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Back To The Future 3 - How Much Track?
Okay... when Doc is outlining his plan to Marty,
he shows him on the map the stretch of track he plans
to use, and says "This stretch of track runs three miles out
to Clayton... Shonash Ravine" Did you catch that??
THREE miles. Now.. feel free to flame me if I'm wrong,
but an object moving at 60 miles per hour, will
travel 3 miles in 3 minutes. A train under constant
acceleration up to and indeed *past* 60 miles per
hour will need a hell of a lot more than three miles
of track to reach 88 mph. I'm sure if you time
the whole sequence and work it out mathematically,
you'll find the train should have plunged into the
ravine after about 5-6 minutes, without ever
reaching 88mph!
Special Requirements:
VCR and Stopwatch
Avg. Rating:    4.1 of 10 - (53 votes cast)
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Contributed By:
Spike on 02-01-2001
Reviewed By:
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Comments:
andrea writes:
At first it may seem like you need to accelerate at a very high speed to go from 0 mph to 88 mph within 3 miles... Some conversions first, 0 mph = 0 miles/s and 88 mph = .024444 miles/s Let's apply a good old physics formula: (final velocity)^2 - (initial velocity)^2 = 2(acceleration)(distance traveled) (.024444)^2 - (0)^2 = 2(acceleration)(3 miles) acceleration = approximately .0001 miles per second squared, or about .5 ft/s^2. The average modern car can accelerate from 0 to 60 in about 10 seconds, yielding an acceleration of 8.8ft/s^2 (You can probably go from 0 to 88 in a matter of 15 seconds). While a Nineteenth Century locomotive is not exactly your average modern car, I'm certain that it could accelerate at a minimum of .5 ft/s^2. Think about it, after a minute, it would be traveling at 30 ft/s (20 mph), but only traveled about 1800 ft or .341 miles... I could be wrong about the capabilities of a Nineteenth Century locomotive, but the actual event would have taken about 4-5 minutes to take place, and they wouldn't have run out of track unless they accelerated too slowly.
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Treb writes:
Math doesn't lie, true, but it can certainly be wrong if an equation is incorrectly quoted. You left a factor of 1/2 out of the equation for distance as a function of time -- d = 1/2 a t^2 -- which would change the time taken to just over FOUR minutes.
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m2morgan writes:
OK, I have worked out this math. I do assume a constant acceleration (which we all know is not correct). I am not trying to determine how long it would take for a locomotive to accelerate to 88 miles per hour (although I will get this value). I am trying to calculate the acceleration necessary to reach a velocity of 88 miles per hour, from a standstill, in a distance of three miles:

d=at^2+V(sub i)t+d(sub i)

V(sub i) and d(sub i) are the initial Velocity and position of the train. For the purposes of this discussion, we will set both of these values to 0. so:

d=at^2

a=(V(sub t) - V(sub i))/t or a=V(sub t)/t: V(sub t) is 88 Miles per hour, d is 3 miles and t is what we are trying to calculate. so:

3=(88/t)t^2 or 3=88t

From this we can calculate that the entire trip should take approximately 0.0341 hours or a just 2 minutes.

Plug this value in for t in the acceleration calculation and it comes to just over 2,581 m/h^2

Now for the power requirements: Force= Mass Times Acceleration or F=ma

m=600 Tons or 450,000 Kg (based on built in the 1940s)
a=2,581 m/h^2 or 1,147 m/s^2
F=450,000 kg * 1,147 m/s^2 = 516,266,666 N
W=F*d
d=3 mi or 4800 m
W=516,266,666 N * 4800 m = 2,478,080,000,000 J
P=W/t
P=2,478,080,000,000 J/ just over 2 minutes or 122 Seconds=20,191,762,962 W or approximately 20,191,763 kW

Big Boy could run at 6000 HP or a measly 4,474 kW.

In Conclusion. The first tip is that it only takes 2 minutes to do this acceleration. In the movie, even accounting for temporal compression, the trip takes much longer. The second problem is that the power requirements are not present to push such a missive object that quickly. I was using data from a locomotive that was not a classic western steam locomotive, but the math doesn't lie. We can also assume that Doc's logs can help to accelerate the locomotive, but I am not sure that could account for an increase of 15,00 kWs.

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