I decided to do some physics to back up my opinion since some felt otherwise.
The following extract was taken from my blog post which has the complete derivation. Vex Note: How a single flywheel ball shooter minimizes the effect of ball mass variations
R is the range of shooter, E_wheel_final is the final rotational energy of the flywheel which depends on w_wheel^2 which in turn determines V^2 speed of the ball at release. A 5" Vex wheel is assumed for m_wheel.
So if I did the math right, the 10% potential range error with a catapult design caused by ball mass variations could be reduced to about 1% with 2 vex 5" wheels acting as the flywheel mass in a single wheel shooter.
5/29 edit note: corrected error in ball energy. Eb = 1/2m_ballV^2( 1+2/5) . (corrected 5/29 Was 1/2m_ballV^2( 1+4/5) Percentage ranges change slightly but basic conclusions are the same.
vamfun, thank you for quantifying the range errors! I suspected that exit velocity stability will go up with the increase of flywheel mass, but never realized it could be 1 to 10 with just two wheels!
However, I have a question. You think that the best design is a single side wheel shooter, but it “loses” 80% of the energy to spin the ball. Is there any consideration, other than the increased time of interaction of flywheels with the ball, to go with single side? Since we need multiple flywheels anyway it is tempting to put them on both sides of the ball and save some motors, by not giving balls so much spin.
The spin energy doesn’t bother me much since there is plenty of energy available from the motors. The spin energy is about 1 joule per ball. If you shoot 2 balls per second then the average extra power devoted to spin is 2 watts. Each motor can provide about 2 to 3 watts continuously so we are talking about one extra motor worth of energy.
The main drawback of the two wheel shooter is that the system energy is 1/4 of the single wheel system because the wheel speed is 1/2. So the extra energy that a ball requires when its mass is increased will have a bigger effect on the final speed of the wheel at exit. One could add extra wheels to each side to help out on this.
Contact variations are magnified due to the short time the ball has to stabilize its velocity before release. With a 5 inch wheel and a semi-circle cowling the ball rolls pi*(r_wheel + 2*r_ball) distance before being released. (about 21 inches or 1.7 ball rotations). The two wheel shooter might be in contact for closer to .1 ball rotation and squishiness comes into play a lot more on the contact accuracy. Both sides must have uniform contact or else side errors will occur. I suspect that as the balls get worn with use this will be a non-trivial source of errors. One might think of the single wheel shooter as a rifle having a longer barrel to help with accuracy. The two wheel shooter is more like a short barrel pistol.
As a final note, it is important to control the initial wheel energy in both systems since errors in this are directly translated to range errors. To control the V^2 to 1% you would need V to be controlled to .5% . This kind of accuracy means feedback control with forward path integration is probably needed to compensate for errors from sources like battery voltage variations and biases due to electronic controller and friction variations.
Here is a plot I made in MATLAB real quick. I think it backs the basis of your theoretical musings (the 10% range difference you mentioned). The ball is launched from an initial height of 0m to a final height of 1.03378m. The points were calculated with a constant kinetic energy of 1.225J (7m/s shot with a 0.05kg ball). Model includes smooth sphere drag that takes into account a variable coefficient of drag. Sea level conditions were used for gravity and air properties. Balls have no spin. https://vexforum.com/attachment.php?attachmentid=9384&stc=1&d=1432860585
There are a few interesting things I saw. Firstly, the lines are roughly linear. However, you can see that the percent difference is not the same when the ball is 10% more massive compared to 10% less massive. Also, you can sort of see how the 45 degree line curves on the right side. The launch angle clearly has an affect as well.
Take from it what you will. If anyone wants to see any other plots let me know
Also, does anyone have any recommendations for how I should post these plots? Uploading to the forums pretty much restricts me to .jpg because of how the others are limited in size. Personally, I think the compression artifacts look horrible and make the plot quite hard to read.
I will work on a model for the transfer of kinetic energy from flywheel to ball this weekend.
