Ball Flight Model

[DavisDad]

Care to compare?

Modified spreadsheet model from here: University of Illinois (http://baseball.physics.illinois.edu/trajectory-calculator.html)

http://simhardware.org/img/BallFlightModel_07jan13.jpg

[adlasa]

Very nice. Could you please explain what CD means?

[themccannman]

Very nice. Could you please explain what CD means?

Chief Delphi obviously

/s

actually though, it's probably coefficient of drag.

[DavisDad]

Very nice. Could you please explain what CD means?

Hi adlasa

This baseball model, posted by University of Illinois, accounts for air drag and lift forces from backspin. Cd is coefficient of drag. The data are calculated with modelling for air resistance. This is a comparison of spinning ball vs. no spin.

[Ether]

Care to compare?

What values did you use for launch angle, initial speed, ρ, CD, and CL?

[RSchorr]

The calculation of drag first requires the computation of Reynolds number. With my initial assumption of 30 degree angle the Initial velocity needed to travel 30 ft to the high goal is about 253 inches/second. This makes the Reynolds number of 2.60E06. My fluid mechanics book is a bit dated (1977) but this would result in a drag coefficient of 0.4 to 0.42.

Using the density of air at 70F of 0.00238 slug/cubic ft and the 24"diameter

Fd = Cd*Rho*V^2/2 *(projectedArea) the initial drag force is 0.65 lbs.

As speed decrease the drag force falls, but this seems to show that drag forces will not be "insignificant".

Can someone confirm please?

[DavisDad]

What values did you use for launch angle, initial speed, ρ, CD, and CL?




Hi Ether,

Sorry for the illegible chart.

Launch angle- 60deg
Init. speed- 22 f/s
Cd - 0.5
Density- 0.074 lb/m^3
Cd is a calculated function; see U of IL spreadsheet: http://baseball.physics.illinois.edu/TrajectoryCalculator-v2.xls

I haven't worked through the calculations completely and still need to adjust Re (spreadsheet is 100 mph,147 f/s)

[NotInControl]

I have developed my own model which only accounts for air resistance, I currently ignore spin because I don't think we will be adding spin to our ball.

But I pluged your numbers into my model: using your launch angle of 60 degrees and inital velocity of 22ft/s the results seem very comparable. Note: I DID NOT change my air density of 1.2 kg/m^3, mass of 1.2474kg or Cd which is 0.47 to yours so that is why the number differ slightly.

But here are my results. The Green line represents the center of the top goal, and the red line represents the white zone.

Note again this is only with air drag, not spin.

Remember this is only a simulation, and while our simulations may agree, I would urge you to do some data validation experiments, before you use the results of the simulation to drive your design.

Hope this helps,
Kevin

[DavisDad]

I have developed my own model which only accounts for air resistance, I currently ignore spin because I don't think we will be adding spin to our ball.

But I pluged your numbers into my model: using your launch angle of 60 degrees and inital velocity of 22ft/s the results seem very comparable. Note: I DID NOT change my air density of 1.2 kg/m^3, mass of 1.2474kg or Cd which is 0.47 to yours so that is why the number differ slightly.

But here are my results. The Green line represents the center of the top goal, and the red line represents the white zone.

Note again this is only with air drag, not spin.

Remember this is only a simulation, and while our simulations may agree, I would urge you to do some data validation experiments, before you use the results of the simulation to drive your design.

Hope this helps,
Kevin

Thanks Kevin for the data, much appreciated!

I am still brushing up on my fluids (humbling exercise) and going through the U of Il theory. My thinking is that simple calcs (no drag) would provide a good enough estimate of velocity required to design shooter. I'm interested in aerodynamic effects for two reasons:


Is there much benefit in back-spin for increasing apex height
Will back-spin reduce any knuckle-ball effect? I.e., is there enough benefit to invest in more complex back-spin design?


For Rebound Rumble (Nerf basketball), backspin had a significant advantage for range and accuracy.

[NotInControl]

No problem,

I will try to post some datasets from my model later so that you can compare more than just one data point.

In terms of design, it will be hard to ignore drag and obtain meaningful numbers. The ball is relatively large in size, and if you reduce your Cd in your model to 0, you should be able to see the numbers reported for the "No Drag" case.

At shallow launch angles the No drag model will report the ball traveling a lot further then it will in reality.

As for spin, the magnus effect on a ball is very real, and if you plan to put significant spin on the ball then you should at least consider modeling it if your model will be used to help narrow your design parameters.

For my case all the ideas my team has come up with so far are of the type of sling-shot, or catapult launcher, and in most cases will not impart spin on the ball.

For any object traveling though a fluid, spin does offer greater stability which leads to better accuracy. So without spin repeatability will decrease. By how much I can not say cause we are still testing the flight behavior of this ball on our prototypes.


