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mytqik



Joined: 26 Jun 2004
Posts: 127


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All very true what you say, but what are you going to do with the energy calculation? As i said it is very hard to calculate the rate of absorbion of energy.

My point is that if you know the force answer, you can the design motor mounts, gearbox mounts, frame etc to take the 9000kg of shear/tension/compression force so that your gearbox isnt ripped out when it lands after being flipped 3m in the air.

As we have "strict" weight limits there will always be a trade off between strength v's weight, but at least knowing this answer we can have a starting point for the design of the robot.

Post Sat Jul 10, 2004 10:09 am 
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colin



Joined: 16 Jun 2004
Posts: 102


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I've never seen so much mangled physics in my life Shocked

For a start Gravitational constant is different to gravity. and potential energy = mgh (mass X 9.8 X height) (using standard SI units, to get a standard SI answer in Newtons)

After that I just gave up on the rest of the thread.

also m is for milli as well as mass and metres.

Post Sat Jul 10, 2004 12:26 pm 
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kkeerroo
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Joined: 17 Jun 2004
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quote:
Originally posted by colin:
I've never seen so much mangled physics in my life Shocked

For a start Gravitational constant is different to gravity. and potential energy = mgh (mass X 9.8 X height) (using standard SI units, to get a standard SI answer in Newtons)

After that I just gave up on the rest of the thread.

also m is for milli as well as mass and metres.


Huh?
Energy = Joules (J)
Force = Newtons (N)
Mass = Kilograms (Kg)
Weight = Newtons (N)
Displacement = Meters (m)

I did tutor high school physics for a bit.

And though I liked the Kinetics lectures, I perfered the Mechanism Design lectures over the Mechanics of Solids lectures.
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Post Sat Jul 10, 2004 9:43 pm 
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colin



Joined: 16 Jun 2004
Posts: 102


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yeah my bad, I always mix my energy force units Razz

I think you started doing it right, but i gave up reading it.

Post Sun Jul 11, 2004 11:55 am 
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Spockie-Tech
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Joined: 31 May 2004
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Well, I'm not hanging on my calculator for this thread, but the discussion of energy principles is always an interesting thing. given that our sport basically involves moving large chunks of it around in a violent way.. Laughing

The deceleration thing is the key point I think.

As we well know (goverment warnings aside), speed doesnt kill at all. you can live you entire life at 800km/hr without a problem. its only when some inconsiderate bastartd goes and puts something - 1. slower then you and 2. heavy and/or solidly mounted in the way that it becomes a bit of a problem.

Then we enter the region of physics that has caused more deaths, splats, booms, squishes, ouches and other undesireable experimental phemonena. The point of deceleration (or "now what the heck am I gonna do with all this energy I've been carrying around in my pocket ?").

Speeding doesnt kill you at all.. but Stopping quite often does.

My original comment about using G to calculate the robots potential energy as it peaked in its hypothetical 3m ballistic arc was flippant, but interesting as the robot now has the same problem of an armoured bot about to meet a spinner... Theres a lot of energy about to come its way in a big rush. and how fast it can disperse that energy evenly, or where it can absorb it without damage will determine if it gets to walk away from this landing or not.

If our theoretical stunt-bot takes .1 of a second to decelerate from 7.67m/s to 0ms, then it will have travelled an further (~) 0.76meters from the moment of impact before coming to a halt. I hope its make of rubber, or else I think it just moved into the "splat" catgeory.

I would guess that any imapct that takes more than 1 millisecond (.001) or so to absorb/disperse is going to be trouble.

Post Sun Jul 11, 2004 12:24 pm 
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mytqik



Joined: 26 Jun 2004
Posts: 127


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Spockie,

You are correct about the speed kills bit Very Happy

However there is one major difference between a spinner & landing from being flipped. Generally the ground in not going to move one bit, so all of the force/energy is going to be transfered into the landing robot.

When a spinner hits a bash bot the energy is distributed to a number of places: Frame of bash bot, frame of spinner bot, motor mounts of spinner, the bash bot may move in the direction of the spinner hit so there is friction between the wheels of the bash bot & the ground using up the energy in the spinners blow. All in all I would have thought that landing & stopping suddenly wouold cause more damage than being hit by a spinner.

However at this stage we dont have any robots capable of throwing another robot any distance into the air so the spinner is still the most damaging weapon.

