paper: Parabolic Trajectory Calculations

[Ether]

Originally Posted by Hugh Meyer

Would you more clearly define the launch angle? Is that from the horizon, or a plumb line?

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?

[Hugh Meyer]

Thank you.

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

-Hugh

[Ether]

Originally Posted by Hugh Meyer

How would we measure the terminal velocity?

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

[Ether]

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.

http://www.chiefdelphi.com/media/papers/download/4614

[GeeTwo]

Originally Posted by Hugh Meyer

How would we measure the terminal velocity?

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.

[Jacob Plicque]

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.

[Ether]

Originally Posted by Jacob Plicque

Ether
What coeficient of drag are you using for the ball in the 2014 spreadsheet?

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

[peronis]

does anyone now the terminal velocity of the fuel?

[Jacob Plicque]

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

[Ether]

Originally Posted by Jacob Plicque

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.

[Jacob Plicque]

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=0.78sec*9.8m/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)

[Ether]

Originally Posted by Jacob Plicque

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.

You are misunderstanding the meaning of terminal velocity.

[Jacob Plicque]

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.

[Ether]

Originally Posted by Jacob Plicque

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.

The definition of terminal velocity in the context of this thread is given in the first paragraph of this web page:

https://en.wikipedia.org/wiki/Terminal_velocity

Please read it.

You are using a different definition.

[Ether]

Originally Posted by Jacob Plicque

The Newtonian model equations ignore drag which calculates a velocity only on the basis of gravity.

Models with air drag are "Newtonian" too. The acceleration is still equal to the net force divided by the mass (Newton's 2nd law).

Perhaps what you meant is constant acceleration model equations ignore air drag.