Mythbusters Bus Jump Small Scale

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.
 
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:

[tex]v_{0} = \frac{R}{\sqrt{\frac{2(R tan \theta +y_{0})}{g sec^{2} \theta}}}[/tex]

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 lets 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.
 

Attachments

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.
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?
 
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:

[itex]\theta = 1.2[/itex] 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:

[tex]v_{0} = \frac{R}{\sqrt{\frac{2(R tan \theta +y_{0})}{g sec^{2} \theta}}}[/tex]

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 ...
 
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:

[tex]t_{F} = \frac{R}{v_{0} cos\theta}[/tex]

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

Homework Helper
2,157
6
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.
 
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 lets 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

Gold Member
17,995
1,585
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.
 

Attachments

Last edited:

diazona

Homework Helper
2,157
6
Here's my understanding of what they were describing in the show:
busjump.png
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

Gold Member
17,995
1,585
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.
 

diazona

Homework Helper
2,157
6
Sorry, I don't see where you're having a problem with it...

Out of curiosity I plugged some numbers into Mathematica and it reports that the curve [itex]y(x) = -0.00121 x^2 + 0.0209 x[/itex] satisfies the necessary conditions: [itex]y(0) = 0[/itex], [itex]y'(0) = \tan(1.2^\circ)[/itex], [itex]y(50) = -2[/itex] (all distances measured in feet). I don't know anything about highway construction, but that seems like a reasonable shape for a road.
 

DaveC426913

Gold Member
17,995
1,585
Sorry, I don't see where you're having a problem with it...

Out of curiosity I plugged some numbers into Mathematica and it reports that the curve [itex]y(x) = -0.00121 x^2 + 0.0209 x[/itex] satisfies the necessary conditions: [itex]y(0) = 0[/itex], [itex]y'(0) = \tan(1.2^\circ)[/itex], [itex]y(50) = -2[/itex] (all distances measured in feet). I don't know anything about highway construction, but that seems like a reasonable shape for a road.
Which looks like what? Can you post it?
 

diazona

Homework Helper
2,157
6
Which looks like what? Can you post it?
OK, but I thought you would have been able to plot a function yourself :-/
mbcurvedroad.png
It's not very steep. I know roads with much higher grades and which probably have greater curvature too.
 
The image I recall from the Mythbusters show is a snap shot of the movie frame showing an LA style elevated highway with a section missing. Trying to estimate the actual geometry from such an image is difficult due to parallex error. The launch ramp would be 2 feet higher than the landing ramp (in DaveC sketch it is shown lower). The launch ramp would be at an incline of 1.2 degrees to the horizontal and the landing ramp at 0 degrees in my understanding. This could be a reasonable approximation for a missing section of elevated highway I suppose.

Edit: sketch attached.
 

Attachments

Last edited:

Integral

Staff Emeritus
Science Advisor
Gold Member
7,184
54
2ft in 50ft is a 4% grade, but the ramp is angled up to have continous road for some distance the grade would have to be greater then 4% isn't this pretty radical for a city bridge?
 
Bus Jump Aerodynamics - Simplest Model

Simplifying Assumptions:

[itex] \theta = 1.2[/itex] degrees
[itex] cos\theta = 0.99978[/itex]
[itex] sin\theta = 0.02094[/itex]

Due to the large cosine and small sine most of the launch velocity directs into the horizontal component. To simplify the study of quadratic drag assume motion is purely horizontal.

Full scale bus jump variables:

M - mass.
A - frontal area.
C - drag coefficient.
[itex] \rho[/itex] - air density.
v - velocity in x-direction.

Drag Force:

[tex]D = \frac{1}{2}\rho CAv^{2}[/tex]

Deceleration in Flight:

[tex]a = \frac{dv}{dt} = -\frac{D}{M} = -\frac{CA}{M} \frac{1}{2} \rho v^{2}[/tex]

1/12 Scale Deceleration:

Frontal area at 1/12 scale is A/(12 x 12). Mass at 1/12 scale is M/(12 x 12 x 12). After substitution the factor 12 ends up in the numerator:

[tex]a = \frac{dv}{dt} = -12\frac{CA}{M} \frac{1}{2} \rho v^{2}[/tex]

Comments:

These differential equations need to be solved for velocity as a function of time, and so far, there is no coupling with the increased vertical drop caused by deceleration. However one can see that the model bus is more sensitive to drag. The term (M/CA) is the ballistic coefficient in physics (the term is used differently for bullets). The final term 1/2 rho v-squared is the dynamic pressure. The full scale bus is 12 times more effective at pushing air than its scaled down model. The drag coefficient C is assumed to be the same for both cases.
 
