# paper: Parabolic Trajectory Calculations

Thread created automatically to discuss a document in CD-Media.

Parabolic Trajectory Calculations
by: Ether

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[b]Parabolic vs Air-Drag Trajectory revC

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Parabolic vs Air Drag Trajectory revC is the same as revB except the graph is not auto-scaling. Some folks may prefer this.

Parabolic vs Air Drag Trajectory revB fixes a small error: the “launch height” user input parameter was not being imported into the parabola equation.

The physics and math for the computer numerical simulation (including topspin/backspin)
is explained in this paper.

Parabolic Equations and constants a, b, c, xp, and yp
y = ax[sup]2[/sup] + bx + c
and y = a*(x-xp)[sup]2[/sup] + yp explained:

How to find constants a and b, given launch speed and angle:

How to find constants a and b, given desired scoring range:

How to find constants a and b, given yp and a point (x1,y1) on the trajectory:

parabola.pdf (48.2 KB)

compute a&b from yp,x1,y1 rev02.pdf (16.6 KB)
parabola vs air drag.zip (57.5 KB)
parabola vs air drag revB.zip (51.7 KB)
parabola vs air drag revC.zip (63.6 KB)
Terminal Velocity.zip (26.5 KB)
given Vo and (d,h) find theta.pdf (44.6 KB)

*For those of you who downloaded the Parabolic vs Air Drag Trajectory spreadsheet, please note that I just uploaded revB to correct a small error. The user input “launch height” was not being imported into the parabola equation.
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**I should have mentioned:

The parabola plot with the original version was correct as long as you didn’t change the launch height.

And even if you did change the launch height, the error affected only the parabola, not the air-drag trajectory.

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Ether,

Would you more clearly define the launch angle? A drawing would be nice. Is that from the horizon, or a plumb line? Thanks.

-Hugh

It’s the elevation angle (from the horizontal).

The reason you don’t see the graph appear to visually correspond to the launch angle is because the X and Y axes are not scaled equally, and when you change the launch angle the scaling auto-adjusts to fit the graph.

While you’re here, does your team happen to have any test data to confirm (or refute) the 37 ft/sec terminal velocity number for this year’s game piece?
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Thank you.

We do not have any data, but we have been talking about it. How would we measure the terminal velocity?

-Hugh

Ether,

The assumption appears reasonable with the shooting range we are seeing on our robot.

I just found the thread today. When we unbag the robot on Saturday, I will be attempting to increase our shooting percentage via angle and may be able to provide some data after that point.

We did overshoot the target a bit last weekend at Southfield, MI a few more times than I would have liked.

I think there might be a misunderstanding. Terminal Velocity is independent of shooting range. It is a function of mg, Cd, rho, and A only.

I don’t know for sure; I’ve never done it. Perhaps drop the ball from a sufficient height next to a marked wall in a tall room and take high speed video with a camera that timestamps the frames. Then tweak the value of Terminal Velocity in this spreadsheet until the model matches your data.

I just posted a small revision (revC) to the air-drag spreadsheet. I turned off the auto-scaling in the graph and re-shaped it so the launch angle “looks” more like the real thing. It may make it easier to visualize what’s changing when you change the input parameters. The downside is you lose some resolution.

http://www.chiefdelphi.com/media/papers/2946

*Given launch speed and a desired point (d,h) on the trajectory, show the derivation of and formulas for the launch angles and the equations of the two parabolic (no air drag) solutions.

My first thought is a football stadium and a radar or ultrasonic speed gun. The procedure is pretty obvious.

If you don’t have a speed gun, a strobe of some sort, including the video method suggested by Ether would be next.

As for 3946, we did the air-resistance-free calculation, added about 50%, tested that we had more than we needed to hit the goal at the ranges we wanted, then we’ll back down based on empirical launch data until we hit the goal at the desired range (this year, with our rear bumper in the outer works). Not as elegant as the full-physics solution, but we’ve built several high-percentage launchers using this paradigm.

Ether
What coeficient of drag are you using for the ball in the 2014 spreadsheet? In 2017 the ball has a Cd in the 0.6 to 0.8 range by looking at wiffle ball data which varies with the Reyonds number.

The calculation is based on the terminal velocity (Cell A5), and an assumption that air drag varies with the square of velocity.

does anyone now the terminal velocity of the fuel?

The terminal velocity can be calculated from the peak height and the gravity constant.

OK, so given g=-9.8 meters/sec2 and peak height = 3 meters, please show us how you would calculate the terminal velocity… without using any other information such as launch speed or launch angle.

From a known height of 3 meters the calculation follows a freefall model (Vy=0 @t=0) which uses Vy=gt and Y=0.5gt^2.
Thus t=sqrt(3m/9.8m/sec^2/0.5sec)=0.78 seconds.
Vy terminal=0.78sec9.8m/sec^2=7.66m/sec
The Newtonian trajectory equations do use the initial velocity Voy as follows:
Vy=Voy-gt
Y=Voyt-0.5gt^2 @ Vy=0 the arc is at its peak
X=Voht
Voy=Vo
sin(launch angle from horizon)
Vox=Vo*cos(launch angle)

You are misunderstanding the meaning of terminal velocity.

The Newtonian model equations ignore drag which calculates a velocity only on the basis of gravity. My answer is only valid for a drag free Newtonian calculation. Terminal velocity with drag provide an upper velocity limit for a falling object. For a falling baseball it would be about 33m/sec (~108ft/sec) while a hail stone is ~14m/sec (45ft/sec). A 3 meter drop reaches a peak velocity of 7.66m/sec (~26ft/sec) which is less than the maximum terminal velocity. of either of the above examples.