Post Sun Jul 11, 2004 12:58 pm 
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Spockie-Tech
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Now theres a good question for you math guru's out there..

What is the amount of Kilojoules/Newtons/whatever required to accelerate a stationary 12Kg mass over a powered distance of 30cm (approx ram travel) to a velocity that will result in a ballistic peak of around 2meters ? 3 meters ?

then, Is there an easy way of translating this required impetus into a theoretical requirement (ignoring losses due to valves etc, which will be the big trick) of X cu/ft per min of CO2/Air ?

That should give your calculators something to chew on.

Post Sun Jul 11, 2004 2:27 pm 
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mytqik



Joined: 26 Jun 2004
Posts: 127


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Spockie,

Thats is easy.

First caculate the velocity required when the robot leaves the flipper. This velocity needs to be enough to over come gravity for 3m. At 3m the robot velocity is 0m/s as it is at the peak of its tradjectory.

Vf^2 - Vi^2 = 2 x A x D
0^2 - Vi^2 = 2 x -9.81 x D
Vi = 7.67m/s (which is the sam as we calculated before)

Next calc the acceleration required to move a robot at 7.67m/s from stand still over a distance of 0.3m

Vf^2 - Vi^2 = 2 x A x D
7.61^2 - 0^2 = 2 x A x 0.3
A = 98.1m/s/s (which makes sense as it is a 1/10th of the distance of the previous calc but with the same start & finish velocity, so the acceleration must be 10 times that of gravity)

The force required to generate this velocity is

F = M x A
F = 12 x 9.81 x 98.1 (we need to convert the weight of the robot into newtons to get an answer in Newtons)
F = 11 548N

Next we need to calculate the cross sectional area of the Ram.
Assume 250psi air pressure = 1.1723MPa (megapascal = 1N/mm^2)
Pressure = Force / Area
1.1723 = 11 548 / Area
Therefore Piston area = 92.37mm^2 (thats a 4" Ram)

Next is volume per flip = Piston Area x Stroke
= 92.37x10^-6 x 0.3 (had to convert all units into metres)
= 27.711x10^-6m^3 (cubic metres of air per flip)

assume 5 flips per minute for a 5 minute battle
= 27.711x10^-6m^3 x 5 x 5
= 692.775x10^-6 m^3

Now assume you are using a air tanks (0.3m long x 0.075m in diametre)

Volume = (0.075 / 2) ^2 * Pi * 0.3
= 1.325x10^-3 m^3

Therefore you need 692.775x10^-6 / 1.325x10^-3 = 0.522 of these air tanks.

HOWEVER this assumes that the tank is kept at 250psi, which is not a valid assumption. But I dont have the time to calculate the volume required if the tank reduces in pressure after every flip, as would occur in a normal robot without an onboard air compressor.

Hope this helps, let me know if you think i have made any unreasonable assumptions, or have cocked up a calculation Very Happy

Post Sun Jul 11, 2004 3:12 pm 
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kkeerroo
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Joined: 17 Jun 2004
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I know this is the wrong section, but what about liquid CO2 and regulators for that constant pressure? I might have to give pneumatics a go one day.
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Post Sun Jul 11, 2004 7:11 pm 
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mytqik



Joined: 26 Jun 2004
Posts: 127


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Well then if we are talking about CO2 the problem calculating the tank size required is that you know the rate of expansion as the CO2 changes from liquid to gas. Also temperature plays a big part here too. The force calculations still hold, except you would have to change the pressure from 250psi to what ever the CO2 regulator is set at. Unfortanuately thermodynamics is something I never got my teeth into.

Post Sun Jul 11, 2004 7:19 pm 
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Philip
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Joined: 18 Jun 2004
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[quote="kkeerrooMass = Kilograms (Kg)
Weight = Newtons (N)
I did tutor high school physics for a bit.

And though I liked the Kinetics lectures, I perfered the Mechanism Design lectures over the Mechanics of Solids lectures.[/quote]

Force is measured in newtons (N) which makes weight correct using F=ma and only a pedantic bugger would mention it.

Kilograms, however, is not Kg it is kg.

I hope you didn't charge too much for being a tutor.

Post Tue Jul 13, 2004 1:12 pm 
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kkeerroo
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Hey, I failed english.
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Post Tue Jul 13, 2004 4:50 pm 
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