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Sorry, energy delivered is not purely proportional to head weight, but it is highly dependant on it.
If you have a 10,000 Watt peak output motor attached to a very light axe, so that it only took only 1 millisecond to fire, then there is no way that it can deliver more than 10 Joules of energy.
Whereas, if you connected it to a very heavy axe, so that it took 0.5 secs to fire, then it has the potential to deliver up to 5000 Joules.
Im building a Rexs Robot Challenge style FW for my IVA project at college, and I want an axe robot rather than a flipper, and I am unsure of a mechanism that uses both rams. Any clues ? I can use a motor, but Id need extra batteries for that
A motor would be easier, as you cant get much space and pressure in a fizzy drinks bottle (meaning less potential shots from the rams than Co2). Ive got to admit a Co2 axe would look better though. It would be quite easy to couple two bicycle rams together and use the same rack & pinion meathod as used on terrorhurtz, although the rack and pinion would probably require machining a new rod (the moving bit) for the ram(s).
You could use a crank lever system powered by a motor. Similar to what Mortis has.
for electric featherweight axes, e-mail me on james@jamesbaker.f9.co.uk and Ill send you a pic and info.
the design was entirly from my own knowledge of degree level engineering, and nothing else.
Having an astronomical respect for my knowledge, my friend then asked everyman and his dog for advice, including John Reid, and the solution he came up with using Johns figures and many computer programs was.... exacly the same as my solution.... exactly. So Id say that sticking to basic pricipals is by far the easiest way to build an effectice axe and if you do not have the maths background e-mail me and Ill send you info about mine
cheers
james
Ive had a look at Pauls equations and Ive fed in all the stats for my feather axe robot. I had to do this on paper as I dont have Excel. I ended with an answer that I was expecting, but when I then tried to modify the inputs for the gear reduction which my robot has the answer ended up at less then a 3rd of the energy. Anybody care to help explain this?
The robots weapon is a 1kg hatchet driven from an E.V. Warrior at 24V with a 4:1 reduction through chains and sprockets. The energy calculated before reduction was 1.4kj and after reduction was 0.4kj. A photo of the half completed robot can be found here http://www.robowars.org/forum/album_pic.php?pic_id=346http://www.robowars.org/forum/album_pic.php?pic_id=346
An EV Warrior powered by two 24V 3Ah Nicad sets would have a peak power output of around 600 Watts.
Say the axe sweep lasted 0.2 secs, then the maximum possible energy would be 120 Joules.
To get a crude approximation to the sweep time, use the formula: t=(2 * s^2 * m / P)^0.3333
where s is the sweep distance of the axe, m is the mass of the axe head and P is the power of the motor. Assuming a 300mm long shaft, then this gives 0.14 secs and 86 Joules.
If you doubled the length of the axe shaft to 600mm and doubled the axe head to 2kg, then you could get it up to 0.29 secs and 170 Joules. With this setup, you would want a gear ratio of around 20 to 1.
To get a very crude idea of the gear ratio required, use the time of the swing to get the average rotational speed of the axe. With the current setup 0.14 secs for half a revolution = 3.5 revs/sec = 210 rpm.
The no load speed of an EV on 24V is 5000 rpm, so the speed at which peak power is produced is 2500 rpm. So the gear ratio should be around 10 to 1.
Youll be glad to hear that you are only losing around 10% in energy by running at 4 to 1 rather than 10 to 1. You will however be drawing a lot more current from the batteries.
Ive made a fether with an axe. Its just a thwack bot, using the stinger-like system. I still have to get it a bit more reliable though :)
See my profile for a picture.
It uses 2 drill motors for each wheel. I still have to replace the actual axe, but its more of a show-bot anyway.
a better way to express the sweep time formula would be:
t = (2 * s^2 * m / P)^(1/3)
x^(1/3) is the cube root of x
s^2 = s squared