A simple formula or rule of thumb for determining a number of values associated with the design of Hinged Counterweight Trebuchets.
This writeup assumes that the reader is already more or less familiar with how a trebuchet works, but is interested in pushing it to the limit.
This message was posted to the old version of the Catapults Message Board, (hosted and moderated by Ron Toms) in response to a question about determining the appropriate mass of projectile for a trebuchet of given dimensions..
|Subject Re: The projectile
Posted by phsstpok on 2002-14-05 02:32
Too light a missle leaves more energy in the machine which must be disipateted through the frame. At the far end of too light is a dry fire (no missle at all) which can be very hard on the frame.
You're right that there comes a point where lighter missles won't go much, if any, farther than heavier missles. For my own experiments I used one and two liter soda bottles filled with water and varied the amount of water in them as well as sling lengths and pin angles.
As for what you should throw. All machines are a little different and there is no definite answer. Here is a rough guideline for HCW machines. Multiply the beam ratio by twenty. Divide your counter weight by the result and that will give a good starting point for your missle weight.
(bold emphasis added by me)
Since the person who posted it uses the handle "phsstpok", without capatalisation, I'll continue to write it that way. His handle, by the way, was borrowed (with a minor spelling change) from the character "Phssthpok", in Larry Niven's science fiction novel Protector.
URL of that posting:
Some Definitions and Clarifications
Refer to ASCII-Art diagram below.
| cw main release
axle axle pin
cw | / \ | sling
hanger | / \ |
[CW] / \ (P) pouch with
/_______\ payload "P"
(If the above diagam does not look quite right, select the text and paste it into a text editor, using a mono-spaced typeface such as Courier)
Significance and Application
Here is "phsstpok's Rule" expressed in two versions of the same simple formula:
MR = BR*20
BR = MR/20
Generally, you design a trebuchet with the idea of obtaining the maximum range or distance at which you can throw the payload. Common sense suggests that if you add more counterweight mass, you will obtain a greater range. You will, but only up to a point where it is no longer practical to add more CW mass.
The reason for that upper practical limit is that as the CW mass increases, (assuming that the mass of the beam is negligible) the mass of the payload becomes negligible, and for all intents and purposes the CW is in "free-fall", accelerating downward at 9.8 meters per second2, as if it was simply dropped off a cliff. It does not matter how much mass you add to the counterweight, it will not fall any faster than free-fall. Since the counterweight cannot fall any faster, the arm will not pivot any faster, the sling won't whip around any faster, and the payload won't be thrown any faster. Since the velocity of the payload does not increase, neither will the range.
As I've said, it's simply a "rule of thumb". The mass of the beam itself does have some effect, because it is not some ideal massless beam, it has inertia. No matter how small the payload mass is, the beam itself still has to be brought from rest to rotation, which takes some energy. I suppose if you found the centre of gravity of the beam and treated that as some additional mass tied onto the beam somewhere between the main axle and the release pin, but unless the beam mass was something like, oh, maybe greater than 10% of the CW mass, it's probably not worth calculating. Suffice it to say that it's not taken into consideration for phsstpok's Rule.
Other applications of the formula
You have selected (or have been given) a particular CW mass, a payload, and a beam length. Where do you place the hole for the main axle?
beam = 8 feet (because you found some inexpensive spruce 2x4 studs)
payload = 0.2 kg (you have a basket of wormy apples to dispose of.)
(Notice that I'm mixing metric and Imperial units here. It doesn't matter, as long as each ratio is calculate with the same units.)
CW = 20 kg (because you have a 20 liter jerry can that you can fill with water and tie to the end of the beam with a rope.)
So, the mass ratio is 20:0.2, or 100:1
That makes the BR (100/20):1 or 5:1
The beam ratio of 5:1 means that the beam is divided into 5+1 parts, and also that the main axle should be 1/6 of the beam length from the CW axle, which is 1 foot and 4 inches, So, CA = 1' 4" and then TA = 6' 8"
Now that you have your TA length of 6' 8", you can calculate the axle height for a 45 degree cocking angle. Conveniently enough, since it's a 45 degree right triangle, you can use a trignometric shortcut and simply multiply the TA by 0.707 to get the axle height of 4.713 feet, or 4' 8.5". This is also the approximate length of the base of the frame from the axle support to the uprange end where the trigger will be. (Downrange means towards the target area, uprange is the opposite way, behind the trebuchet.)
With the axle height of 4' 8.5 inches, you can next calculate that the length of the CW and CW Hanger combined should be just under 4' 8.5" minus the CA of 1' 4", or 3' 4.5". Any longer than that, and it will bottom out on the trough.
Since a typical sling length is about the same as the length of the TA, that means the downrange end of the trough should be at least 6' 8" from the back or uprange end of the trebuchet frame, or 6' 8" minus 4' 8.5', which is 1' 11.5" beyond the main axle support post. Or, make it symmetrical, and have a frame base of 2 times 4' 8.5", or .9' 5". What the hey, since you have all those 8' 2x4 studs, just make it 8 feet long and be done with it.
That is about 90% of what you need to know to build it.
There is a small trebuchet competition, You are allowed a maximum height of 1 meter, when it's in the cocked position, and the supplied ammo will be tennis balls.
Figure out the length of the beam, assuming a cocking angle of 45 degrees.
So, what if I add more counterweight mass anyway? Is there a penalty?
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