Forum: Catapult Message Board

A program I wrote


CarlofET 3 stars
Hi everybody. I thought you might be interested in a program I wrote. We were working on the Freedom Launcher, and we were wondering how much force we are going to need to pull down and cock the throwing arm. Being the Mechanical Engineering student that I am, I got to thinking about how to figure that out. It's a somewhat complicated problem, because our pully will be stationary, so the angle at which we're pulling on the arm will be constantly changing - just as the angle at which the counter-weight is acting on the arm will be. It's doubly a function of the angle. So anyway, after some thinking, I derived this equation:

T=s*w*sin(x)*sqrt(2*h*l*cos(x)-2*d*l*sin(x)+d
2+h
2+l
2)/(l*(d*cos(x)+h*sin(x)))

Here's what all the letters mean:

T: The tension in the cable
x: The angle between the throwing arm and an imaginary vertical line. For example, when the arm is at rest, the angle is 0. When it's cocked, it should be somewhere around 140 or so.
s: The length of the short end of the arm - from the axle to where the weight is attached
l: The length of the long end of the arm - from the axle to wherever the cable attaches to the arm
w: The weight of the counterweight
h: The difference in height between the axle and the pully. For example, the Freedom Launcher's axle is 18 ft off the ground. The cable will be entering the pully at about 1 foot off the ground. So this value for us is 17.
d: The distance from the pully to the upright support. Easily computed using the Pythagoren Theorem: d = sqrt(l
2-h^2)

Now, this is a rather messy equation, so I made a program in MatLab where you input the s, l, w, and h values, and it tells you where and what the maximum tension in the cable is, as well as how much force will be required to hold the arm in the cocked position. This is very helpful for figuring out the load on a trigger - and not to mention figuring what type of cable is needed to winch the arm down. I've attached a screenshot of the program. However, this .exe only works on computers that have some MatLab software installed. So if anybody is interested, I can supply you with the MatLab code or the .exe, but the extra (free) MatLab software required for the .exe has a 75 MB or so install file. Just thought I'd spread the word.

Oh, and visit http://www.freedomlauncher.com (external link) - we've done some updates.



bboy 3 stars
whoa



Kaiser 3 stars United States
CarlofET,

Interesting stuff. I did my own calculations calcuations and came up with the attached excel spreadsheet. My equation is shown in the spreadsheet. It looks different but I calculates the same max load at 70.7569 degrees. But our equations get slightly different results at the "cocked" position. I hand verified that my equation works at a couple different angles, but I don't have time to track down the discrepency right now. Give it a look.



CarlofET 3 stars
Kaiser -

I think I've figured it out. Our equations are, effectively, exactly the same. However, the difference is that my equation takes into account what happens when the arm reaches the position of the pully. In a real situation, of course, this can never happen. If it did, the length of the rope on the pully would be zero, and so there could be no tension. Furthermore, if the arm could somehow pass this point, the tension would become a compression. So the way I calculate the tension in the cocked position is by taking the exact angle in which it's cocked and then subtracing .000001, or however small a number I can in order to still get a result. Again, realistically, you wouldn't even be able to get to this point, so the best bet is to subtract a full degree or so to take into account the distance between the pully and the arm when it's cocked. Know what I mean? I've made and attached a graph in excel to show the results of the two equations. It should clarify everything. The reason the vertical line on my graph isn't exactly straight is because I only took values at even 1-degree values. If you graph the actual function, that line will be totally vertical.



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