# Can normal cars jump bridges

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#### syhprum

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##### Can normal cars jump bridges
« on: 30/04/2014 15:11:23 »
It is a common theme in action films to have cars jump across opening bridges I enclose a picture of such a bridge as illustration.
In the film "The Blues Brothers" a rather ordinary looking saloon car accelerates from stationary about a 100 feet from the start of the half open rising section (that is at 45° and makes a perfect landing on the other section).
Is this possible and what acceleration would be needed.
Sorry no SI units as this is a rather old bridge.
« Last Edit: 30/04/2014 15:15:05 by syhprum »
syhprum

#### CliffordK

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##### Re: Can normal cars jump bridges
« Reply #1 on: 30/04/2014 23:14:11 »
The Dukes of Hazard TV show destroyed about 150, 1969 Dodge Chargers during the filming of the show, undoubtedly increasing the rarity of the vehicles.

I suppose you could look at it in one of two ways.  Real cars were being used in the filming of the jumps (with some added tail weight to balance out the engine), but the landings were hard on the cars.

The calculations should be pretty straight forward.

A car making the jump would be just about the same as a bowling ball or a football making the jump.  Let's ignore wind resistance.  Then at the point of takeoff, you have an the velocity and angle giving you an upward velocity vector, a horizontal velocity vector, Gravity working on the vertical axis, time and distance.

I'm not quite sure what induces the spin on the car, perhaps the front leaving the ramp a fraction of a second before the rear, but the center of gravity also apparently plays a role.  Wind?

Using the formula 1,1, $$\sqrt{2}$$, the gap between the two bridge parts for a 235 feet span at 45° (assuming the hinge at the center of the towers) is a 69 foot opening.

I'll try to calculate the velocity one needs to reach at the top of the span later, then one could try to calculate the acceleration one would need.  I'd think a 1/4 mile run-up would be good.

The other thing, I'd hate to hit a 45 degree angle at the start of an incline at a high velocity without some kind of a gradual slope change.

#### syhprum

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##### Re: Can normal cars jump bridges
« Reply #2 on: 01/05/2014 06:41:22 »
I think a 100 feet run-up would be near impossible even if a wire or steam catapult is used, as you point out the transition from the horizontal run-up to the sloping section is a very large difficulty and the calculation of acceleration must take into account whether or not acceleration can be maintained up the sloping section.
It is no surprise the filmmakers destroy a lot of cars!
I think an angle of less than 45° is commonly used but it does not look so impressive
syhprum

#### CliffordK

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##### Re: Can normal cars jump bridges
« Reply #3 on: 01/05/2014 11:17:14 »
Ok, let's try some calcs..  I think I'll try meters/second.  So, above I calculated 69 feet, or 21 meters.

v = velocity leaving bridge
vh = horizontal velocity (constant) leaving the bridge
vvi = initial vertical velocity leaving the bridge
dv = vertical distance relative to the top of the ramp.
t = time for the jump.
a = the acceleration due to gravity (9.8$$\frac{m}{s^2}$$)

Where
$$\frac{v}{\sqrt{2}}$$ = vh = vvi
t  = $$\frac{21m}{v_h}$$ =  $$\frac{21m}{v_{vi}}$$

For the vertical component of the velocity and distance:

vv = vvi - at

dv = vvi t -  ½ a t2

The midpoint in the flight would be setting vv = 0

So, substituting half the time above:

vv = 0 = vvi - 9.8 * ½ * $$\frac{21m}{v_{vi}}$$

0 = $$\frac{v_{vi}^2}{v_{vi}}$$ - $$\frac{102.9}{v_{vi}}$$

vvi = $$\sqrt{102.9)$$m/s = 10.14 m/s

Set dv = 0 (landing), and substitute in time from above, and one gets:

dv = 0 = vvi * $$\frac{21m}{v_{vi}}$$ -  ½ * 9.8 *   $$({\frac{21m}{v_{vi}}})^2$$

0 = $$21m*\frac{v_{vi}^2}{v_{vi}^2}$$ - 21m* 4.9 * $$\frac{21m}{v_{vi}^2}$$

0 = vvi2 - 102.9

vvi = $$\sqrt{102.9)$$m/s = 10.14 m/s  Whew, the same as above when calculating with velocity.

t = $$\frac{21m}{10.14 m/s}$$ = 2.07 seconds.

Now, to recover the initial velocity, v = 10.14 m/s * $$\sqrt{2}$$ = 14.3 m/s = 32 mph

That is actually a lot slower than I had expected, but it makes sense for about 2 seconds of air time.

On the level, a good car should be able to reach 32 mph in a little less than 1/8 mile, or about 200 meters.  Perhaps a lot less as the low speed acceleration is always easiest.

So, from your drawing, assuming a safe, gentle transition from horizontal to the 45° angle, one may be able to do the run, starting just off of the bridge, just before the first tower.

#### syhprum

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##### Re: Can normal cars jump bridges
« Reply #4 on: 01/05/2014 12:11:22 »
I did a rough calculation and was surprised by the relatively low speed required, powerful cars can produce at least 1g acceleration up to 100km/s  the largest problem I see is the transition from horizontal to 45° and the backwards rotation produced by this transition, if I had to design a car to do this I would incorporate some flywheels such as are used to despin satellites.
syhprum