Hope that helps,
Kevin

[DavisDad]

No problem,

I will try to post some datasets from my model later so that you can compare more than just one data point...
Kevin

That'll be great! I'll be working on sorting out the baseball theory...

We worked on this catapult design for Rebound Rumble:

Rolling Release Catapult (http://www.chiefdelphi.com/forums/showthread.php?t=102977)

We got it shooting 35' fixed hard to the ground, but ran out of time and the design was questionable re rules and safety. If interested I could post some videos of the last prototype (not in thread).

Craig

[Mike Schreiber]

I've attached a simple to use iterative model that can be edited including comments on how everything is calculated and a FBD. I'm confident the no drag calculation is right because it is simple kinematics, I could use a reality check on the drag included portion because it seemed a little drastic to me, but I checked all my calculations.

I assumed the drag coefficient to be constant, I calculate the Reynolds number but I could not find a look up table online for Re versus Cd of a sphere, if someone knows of one I can easily add it in to recalculate and interpolate to this table on each iteration (probably won't make a significant difference).

I did not include spin because I don't believe many teams will intentionally put spin on the ball (I could be wrong).

Hopefully this can be helpful to someone.

[pfreivald]

It jives well with my Mathcad model.

[Ether]

- Free-body force diagram

- Derivation of differential equations of motion

- C pseudo-code for the difference equations for 2nd order numerical integration

- Includes drag and spin

- Example graphs

http://www.chiefdelphi.com/media/papers/2725

[DavisDad]

pfreivald, Ether & Mike,

Thanks for the replies, much appreciated!

[NotInControl]

So as promised, I have included some additional data sets to test for your model.

I have yet to confirm the results to actual measured results from the balls flight trajectories.

Once again my model only includes air resistance no magnus force. All Paramters needed should be included on the plots.

The ball in the plot is to scale.

The only trouble I am having is determining the required force based on the lauch velocity. The numbers are way to small in my opinion. Maybe it is because using Vf = Vi + at assumes accel is constant over time and because I assume the force is applied over t=0.5 seconds which may be too long.

I don't know, I haven't spent time looking into it. But I would expect the required force to be in the 100lbf+ range for quick shots. Any thoughts?

Hope this helps,
Kevin

[DavisDad]

So as promised, I have included some additional data sets to test for your model.

I have yet to confirm the results to actual measured results from the balls flight trajectories.

Once again my model only includes air resistance no magnus force. All Paramters needed should be included on the plots.

The ball in the plot is to scale.

The only trouble I am having is determining the required force based on the lauch velocity. The numbers are way to small in my opinion. Maybe it is because using Vf = Vi + at assumes accel is constant over time and because I assume the force is applied over t=0.5 seconds which may be too long.

I don't know, I haven't spent time looking into it. But I would expect the required force to be in the 100lbf+ range for quick shots. Any thoughts?

Hope this helps,
Kevin

Hi Kevin,

thanks so much for the charts!

I'm working on modelling a mechanical linkage to pneumatic cylinder with:


Force of accelerating ball at end of long member
Calculation of airflow to cylinder using 0.3 Cv
Rate of travel of cylinder rod (piston)


I'm thinking that the limitation of the pneumatic's flow capacity can be addressed by the progressive increase in ball speed vs. piston speed via a linkage design. See video below:http://youtu.be/DaTn6ZMzihw

I'm not sure I follow your question regarding acceleration. Could you provide more detail?

Regards,

Craig

[NotInControl]

Sorry for taking so long to reply. Unfortunately I had to take a brake from Robotics this weekend and do other work.

In regards to my last post about Force, I was trying to do a sanity check to see if the model was accurate. So one way to do it was to calculate the force required to impart on the ball to give it its initial velocity, but the numbers were too small in my opinion. I only spent 5 minutes on it and then I realized I was making the wrong assumption using Vf = Vi + at, because that assumes acceleration is constant, which is incorrect.

So here is what I did which didn't work, and then below is what I did to fix it.

(Incorrect Calculations)

F = ma;
Vf = vi + at;

In the above equations, we know Vf, and m. Vi is 0 because ball is at rest, and t we can define based on how fast we want to shoot.

So if we solve eq. 1 for a and plug into eq. 2 and solve for F we get.

F = Vf*m / t

lets say we defined t to be 0.5 seconds the above would yield force required, assuming acceleration is constant.

But this is a wrong assumption because acceleration is not constant throughout the launch.

So the better way to do it is use the Law of Conservation of Energy, and in this case for my team we were thinking about springs, so I wanted to calculate the spring force we would need.