I love all the analysis you and some of the others have been doing with these balls and flywheels and such, so I’m really reluctant to ask any questions. But I’m not sure that your approach here provides the best way to think about this. You are presuming that each ball is getting the same amount of kinetic energy when launched, whereas I think what you are most likely to see is that each ball would get launched with approximately the same velocity since its velocity is (within limits) geometrically constrained by the way the flywheel(s) speeds up the ball. So while, in reality, each ball would start out with about the same velocity, they would possess different values of kinetic energy because of their mass differences. Or am I confused about what these plots are showing?
You are reading the plot right. And yes, it is wrong to assume that the balls will be launched with the same kinetic energy as their masses change. This plot is simply here to validate vamfun’s initial calculation that gave him 10% variation. He had used a model without drag to come up with this value that related the change in mass to the change in range with a constant kinetic energy, so I wanted to verify it with a model that included drag.
If we assume a no-slip condition, then the balls will not be launched with the same kinetic energy or velocity because as energy is transfered to the ball the flywheel(s) will slow down, and the amount of energy tranfered is dependent on the ball mass. Vamfun went on to calculate how much kinetic energy is transfered to the ball assuming no slip. I was going verify this and make some plots later on when I have some more free time.
So is this plot for the flywheels or for a catapult? I thought Vamfun got a 10% variation only from the catapult design. His flywheels got only about 1.8% variation or less, I think. And 1.8% of 15 feet is only about 3.24 inches, so I’m not sure why anyone would care to add more flywheel mass, if those calculations are correct, since I’m guessing drag differences caused by seams and so forth would be greater than that. Intuition tells me that more flywheel mass is better but I’m not sure what’s happening with these numbers now.
If i’m understanding this correctly, that was a chart for the ball, and all that they are doing is quantifying the effects of conservation of energy, or conservation of angular momentum (if it’s easier to conceptualize)
The physics is that both rotational energy and angular momentum are dependent on rotational inertia and angular velocity. Angular velocity is what we need to keep constant to maintain the proper range. So each time we shoot a ball, some of the angular momentum/velocity is given to the ball, and taken away from the flywheel.
The engineering is that since playing with velocity isn’t optimal (we would have to play with the flywheel speed while tracking the ball differences for this), we can play with inertia. And to play with inertia, this is where the mass and radius of the flywheel comes in. However, radius is much more important because it’s squared. In this case, we can play with the inertia of the wheel by adding more mass like you said, but it’s a lot more effective to play with the radius of the wheel. In the end, the increased inertia will affect the total momentum/energy, allowing you to maintain a constant speed so much better, making changes in the mass/size of the ball negligible
The plot can only be for catapult since the kinetic energy is constant which is what you intended to check on. The slope variations are a little misleading since even though they are at the same height, the nominal 0 percent error will have a different range for each launch angle. Typically, the kinetic energy needs to be adjusted for different launch angles. i.e. Rmax = v^2/gsin(2theta). Maybe run again but adjust the kinetic energy to keep the range the same for all thetas.
BTW folks, I had an error in my ball energy which I corrected and reran the errors. The ball rotational energy is only an extra 40%, not 80% as I had stated. This actually helps the range variations. I posted the corrected results 5/29.
You can always post them to the media section of your blog (or flickr or dropbox or wherever is publicly accessible) and reference the image via the img tag here and use a GIF or PNG file.
Also, cool analysis. Is the target location the full plane of the triangle or a specific spot described in X,Y coordinates? The goal plane is a sloped triangle shape but the sweet spot of staying in the goal is probably less (unsure of that sweet spot shape).
Vamfun, I will post a plot where the different angles have the same range for m=0.05kg tonight when I am home. Are there any particular launch angles you are interested in? I can create a 3D surface plot of the angle-mass-range if you’d like.
Giraffes, the range solver is set to stop when the centet of the ball reaches a predefined final height and is traveling downwards. The height used in my last plot was around the middle of the high goal triangle I think, 1.03378m
I would like to get some feeling for how just the air drag affects these balls vs. their mass (ballistic coefficient effects). Could you make a plot that assumes each ball gets launched at the same speed and angle (~45 to 70 degrees) with zero spin and the ball masses varying by +/- 10% of 0.050 Kg?