Last edited:

DaveC426913

Gold Member
17,995
1,585
OK, but I thought you would have been able to plot a function yourself :-/
View attachment 22446
It's not very steep. I know roads with much higher grades and which probably have greater curvature too.
Hm. So the bus is leaving the ramp at almost horizontal and it has to fall less than 2 feet in the time it takes to cover a 50ft horizontal distance.
 
I need to visit my calculations again. However it appears drag may have a negligible effect on the time to travel the jump distance on both scales. Using rough numbers I'm getting 1 millisecond at full scale and less than that at small scale. Next time I'll post a formula for time and horizontal distance for review.
 
I'll post my draq equation if anyone's interested, but it appears the deceleration caused by drag can be neglected over the jump distance due to the bus having significant mass.

One factor that could cause the nose of the bus to drop is a torque impulse imparted as the front wheels of the bus catch air before the rear wheels. The weight W acting at the center of mass tends to impart a torque about the rear axle. This would be counteracted by a torque in the driveline, however, for a bus, the counteracting torque is probably insufficient. A motorcycle could provide more countertorque and the rider could adjust the center of mass rearward and use the throttle in mid air to help loft the front end.

An estimate could be made of the moment of inertia J and the the impulse torque, but at this point I lack good data so I don't plan any calculations.
 

Attachments

I am glad to see someone else being "driven nuts" (pun intended) by mythbusters. their "myth-perceptions" on pressure drive me nuts
We give a basic pressure school at our facility, and myself and another instructor have threatened to send them a couple free passes
and they have even "myth-used" our products, too

ok, (breath deep) will not rant, will not rant

dr
 
dr dodge,

I'd like to model some air cannons in an engineering simulator. Would your materials cover building the system model and writing the differential equations, or is it more basic stuff you offer?
 
System, I am a bit unsure exactly what you need. I have my own fab shop at the house. No mill or lathe (yet) but I know quite a few "job shop" guys around here. For my "day job" I deal in high accuracy high and low pressure metrology. ....but, at my house, for "personal R&D purposes" I can generate 40,000 psi (in less than 20 seconds) hydraulic pressure, and 15,000 psi gas pressure. I think we could very easily build some air cannon stuff.

I was originally going to post that I have found a "multitude" of used buses for under $2500.... ranger probably knows a driver or two...
The ramp approach could easily be simulated in gravel
and we videotape it and start the new series
"engineering de-myth-defied"

dr
 
I'm looking to do personal computer (silicon) simulations that explain and predict Mythbuster results with reasonable accuracy. This is engineering modeling problem with components such as fluid capacitor (air tank), fluid inertor (gun barrel), fluid resistor (heat transfer losses), ideal power transformer (pressure-volumetric flow power to force-velocity power), and mass of the projectile. I would "build and fire" this gun in a simulator on the computer but I don't have good data to compare to exit velocity from an actual air gun. Anyway I am not up for actual build and filming efforts as it is not along my interests or talents.

Does your company provide fluid power meters, which measure both pressure and volumetric flow rates? If so, please point me to a link or reference that helps with those measurements.
 

diazona

Homework Helper
2,157
6
I finally got around to doing my usual writeup on this Mythbusters episode:
http://www.ellipsix.net/blog/post.80.html [Broken]
 
Last edited by a moderator:
Great writeup. Thanks diazona. One thing I noticed in the episode was the short run to accelerate the model bus to 23 mph, and no clear reading on the radar gun. The radar gun could be off by +/- 1 mph according to some specs I've read. Also, once the front of the bus starts downward it is possible an aerodynamic downforce develops as well.

The takeoff drag force on the front of the full scale bus is > 1,000 lbs and on the scale bus at least 1/2 pound force (rough numbers). This doesn't slow the bus much over the short jump but I neglected the possibility of aero downforce.
 

Physics Forums Values

We Value Quality
• Topics based on mainstream science
• Proper English grammar and spelling
We Value Civility
• Positive and compassionate attitudes
• Patience while debating
We Value Productivity
• Disciplined to remain on-topic
• Recognition of own weaknesses
• Solo and co-op problem solving
Top