# Accelerating onto and over inclined plane

Consider an RC car going off a jump. What angle is optimal to achieve the greatest distance? The height of each jump is the same. The angle of the jump and the related length of the jump changes. This is an experiment my son is doing for his 6th grade science fair.

The RC car used is a hobbyist RC car with a top speed of approximately 30 mph. It has shocks, and the tires are rubber with foam inside.

We've used a Netduino to control acceleration so that the each run is identical.

1. The optimal angle for ballistic trajectory is 45 degrees. This is based on his research as he hasn't quite reached calculus yet. http://en.wikipedia.org/wiki/Range_of_a_projectile

2. My understanding is that the Work to climb the inclined is identical for each jump. They are all the same height, and climbing them produces the same potential energy. Am I completely misunderstanding this?

3. The tires do not slip.

4. The car starts from 20 feet from the high end of the jump.

Given this information, would the 45 degrees be the optimal angle of the jump? We haven't completed the test runs yet, but initial tests show the 45 degree jump performs poorly.

We reduced the test speed such that the car does not scrape the bottom when it hits the ramp.

Other than the trajectory and the Work required to climb the jump, what aspects are we missing? Where is the energy going when it hits the 45 degree jump?

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How long and high is the jump? Is it a small jump at the end of the track, or a large curved one? Like you say, a lot of the energy could be lost in the climb, but if it's a 12-inch jump after a 20-foot acceleration, then 45° is probably your best best. –  krs013 Mar 12 '13 at 4:15
Also, if the jumps are higher than the plane your car will be landing on, 45° isn't necessarily optimal. You would have to calculate that more specifically. In any case, I wouldn't expect RC cars going off of jumps to go very far. –  krs013 Mar 12 '13 at 4:23
The track is straight. The jumps are 12 inches high. The 45 degree jump is 17 inches long. I taught him some basic trig to calculate the length using the angle and the height to find the hypotenuse. The landing is back at ground level, so your point is valid. You'd be surprised how far these RC cars can jump. At full speed of approximately 30 mph, it can go quite far. However we had to drop the speed considerably to eliminate the car bottoming out on the 45 degree jump. –  Robert Graves Mar 12 '13 at 11:41
After doing a bit of testing using a trajectory calculator which takes into account an initial height (had2know.com/academics/…), the optimal angle would be closer to 43 degrees. However this still doesn't account for the poor performance of the 45 degree angle jump. Next week we'll run the full tests and we'll post the results. –  Robert Graves Mar 12 '13 at 12:11
Something else to consider is the mass distribution, though that might be alright in the case of hobbyist RC cars. @ja72 also touches on an important point: the transition from a flat to an inclined plane should be smooth for minimal energy loss and optimal control (reproducibility). –  Wouter Mar 12 '13 at 22:55

If you are limiting the takeoff speed to prevent it bottoming out, then I suggest you lower the ramp. $45^\circ$ gives optimal range for a given takeoff speed (ignoring friction), but only if you don't care what the vertical component of the velocity is on landing.

At $30\: \mathrm{mph}$, with a $45^\circ$ jump, you say it doesn't bottom out. The vertical component of the velocity on landing has approximately the same magnitude as on take off (air resistance losses), which would be $30 \; \cos(45^\circ)$ = about $26\: \mathrm{mph}$.

To maximise the range, then, we need to keep this vertical component ($v_y = 26\: \mathrm{mph}$), whilst letting the overall speed ($v$) increase to $45\: \mathrm{mph}$.

$v^2 = v_x^2 + v_y^2$

$v_x^2 = v^2 - v_y^2 = 45^2 - 26^2 = 1349$

$v_x = \sqrt{1349} = 36.7 \mathrm{mph}$

So we can calculate the angle from the $x$ and $y$ components of the velocity:

$\tan(\theta) = v_y / v_x = 26 / 36.7 = 0.71$

$\theta = \arctan(\theta) = \arctan(0.71) = 35^{\circ}$

So based on that information, try $35^\circ$, with a speed at the ramp of $45\: \mathrm{mph}$.

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