Mythbusters Bus Jump Small Scale

[SystemTheory]

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 traveled 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.

 

[diazona]

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.

 

[diazona]

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.

 

[SystemTheory]

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.

 

[msakkas]

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.

 

[dE_logics]

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 traveled 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.

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.

 

[diazona]

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.

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.

 

[nasgold]

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.

 

[DaveC426913]

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.

 

[SystemTheory]

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:

$$a = \frac{W}{m} = \frac{mg}{m} = g$$

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

Range R in the scale down study is:

$$R_{S} = \frac{R_{B}}{12} = \frac{50}{12} = 4.17$$ 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.

 

[msakkas]

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.

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?

 

[SystemTheory]

In other words, how does simulating 1/12 gravity add to the realism of the scaled down bus jump?

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}.

 

[Integral]

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.

 

[SystemTheory]

So neglecting air resistance and the 2 foot elevation this formula applies:

$$v_{0} = \sqrt{ \frac{Rg}{sin 2 \theta}}$$

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.

 

[Integral]

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.

 

[SystemTheory]

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 ...

 

[SystemTheory]

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.

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.

 

[diazona]

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

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

If we were to scale down $y - y_0$, $x$, $g$, and $v$ by 12, the ratio $x/v$ 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 $v$ to compensate for scaling down $y - y_0$ and $x$. Increasing $v$ by $\sqrt{12}$ 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, $y - y_0 = 0$ at the landing point, so
$$x \tan\theta = \frac{g}{2} \biggl(\frac{x}{v}\biggr)^2\sec^2\theta$$
or
$$\sin\theta\cos\theta = \frac{g}{2} \frac{x}{v^2}$$
or
$$v = \sqrt{\frac{xg}{\sin 2\theta}}$$
as described on the website.

 

[Integral]

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 ...

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 their error.

 

[SystemTheory]

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.

 

[SystemTheory]

When the modle missed as badly as it did they should have gone back to the drawing board to find their error.

Maybe some fans will complain. They'll go back to the drawing board, and we'll get to see the little bus jump the gap in super-slow motion. I study RC car drivelines as a hobby so to me its an interesting waste of $$$ (which they can turn into profit via entertainment value).

Why do you say the drop is .16' in one place and 2' in another

The drop is 2 feet full scale and windows calculator runs 2/12 = 0.167 foot at 1/12 scale.

Where did the .68 multiplier come from?

In google search field I type: 1 ft/s in miles per hour

Result: 1 (foot / second) = 0.681818182 miles per hour

 

[SystemTheory]

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.

Could you list the exact launch angle, vertical drop, and range used to find these velocities? I'll compare to the Hyperphysics website calculator and my own code model in Numerit Pro.

 

[diazona]

Could you list the exact launch angle, vertical drop, and range used to find these velocities? I'll compare to the Hyperphysics website calculator and my own code model in Numerit Pro.

launch angle:
$\theta = 1.2^{\circ}$ in both cases
vertical drop:
$y_0 = 2\,\mathrm{ft.}$ for the bus in the movie
$y_0 = \frac{2}{12}\,\mathrm{ft.}$ for the model bus
range:
$x = 50\,\mathrm{ft.}$ for the bus in the movie
$x = \frac{50}{12}\,\mathrm{ft.}$ for the model bus

What didn't you follow in the substitution? I thought it would be pretty easy...

P.S. Just realized, those numbers came from my calculator, not Mathematica. Mathematica's handling of units is pretty terrible.

 

[SystemTheory]

Thanks. I'll use those numbers to build a digital simulator and post a graph of the result here.

Next if anyone want to look into the impact of air drag here are resources.

Quadratic drag equation:

http://hyperphysics.phy-astr.gsu.edu/hbase/airfri.html#c1

A closed formula approximation published by NASA is here:

http://www.grc.nasa.gov/WWW/K-12/airplane/flteqs.html

although it is a little tricky to read the NASA pages. An estimate of the air density must be made and the coefficient of drag for the bus. The width and height must be estimated too.

The bus might have a drag coefficient estimate Cd = 0.82 similar to the long cylinder:

http://en.wikipedia.org/wiki/Drag_coefficient

as a first estimate to see how significant this is at full size and at small scale.

