# Trebuchet Help/Understanding

1. May 14, 2007

### SirAwesome

I was hoping someone out there could offer me much needed insight on my high school Physics project. I had to build a trebuchet and now I have to write a report about it. I have struggled with this and put much time into try to figure the velocity of the projectile (tennis ball) as it leaves the trebuchet (would this also be the velocity in x dir. b/c its constant?) I was using the equation mgh= 1/2mv^2 and solving for v. (Since m cancels, not needed, and I was using h as the distance the counter weight falls)(All the hint that was given by my teacher was that something might need to be added to the equation, but then how could figure in mass c.w. , mass of ball, h of c.w. etc.)

However this velocity is way too slow to give me the estimated distance that I achieve. I have an actual distance of about 14 m. I know I could reverse solving it using my time but I need to use the energy equation to show my teacher that I can estimate the distance it goes without actually measuring the distance from the launch. I also know that from the energy equation above the velocity should be higher than the actually velocity because of energy lost in various aspects and in flight.

Variables

t(flight time) =.9 s
h(how far c.w. falls)= .406 m
g(gravity)= 9.8 m/s2

Using stuff above I calc. a v of 2.82 and from that get a *clalculated* distance of 2.53 m by multiplying t by v (see the problem  )

Sorry for the long post, but hopefully you can see my frustration with the matter.
(P.S. I am in just basic/normal high school Physics, so please keep it below college level)

Thank you to all who read and/or respond

2. May 14, 2007

### DyslexicHobo

Knowing the mass of both the counter-weight and the tennis ball along with the height the counter-weight drops before the tennis ball leaves, you can find out a lot of information.

Try using conservation of energy equations, also.
U = mgh
KE = (1/2)mv^2

You know the height of the counterweight, the mass of the counterweight, and the mass of the ball. This equation easily solves for the velocity of the tennis ball. The mass on the left side would be the mass of the counter-weight, and the mass on the right side would be the mass of the tennis ball.

Now that you know the velocity of the tennis ball, you need to find the x and y velocities in order to have an accurate prediction of displacement. You could do this by finding the length of the lever arm of the tennis ball. If treated like a triangle (at its fired position), you can find the angle of projection.

Then we can treat it as a basic projectile motion problem. You can find the x and y motion, solve for how long it will take the projectile to hit the ground using basic kinematic equations, and then use that time and plug in for x = v_x*t.

Please note: This does not take into account the mass of the actual beam which is firing the tennis ball. You may or may not decide to include this in your calculations. It also does not take into account any rotational energy introduced (depending on how exactly your trebuchet is designed, this probably won't effect your predictions much).

I hope this helps, and please ask any questions if I misunderstood or you don't understand what I'm saying. And anyone else, please correct me if any of my suggestions are incorrect. Good luck!

Last edited: May 14, 2007
3. May 14, 2007

### Staff: Mentor

Just use the PF Search feature, and fill in your keyword Trebuchet. You'll get the earlier threads where they have been discussed -- hopefully that will help out. Welcome to the PF, BTW.

4. May 17, 2007

### SirAwesome

One final Quesiton

Ok thanks a lot guys. But one final question. Do I take the whole throwing arm length or just the length from the launch end to the axel (where it rotates) when its locked back and ready to be launched ? Or am I getting the wrong picture and should take the length of the arm when it’s fully vertical (at rest just after it launched, but then how would I calculate other sides then?)

And thanks again to all who help !

5. May 17, 2007

### DyslexicHobo

You should use the length from the center of the tennis ball to the axis of rotation. The angle you should use to find the x and y velocity vectors is the angle at the position at which the tennis ball leaves the launching mechanism.

Ask any other questions if you need help.

6. May 21, 2007

### SirAwesome

Hey, guys I’m back with one final problem/ issue (last I swear…. I think :) )
Using the info you guys so kindly gave I was able to find the angle at which the arm makes when locked and loaded (correct right ?) I believe this is right as my trebuchet looks like this (http://www.ripcord.ws/plans/tennistrebuchet.jpg ) and I’m not sure it’s correct to measure the ball/sling fully extended from the arm so I measured from the prong end to the center of the axel and found the angle via that way. I broke the velocity down of the ball from the kinetic and potential energy equation down into x and y components. The velocity in the x direction is perfect (slightly higher than actually, again perfect, calc out to 16.5 m/s) However I get Viy to be 24.56 m/s (fast?) Now I wanted to calculate Vfy and to do this I used Vfy =Viy + at. From that I was able to calculate Vfy to be 33.38 m/s. Now what I’m getting to is that I wanted to find the energy lost (and verify that I’m thinking correctly). So to calculate the velocity the ball has right before impact I used Pythagorean id to added Vfy^2 and Vx^2 together and then took the square root of that to get an answer of Vf= 37.23. Plugging that back into mgh-(.5mv^2) = E lost.
I get a negative number. Am I not doing something right (what are the chances?) or is the something I’m missing or am I calculating a velocity wrong?

I did play around with the numbers and angle, If I included the sling in the length that would give me a longer hyp. and a smaller angle, this gives me a good velocity in the y however in turn messes up my velocity in the x direction giving me a velocity that is way too high. I also know that I should lose energy not gain energy (or is this because of gravity action upon it , but then how would I find the energy lost ?)

Info I calculated

Vball=29.59 m/s
Vxball=16.5 m/s
Viyball=24.56 m/s
Vfy = 33.38 m/s
T total = .9 s
m of c.w. =6.33 kg

Heres my measurements for prong to pivot (center) and pivot (center) straight down

(See attachment)
In case attachemnt doesnt work
lenght from prong to pivot center = .985 m
center of pivot straight down = .817
Angle using sin^-1 = 56.1 degrees

And again much thanks to all who read and help, I truly appreciate it!

Last edited by a moderator: Apr 22, 2017
7. May 22, 2007

### SirAwesome

After some thinking I was able to determine that I believe that I do in fact need to measure/ calculate the angle at which the ball is launched. But now that brings up another question of mine, what do I use the potential energy/ kinetic energy velocity calculated for. Because originally I had used it to break down the components velocity components as it initially left the trebuchet, not as it was initially launched and from there calculate theoretical velocity. Are all these numbers now screwed up? Or do i still need them for something? And I am sorry I am just very confused as to how exactly I calculate the release angle and what to do to break the velocity down into the x and y components.

Thank for your much needed help

8. May 22, 2007

### DyslexicHobo

I'm sorry, I was envisioning a trebuchet as having a fixed launcher, such as a cup like a catapult. I didn't think that it was a sling. This could make it a little bit harder to calculate the launch angle (based on the elasticity of the rope).

However, if you seem that your x-component of velocity gives you the correct distance when plugged into $$d = vt$$, then it leads me to believe that your measured angle is correct. If you're using conservation of energy to measure the mechanism's energy loss, then you will need to calculate the ball's initial theoretical $$V_o$$, and add the ball's initial potential. The ball's total energy at the time of release is it's kenetic + potential. You can then compare it with it's actual mechanical (PE + KE) energy by finding the actual $$KE_f$$ at the point at which the ball hits the ground.

Please let me know if you understand. I'm not always the best at explaining.

Edit: Sorry, didn't even answer your main question. To find the angle of release, find the angle at which the sling stops accelerating. This is easy for a cup, but for a sling it may be difficult, depending on how elastic it is.

Last edited: May 23, 2007