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If your robot is running, you also need to account for dynamic affects -> force = mass * acceleration. For example, if your robot was to jump, acceleration must be greater than gravity. So you'd add static + dynamic force calculations together.Use this:http://www.societyofrobots.com/mechanics_statics.shtmlAnd never ever ever do a calculation without a force body diagram!
So for my robot when it's running, it is still static+dynamics?
Also, is acceleration the speed that the whole mass is moving (not just one leg)? or how fast the servos are moving?
And the sum of static+dynamic is equivalent to the amount of weight my robot's servos must be able to bear...right?
then now I need to figure out dynamic...but that's where I get confused with the acceleration...
QuoteAlso, is acceleration the speed that the whole mass is moving (not just one leg)? or how fast the servos are moving?No, acceleration is not the same as speed.
force= mass * accelerationforce= .5kg * 4/s^2
force= .5kg * .04m/s^2force= .5kg * .0016
Would this torque unit be measured in Newton-Meters?
If so, then I'd just have to convert that into Kilogram-centimeters, then I could add it to my results for my static calculations..
Quoteforce= .5kg * .04m/s^2force= .5kg * .0016uhhhh what just happened?
force= .5kg * .04m/s^2
300g/2*3.8cm=300g/7.6=39.5 NewtonsConvert that to kilogram-force (kgf):39.5/9.80665=~4kgf
0.3/2 kg * .038 m * 9.81 m/s^2
Why would I need to multiply by 9.81m/s^2? Unless that's because of gravity.
Ahh...ok then, I get it!So it is .056kgf, or is it .056kgf-m?
Quote from: Canabots on May 30, 2009, 08:00:17 AMAhh...ok then, I get it!So it is .056kgf, or is it .056kgf-m?Well, you multiplied by meters, so you need to have meters in the final solution. Silly Always pass down the units with each calculation, don't try to guess the units after you do all the calculations. Another advantage of keeping track of units is that it acts as a 'dumb mistake' checker. If the units don't match what you expected to have, you messed up. Units also help let you know if you have a solution yet or not.For example, torque requires lb-in, but if you just have lbs, then you did something wrong.
I got 0.0057 kg*m, or 0.056 kgf*meters.0.3/2 kg * .038 m * 9.81 m/s^2
Torque : 1.6 kg-cm
The reason I say this is because QuoteTorque : 1.6 kg-cm doesn't make sense. Mass multiplied by distance isn't torque . . .You could email them, but doubt they'll know
does anyone know where to buy micro-servos with atleast 5kgf-cm of torque for a very low price?
If I only substituted out the servos that support the body, do you think that the robot would be a little awkward
weight = .42kg*2 (2 being the minimum number of legs on the ground)weight = .84kg