What didn't you follow in the substitution? I thought it would be pretty easy...

In the final term you go from (x/v_x) to x/v with the secant square appearing. I would have to crack my 25 year old textbook on college algebra to see how you do it.

 

[diazona]

$$\frac{x}{v_x} = \frac{x}{v\cos\theta} = \frac{x}{v}\frac{1}{\cos\theta} = \frac{x}{v}\sec\theta$$
Then just square that.

I'm planning to write up something on the effect of air resistance on this experiment - I probably would have done it today if I had time. The Mythbusters do a lot of stuff where air resistance is a factor so I'm getting used to it ;-) but that NASA link looks interesting, thanks.

 

[SystemTheory]

One interesting detail of scale down studies is the ratio of mass to aerodynamic surface area makes the small scale object much more sensitive to air drag, lift, etc.

This is because mass reduces by the scale fraction cubed but area only goes down by the scale fraction squared. Anyone who does aerodynamic studies for a living (not I) learns how to compensate for this and other adjustments that do not scale smoothly.

 

[SystemTheory]

Derived this formula from diazona's equation in post #18. Note a slight change in the use of terms per the attached problem sketch to allow proper landing height yF = 0.

Launch velocity from elevated ramp assuming zero air resistance:

$$v_{0} = \frac{R}{\sqrt{\frac{2(R tan \theta +y_{0})}{g sec^{2} \theta}}}$$

where jump distance R = 50 {ft}, ramp elevation y0 = 2{ft}, ramp angle theta = 1.2 degrees, and standard gravity g = 32.174{ft/s/s} for the bus in the movie.

This code let's me set the Scale Fraction SF = 1/1 or SF 1/12. It multiplies three input variables by SF, namely, g, R, and y0. When I press run, it converts the angle A to radian, solves the equation above, and prints vmph to a variable viewer.

SF = 1/1 `Scale Fraction
g = 32.174*SF `{ft/s/s}
R = 50*SF `{ft}
y0 = 2*SF `{ft}
A = 1.2*(pi/180) `{rad}

num = 2*(R*tan(A)+y0)
den = g*sec(A)*sec(A)
v0 = R/(sqrt(num/den))
vmph = 0.68*v0 `{mph} launch velocity

v0 = 6.51{mph} when the scale fraction is 1/12 (with g/12, R/12, y0/12).
v0 = 78.12{mph} when the scale fraction is 1/1 (with g/1, R/1, y0/1).

The ratio 6.51/78.12 = 1/12 which shows the concept of "scaled gravity."

However we all know gravity does not scale on Earth, so I remove SF from the value of g and run at SF = 1/12:

SF = 1/12 `Scale Fraction
g = 32.174 `{ft/s/s}
R = 50*SF `{ft}
y0 = 2*SF `{ft}
A = 1.2*(pi/180) `{rad}

num = 2*(R*tan(A)+y0)
den = g*sec(A)*sec(A)
v0 = R/(sqrt(num/den))
vmph = 0.68*v0 `{mph}

v0 = 22.6{mph}

These results with 1/1 g can be confirmed on the Hyperphysics tool if you convert v0 to feet per second (1.47 * v0), keep launch angle at 1.2 degrees, and set y = -y0 for each case. Read the result in t2 and x2 boxes. The Hyperphysics link is reproduced here:

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

It appears to me the air resistance is significant in the small scale model and must account for the bad stunt result, unless there was some other error introduced by the Mythbusters execution of the mini-stunt.

 

[msakkas]

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

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

If we were to scale down $y - y_0$, $x$, $g$, and $v$ by 12, the ratio $x/v$ 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 $v$ to compensate for scaling down $y - y_0$ and $x$. Increasing $v$ by $\sqrt{12}$ 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, $y - y_0 = 0$ at the landing point, so
$$x \tan\theta = \frac{g}{2} \biggl(\frac{x}{v}\biggr)^2\sec^2\theta$$
or
$$\sin\theta\cos\theta = \frac{g}{2} \frac{x}{v^2}$$
or
$$v = \sqrt{\frac{xg}{\sin 2\theta}}$$
as described on the website.

