

A good engineer will never stop learning. From his peers, technicians, and competitors. The field is just to vast for any one person to know it all.  Mr. Mike [more] 



Parabolic vs Air Drag Trajectory revC is the same as revB except the graph is not autoscaling. 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 = a*x^{2} + b*x + c and y = a*(xxp)^{2} + yp explained:
http://www.chiefdelphi.com/media/papers/download/3868
How to find constants a and b, given launch speed and angle:
http://www.chiefdelphi.com/media/papers/download/3871
How to find constants a and b, given desired scoring range:
http://www.chiefdelphi.com/media/papers/download/3872
How to find constants a and b, given yp and a point (x1,y1) on the trajectory:
http://www.chiefdelphi.com/media/papers/download/3877
Terminal Velocity calculator spreadsheet:
http://www.chiefdelphi.com/media/papers/download/3900
parabola.pdf
download filegiven speed&angle, find a&b.gif
download filescoring_range.gif
download filecompute a&b from yp,x1,y1 rev02.pdf
download fileparabola vs air drag.zip
download fileparabola vs air drag revB.zip
download fileparabola vs air drag revC.zip
download fileTerminal Velocity.zip
download filegiven Vo and (d,h) find theta.pdf
download file02272014 06:23 PM
Ether
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.
02282014 12:07 PM
Ether
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.

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 airdrag trajectory.
03042014 09:55 AM
Hugh MeyerEther,
Would you more clearly define the launch angle? A drawing would be nice. Is that from the horizon, or a plumb line? Thanks.
Hugh
03042014 10:19 AM
Ether
Would you more clearly define the launch angle? Is that from the horizon, or a plumb line?

03042014 10:36 AM
Hugh MeyerThank you.
We do not have any data, but we have been talking about it. How would we measure the terminal velocity?
Hugh
03042014 12:23 PM
marccenterEther,
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.
03042014 12:28 PM
Ether
The assumption appears reasonable with the shooting range we are seeing on our robot.

03042014 02:00 PM
Ether02172016 07:15 PM
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
02172016 11:44 PM
GeeTwo01112017 04:06 PM
Jacob PlicqueEther
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.
01112017 04:25 PM
Ether
Ether
What coeficient of drag are you using for the ball in the 2014 spreadsheet? 
01132017 02:49 PM
peronisdoes anyone now the terminal velocity of the fuel?
01132017 04:29 PM
Jacob PlicqueThe terminal velocity can be calculated from the peak height and the gravity constant.
01132017 05:43 PM
Ether
The terminal velocity can be calculated from the peak height and the gravity constant.

01142017 09:57 AM
Jacob PlicqueFrom 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=Voygt
Y=Voyt0.5gt^2 @ Vy=0 the arc is at its peak
X=Voht
Voy=Vo*sin(launch angle from horizon)
Vox=Vo*cos(launch angle)
01142017 10:34 AM
Ether
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. 
01142017 03:51 PM
Jacob PlicqueThe 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.
01142017 04:02 PM
Ether
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.

01172017 07:13 PM
Ether
The Newtonian model equations ignore drag which calculates a velocity only on the basis of gravity.

01192017 09:45 PM
Ether
Thread created automatically to discuss a document in CDMedia.
Parabolic Trajectory Calculations by Ether 