Which ramp would be better, taking air resistance into account?

Question

Everything else is the same. 40 degrees and 50 degrees. Which one would launch a motorcycle farther?

http://s226.photobucket.com/albums/dd11/jackphotos123/?action=view¤t=4050.jpg

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

http://formularium.org/en/10.html?go=88.114

Answer 1

Range = v^2 (sin 2Θ) / g
All things the same: sin (2 * 40) = .98
sin (2 * 50) = .98
So the range is the same for either angle, in a vacuum.

The 40 degree angle seems better for avoiding wind interference.

Answer 2

Air resistance creates a drag force on the object proportional to the square of its velocity, and in the opposite direction. This creates a non-linear situation that cannot be solved analytically, and must be solved using numerical methods.

I did such a numerical analysis, assuming the object had equal drag for a given velocity in any direction, which may or may not be true (it's definitely not true for an object like an airplane).

It turns out that the higher the drag on the object, the smaller the ideal angle for maximum horizontal distance becomes. For zero drag, the ideal launch angle is 45 degrees. However, for a fluid density of 1 (ignoring units), the ideal launch angle becomes, say, 36 degrees. When the fluid density is increased to 5, the ideal launch angle may be only 22 degrees.

In short, the larger the drag forces are compared to the object's weight, the smaller the ideal launch angle becomes.

Answer 3

The two are equal for sure. Consider the kinematic equations in this form: (use Q for theta)
y = v*t*sinQ - 4.9t^2
x = v*t*cosQ
They should both make sense to you if you think about it. Now, when the projectile gets to the end of its path, y=0, but y=0 at the beginning of its path too. So if you plug in y=0 for the first equation and solve for t, you get two solutions. One of them is t=0 since y=0 at the beginning. The other solution is
t=(v/4.9)*sinQ
So this is the time that the projectile lands, and therefore the range is now given by our x equation:
x = v*t*cosQ
=> R = ((v^2)/4.9)*sinQ*cosQ
So all we need to know is, which is higher? sin40cos40 or sin50cos50? The answer is that they're the same, so neither angle is better.

BTW 2sinQcosQ=sin(2Q) so the dude above with the range formula is telling you the same thing as me.

Answer 4

to the guy above
actully 40 degrees,
45 only works if you dont factor in air resistance/gravity...

the best angle to fire something at for distance i beilve is somewhere around around 15°-20° if you factor in gravity, with air resistance... probably 30-40.
I used to study ballistic trajectories a few years back.

So with that id say 40 degrees is the better choice.

IM guessing you dont want the math (which i cant seem to locate my notes from back then anyway...) so i wont include.

Answer 5

40 degrees.

When launched at 40, the motorcycle will spend let time in the air, so air resistance is less of a factor.

Answer 6

They are the same - the best pitch for a ramp is 45 degrees - making it steeper or shallower from there decreases the distance.

Answer 7

40 Degrees will definately do good for ur bike :) lesser the angle larger the distance covered as u get more acceleration which in turn imparts maximum speed to ur bike.