Diazona:
Very interesting and thank you for your reply. I can (sort of) follow the mathmatics. But I'm still having trouble with the basic concept. If gravity has the same effect on a big bus and a little bus, why do we have to "alter" gravity for the 1/12 sized simulation? It seems that the simulation should be accurate without having to scale gravity. What am I missing?

 

[SystemTheory]

If gravity has the same effect on a big bus and a little bus, why do we have to "alter" gravity for the 1/12 sized simulation? It seems that the simulation should be accurate without having to scale gravity. What am I missing?

Nothing. But your question is the same one I had when opening the thread. Although you may be asking diazona directly I offer this comment.

In the movie, according to the Mythbusters, the givens are as follows (thanks to diazona):

Constant in Both Cases:

$\theta = 1.2$ degrees launch angle
g = 32.174{ft/s/s/} standard acceleration of gravity

Movie Bus Jump:

R = 50{ft} horizontal range between launch and landing ramp
y0 = 2{ft} elevation of launch ramp above landing ramp

Model Bus Jump:

R = 50/12{ft} horizontal range between launch and landing ramp
y0 = 2/12{ft} elevation of launch ramp above landing ramp

Neglecting air resistance, bus angular rotation, bus length, etc. the initial velocity to make the jump is given by:

$$v_{0} = \frac{R}{\sqrt{\frac{2(R tan \theta +y_{0})}{g sec^{2} \theta}}}$$

where gravity g and angle theta are the same in both calculations and we're getting the same numbers as the Mythbusters mention in the episode.

This equation gives v0 in feet per second with the above units. Multiply by 0.68 to get mile per hour. It is not clear yet why the model bus misses the jump in the episode ...

 

[SystemTheory]

Is Scaled Gravity a Reasonable Concept?

Based on two equations I think the concept of scaled gravity makes theoretical sense. The equation above gives launch velocity v0 as a function of launch elevation y0, jump range R, ramp angle theta, and gravity g. The second equation gives total flight time as:

$$t_{F} = \frac{R}{v_{0} cos\theta}$$

Run the movie bus jump. Launch velocity is 78{mph} and time of flight is 0.435{s}.

Run model with g/12 scaled gravity. Launch velocity would be 6.5{mph} and time of flight is 0.435{s}.

Run the model with g. Launch velocity is 23.6{mph} and time of flight is 0.126{s}.

The flight times would match if gravity could be scaled down to 1/12, and the launch velocity would be 1/12 of the full scale value. We can't scale gravity when building such models, so we don't usually think about it that way. I think the idea of scaled gravity is reasonable now but you can build the model without resorting to this concept.

 

[diazona]

Theoretical sense? Well of course you can theoretically set gravity to whatever you want, but it's just a theory (unless you have rockets). I'm not sure what you're getting at.

 

[SystemTheory]

If one wants to graph the scale down motion as it corresponds point for point with the large scale motion over time, one would scale down the gravity to draw the comparison.

So let's say one films a bus jump at small scale, to insert in a movie as a special effect representing full scale. The time of the video playback would then have to be adjusted to be faithful to the full scale physics. If one could film at scaled down gravity, the time and motion would be faithful to the full scale physics without any adjustments. (Air resistance is a separate concern).

 

[DaveC426913]

Would someone mind posting a sketch of what y'all are trying to model?

I am still trying to reconcile how a road with any semblance of continuity (i.e that's supposed to be an actual road) can be made to act as a launch ramp.

 

[diazona]

Here's my understanding of what they were describing in the show:

I can see how it could be part of a continuous road, if the missing part happened to be a small hill... but realistic or not, that's the model I'm using. I haven't seen the movie so all I have to go on are the parameters the Mythbusters reported in the show.

 

[DaveC426913]

Here's my understanding of what they were describing in the show:
View attachment 22439
I can see how it could be part of a continuous road, if the missing part happened to be a small hill... but realistic or not, that's the model I'm using. I haven't seen the movie so all I have to go on are the parameters the Mythbusters reported in the show.

The key is to now try to add in the missing piece of road. That's where I'm having an issue.