Sure thing, I’ll be home and get yours and Vanfum’s plot posted in probably 2 or 3 hours. Is there any particular speed you are interested in? If I remember right, a 7m/s shot at 45 or 60 degrees (I can’t remember which) gives a range a bit longer than the diagonal of the field.
Sorry it’s a bit later than I said. Here is the plot for Vamfun. I arbitrarily chose 5m for the range. https://vexforum.com/attachment.php?attachmentid=9385&stc=1&d=1432959040
Looking at it now, I can’t tell which line is which based on the plot (I’m color-blind). For those having difficulties, looking at the left side of the graph the top line is for 45 deg, the middle 60 deg, and bottom for 75 deg I think
Here is your plot FullMetalMentor. The first one includes all three launch angles. I used a different line style this time. Which is easier to read? I can make the color plots lines thicker if that helps.
I forgot to include a legend. The top dashed line is 60 degrees, the middle solid line is 45 degrees and the bottom dotted line is 75 degrees. I am really sorry I forgot that. I can fix it if anyone needs it. https://vexforum.com/attachment.php?attachmentid=9386&stc=1&d=1432960351
I also made a plot of just the 60 degree launch so you could see the curve a bit better. https://vexforum.com/attachment.php?attachmentid=9388&stc=1&d=1432960423
It is interesting to see that the relationship is not quite linear, however a linear approximation would probably work fine in this case if you were able to measure the mass of the ball. It may be more interesting to see what it looks like when we include spin. I can try that if anyone is interested.
This is excellent! So it looks like if speed can be accurately controlled, then the ballistic coefficient differences won’t necessarily be a dominating factor.
Yes, if you think you’ve zeroed in on a good way to simulate spin, that would be great. I’d especially be interested in knowing at what spin speed do you start to see significant effects.
Fun fact: you can aim a leaf blower straight up and place one of these balls in the air stream and they hover quite nicely. My kids haven’t got any useable data off of doing that (I guess somehow you’d have to know the air speed, etc.) but it looks kinda cool.
Remember that the plot uses a constant initial velocity of the ball right after launch. Vamfun has shown logically that if you were to keep the flywheels at a constant speed and vary the ball mass, they would have different exit velocities. I’ll try to incorporate this aspect into the model and give a plot soon. So far Vamfun’s calcuations have looked solid so I don’t think I’ll find anything interesting
So I think I have the theory down well. However, considering the assumptions I had to make, I wouldn’t be surprised if empirical data doesn’t match up nicely with theoretical data. Don’t get me wrong, I think the model is better than not accounting for spin. But until I get my hands on some real-world data I will hold my tongue on the validity of my work.
EDIT: So I made a serious mistake. Every time it says rpm in the following bit should instead be rotations per second. Sorry! And thanks to Vamfun for pointing it out.
Here is a plot I made that incorporates spin. The left-most loop has +200RPM (backspin) and the shortest shot has -200RPM (topspin). https://vexforum.com/attachment.php?attachmentid=9392&stc=1&d=1433017844
I noted the interesting points where you see the ball starts to loop back around 130ish RPM and how we see our range maximized around 50ish RPM. One thing to note is that I assumed the ball had a constant rate of rotation through the entire flight.
I spent the past half hour playing with my air compressor and a golf ball in the garage. I went until the air started to condensate :rolleyes: I’ll let it cool off some and then get a video posted. If you want to blow your kids minds a little show them the vertical blower and ask them what’s keeping the ball up. They will say the air is pushing on it. Then tilt the blower at an angle of 15 or 20 degrees and show them that the ball will still stay up and ask them the same question again. They will probably say that the air is now pulling on the ball. These things are often called Bernoulli Blowers due to his aerodynamic principal which lends to the horizontal stabilization of the ball (the lift force is an entirely different topic). When it is at an angle Bernoulli helps keep it up, but you will also notice with smaller balls that they start spinning fast and get a little help staying up thanks to Kutta-Joukowski. Although I doubt your students will really care who’s theorems or principals are making the ball float
Anytime! I love any excuse to break out MATLAB and solve some numeric problems.