Kinectic Energy = 1/2 * m*v^2
Spring force = 1/2 * k * x^2

If we assume 100% energy transfer from spring to ball we can set these equations equal. we know m and v, and we can choose x based on our design. So lets say we only wanted to compress a spring 5 inches. We solve the below for k;

1/2 mv^2 = 1/2 kx^2

K = mv^2 / x^2

I have plotted the lb/in spring constants while varying the max displacement for the points I gave you. This should give an idea of the force required.

The numbers do align with what I think they should be, so my sanity is back in check.

Hope this helps,
Kevin

[DavisDad]

...So here is what I did which didn't work, and then below is what I did to fix it.

(Incorrect Calculations)

F = ma;
Vf = vi + at;

In the above equations, we know Vf, and m. Vi is 0 because ball is at rest, and t we can define based on how fast we want to shoot.

So if we solve eq. 1 for a and plug into eq. 2 and solve for F we get.

F = Vf*m / t

lets say we defined t to be 0.5 seconds the above would yield force required, assuming acceleration is constant.

But this is a wrong assumption because acceleration is not constant throughout the launch.

So the better way to do it is use the Law of Conservation of Energy, and in this case for my team we were thinking about springs, so I wanted to calculate the spring force we would need.

Kinectic Energy = 1/2 * m*v^2
Spring force = 1/2 * k * x^2

If we assume 100% energy transfer from spring to ball we can set these equations equal. we know m and v, and we can choose x based on our design. So lets say we only wanted to compress a spring 5 inches. We solve the below for k;

1/2 mv^2 = 1/2 kx^2

K = mv^2 / x^2

I have plotted the lb/in spring constants while varying the max displacement for the points I gave you. This should give an idea of the force required.

The numbers do align with what I think they should be, so my sanity is back in check.

Hope this helps,
Kevin

Hi Kevin,

Thanks for the thorough write-up. Will you include the mechanism in the model?

I'm currently plodding through this document: Dynamic Analysis of Pneumatically Actuated Mechanisms
(http://www.redalyc.org/pdf/768/76818412002.pdf) and scratching my head about how to use the formulas.

[DavisDad]

Putting all the mathematical modelling aside, I think team 1726 has demonstrated that the ball can be launched with pneumatic cylinders, and without springs or other energy storage:

2014 FRC 1726 Shooter Test 1 NERDS Pneumatic Catapault (http://www.youtube.com/watch?v=akZQIIDi12I)

Here's what I see:


This is a "rolling release" (http://arxiv.org/ftp/arxiv/papers/1005/1005.0176.pdf) catapult. An improvement over cup type ball holder.
The linkage starts with a reverse leverage (better term ??) where the mechanical advantage decreases with rotation; as the piston extends, the moment applied by the rod decreases.
The small initial moment with the inertia of the ball/catapult-arm, allows the pressure to build in the cylinder.


Nice work N.E.R.D.S. !

[Ether]

The linkage starts with a reverse leverage (better term ??)

3rd class lever.

where the mechanical advantage increases with rotation; as the piston extends, the moment applied by the rod increases.

Sort of like this:

http://www.chiefdelphi.com/forums/showthread.php?p=1327180#post1327180

... except I can't tell from your video if your piston is over-center in the loaded position. If not, how are you holding (and releasing) it?

[DavisDad]

... except I can't tell from your video if your piston is over-center in the loaded position. If not, how are you holding (and releasing) it?

Hi Ether,

Not over-centered, hadn't thought of it; great idea for "cocking" mechanism. Thanks for that!

In the sketch, the loaded position would be determined by hard stops and/or rod fully retracted. I was trying to evaluate the feasibility of shooting the ball with cylinder stroke only (no spring storage, no trigger release).

After seeing the video of the team 1726 catapult prototype, I'm thinking I have missed an important pneumatic dynamic: the inertia of the ball/launcher and the pressurization of the cylinder during the initial acceleration. We struggled with the Cv limitations of the solenoid valve when working on Rebound Rumble ball shooter prototypes. Your over-centered system looks like an even better way to deal with CA flow restrictions.

Your feedback, as always, is much appreciated.

NOTE: fixed error in previous post re "increased moment"

[Ether]

Hi Ether,

Not over-centered, hadn't thought of it; great idea for "cocking" mechanism. Thanks for that!

I'm just the messenger who drew up a sketch:). Kudos to Billbo911 for sharing the details of what his team was doing.

[DavisDad]

Ether- thanks so much for pointing me toward that. Here's our 2nd generation prototype; the lead-screw positions the ball-stop. The shooter captain would like ball position and proportional cylinder/accumulator pressure control.

http://simhardware.org/img/ShooterProto-2_a.JPG

http://simhardware.org/img/ShooterProto-2_b.JPG

http://simhardware.org/img/ShooterProto-2_c.JPG