. Integral

    Integral 7,345
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    I am not sure what is up with scaling gravity. If you want to scale down the problem you need to pick the distances for your model, then compute the speeds for that set up with g as it stands gravity does not change just because I am doing scale model.
     
  17. I used the Hyperphysics calculator:

    http://hyperphysics.phy-astr.gsu.edu/Hbase/traj.html#tra9

    to run the scale jump with v0 = 35{ft/s}, launch angle theta = 1 degree, and drop height y = -0.167{ft}, and then the scale bus should jump the range R = 4.3{ft} if x2 is the total range (which is unclear). Notice 35{ft/s} *0.68 = 23.8{mph} which is Grant's number, so why did the little bus miss the jump so badly?

    Either the Hyperphysics calculator is in error, or I'm interpreting it wrong, or wind resistance at the small scale is very significant factor (or some combination).

    Anyway that little 2 foot drop probably does help out quite a bit in making the leap at lower launch speed assuming zero atmosphere conditions. I'll have to derive an equation or two to be convinced of the accurate numbers ...
     
  18. Of course it is impossible to scale gravity on Earth! But as an interesting thought problem I would run the calculation at 1/12 g just to see what numbers change or remain the same.

    I pretty much described the process of running a super simple scale down study in my post above if you follow the details.
     
  19. diazona

    diazona 2,156
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    I calculated this last night during the show, I wasn't going to post it right away, but why not ;-)

    The bus takes off, with velocity [itex]v[/itex], from a launch ramp inclined at an angle [itex]\theta[/itex] above the horizontal. That means the horizontal and vertical components of its initial velocity are [itex]v \cos\theta[/itex] and [itex]v\sin\theta[/itex] respectively. Also, let's designate the height of the cusp of the landing "platform" as [itex]y = 0[/itex], so the edge of the ramp is at some height [itex]y_0[/itex] (+2 feet in the movie, according to Mythbusters); and let [itex]x=0[/itex] be the horizontal position of the edge of the ramp.

    The equations for motion accelerated by gravity (without air resistance) are
    [tex]x = v_x t[/tex]
    [tex]y = y_0 + v_y t - \frac{1}{2}g t^2[/tex]
    Putting them together gives
    [tex]y = y_0 + \frac{v_y}{v_x}x - \frac{1}{2}g \biggl(\frac{x}{v_x}\biggr)^2[/tex]
    and substituting in for the velocities gives
    [tex]y - y_0 = x \tan\theta - \frac{1}{2}g \biggl(\frac{x}{v}\biggr)^2\sec^2\theta[/tex]

    If we were to scale down [itex]y - y_0[/itex], [itex]x[/itex], [itex]g[/itex], and [itex]v[/itex] by 12, the ratio [itex]x/v[/itex] would remain the same and the factor of 1/12 in everything else would cancel out. So a perfect 1/12 scale model with 1/12 gravity should work. But if you can't change gravity, you need to change [itex]v[/itex] to compensate for scaling down [itex]y - y_0[/itex] and [itex]x[/itex]. Increasing [itex]v[/itex] by [itex]\sqrt{12}[/itex] seems like it should do the trick.

    I plugged some numbers in using Mathematica and calculated that the real bus would need a velocity of 78 mph to clear the jump, and a 1/12 scale model bus (with full-scale gravity) would need 22.6mph to clear a 1/12 scale model jump. (I guess that assumes the bus is a point particle, but whatever) I would have thought those numbers should be higher :-/ I suspect air resistance.

    P.S. if we neglect the 2-foot elevation, [itex]y - y_0 = 0[/itex] at the landing point, so
    [tex]x \tan\theta = \frac{g}{2} \biggl(\frac{x}{v}\biggr)^2\sec^2\theta[/tex]
    or
    [tex]\sin\theta\cos\theta = \frac{g}{2} \frac{x}{v^2}[/tex]
    or
    [tex]v = \sqrt{\frac{xg}{\sin 2\theta}}[/tex]
    as described on the website.
     
  20. Integral

    Integral 7,345
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    Where does the 35ft/s come from? Why do you say the drop is .16' in one place and 2' in another? Where did the .68 multiplier come from?

    They did not say how they knew the speed of their model so we have more questions then there are answers. When the modle missed as badly as it did they should have gone back to the drawing board to find thier error.
     
  21. diazona,

    That's very close to the derivation I had in mind. I don't quite follow all the trig substitutions yet but it looks pretty rigorous. I may write a little code to plot the function and see how that squares with your work when I get some time. Thanks for the insights.
     
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