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/diy/ Do-It-Yourself

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Anonymous 2014-10-12 13:45:07 No.710017

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hey /diy/, please help me understand torques relevance for weight.
Let's say i have 4 electric motors that each have a torque rated at 7.8Kg/cm. The wheels are 9 cm in diameter.
how much weight would these motors pull?
The way i understand it it's 4*7.8/(9/2)=~7Kg
if that's true i'll be very sad.

Anonymous 2014-10-12 15:13:41 No.710043

Assuming the wheels are ideal and the friction coefficient is extremely high the force is equivalent to do leverage. Not sure if that equation is correct, haven't checked.

Anonymous 2014-10-12 15:27:25 No.710049

Anonymous 2014-10-12 16:05:50 No.710057
force is not weight
kg is mass, weight depends on gravity.
anyway you can find the force through the wheels easy enough, you seem to have managed.
from here you can calculate the acelleration of an object if you know its mass and the force applied with a = f / m.
It's been a while since high school physics but i think in theory any force will move any mass. in theory you have losses to overcome and at some point it stops moving. this is your coefficients of friction between materials and probably what you want to know.
to find the force required to drag an object you need to know the coefficient of friction between the object and the ground that you need to overcome to allow motion. if it has wheels you need to compound all of the various CoF in all the bits of bearings and things.

the easier way is to use a weighing scale (you get them for aeroplane luggage) and pull whatever you are trying to pull with it. the value on the scale (in kg's) will indicate the force you are looking to generate from your motors.

the key however is that the mass of the object is not the force you require to move it it.

one last thing, don't feel to comfortable relying on 4 motors giving 4 times the force, if one isn't gripping properly or slips then you are up shit creek.

Anonymous 2014-10-12 17:03:08 No.710080
>force is not weight
>kg is mass, weight depends on gravity.
Remember this, it will be useful to know when you travel somewhere where the gravity is different.

Anonymous 2014-10-12 18:38:28 No.710107
F = M*a
M = your car
F = Torque*Radius
a = insane value

Your car will have mad acceleration. 7Kgf is a lot of force that will be accelerating your car forward. If your car weighs 7kg, it will get to ~10m/s after 1 second.

Anonymous 2014-10-12 19:39:30 No.710135
This guy is right.

Kg/cm is not torque. Assuming that what it means is it can support 7.8kg at a radius of 1cm, the torque would be (7.8kg * 9.81m/s^2) * 0.01m = 0.76518nm

From that, the total force supplied by the wheels is 4 * (0.7652nm / (0.09m / 2)) = 68.016n

And then f = ma -> a = f / m -> a = 68.016n / 7kg = 9.7166m/s^2

Then you can use v = a * t + vi and x = 0.5 * a * t^2 + vi * t + xi to find positions and velocities.

Another thing to note, torque is inversely proportional to RPM. The torque given would have been the stall torque, where the motor isn't rotating at all. The specs should also list a "no load speed", which you can use with the stall torque to find the motors torque at any speed.

Anonymous 2014-10-13 03:41:38 No.710330
This thread is relevant to my interests because I suck at maths and there's nothing online that can do it for me (you'd think there'd be a java applet or something)

Lets say you have 100kg object on relatively small wheels and you want to drive at 5km/h up a medium gradient. What power or torque would be required?

Anonymous 2014-10-13 16:44:03 No.710530
>and there's nothing online that can do it for me
It's basic algebra. Anyone who graduated primary school should know it.

Anonymous 2014-10-14 01:39:15 No.710706
We didn't learn anything like this in the shit schools I went to.

100kg * 5 km/h = 500
500 / 10cm radius = 50

50n force? Don't know how to do it uphill though.

Anonymous 2014-10-14 04:37:25 No.710765
>100kg * 5 km/h = 500
>500 / 10cm radius = 50
You need to think about what units you're using.

5km/h = 1.39m/s
Suppose the slope is 20 degrees.
You're then climbing vertically at 1.39m/s * sin(20°) = 0.475m/s

Surface gravity is 9.81m/s^2
Your object has a weight of 100kg * 9.81m/s^2 = 981N

Lifting 981N at 0.475m/s requires 981N * 0.475m/s = 466W

So you need a 500 Watt motor.

Anonymous 2014-10-14 20:28:41 No.711000
Been in engineering too long for this stuff. It is one of those 'Wow, you don't know that?' moment. High school physics everyone!

Look up free body diagrams. There are three rules when dealing with FBDs.

1. Always keep units. If it is km/h*kg, your result is kg*km/h. 100km/(5km/h) = 20h. They never disappear and will always be there. Unless you get to dimensionless numbers like Reynolds number, but don't go there for a few years.

2. Conservation of Energy, Force, and Momentum. Three rules of thought. Energy is not created or destroyed. The sum of forces is equal to mass*acceleration and must be accounted for. Momentum is constant on an object with no forces applied to it.
Energy, Momentum, and Force have different equations, but they are usually integrals and derivatives of each other.
F= k*x
E = 1/2*k*x^2

3. Everything is accounted for. This falls back on the other two. If you don't know what you are doing, look at the units. Force is in N. Work is in Watts. Energy is in Jules. Look at the conversion equations below. If you have a number with a unit, find out where it will fit.

J = kg*m^2/s^2 = N*m = Pa*m^3 = W*s
W = J/s = N*m/s =kg*m^2/s^3
F = m*gravity

Hope that helps. It also assumes you know vector math and can sum up the vectors in the X and Y. If you don't, you need that too.

Anonymous 2014-10-15 01:04:46 No.711070
Thankyou kind mathemagicians!

Anonymous 2014-10-15 01:36:39 No.711078
Very much this. Still tho, in theory, any force with move any object, no matter how small the force or how large the mass of the object, it's just that the acceleration will be smaller the heavier the object is/the smaller the force.
The thing that limits your real life object from moving is frictions and if you ever had to really theoretically calculate frictions, you know it's a pain in the ass.
Being an engineer, maybe you had some analytical mechanics. I once had to solve a two body three dimensional Euler-Lagrange-equation system accounting for -linear- friction using a Rayleigh dissipation function for a homework assignment. That shit was a pain in the ass and it was a grossly simplified model of a real system.
Point being, as >>710057 said, get a scale and pull your load to just measure the force you'll need, take that number time 1.2 (or more, depending on the kind of surface it's gonna drive on) for a safety margin and use that. Calculating this shit is really just the worst for everyday needs (except if you're an engineer and know exactly how to do this. Being a physics student, I basically can't do anything that's not on a piece of paper)
Then again, serious internets to everyone in this thread. Would have expected a crapload of "hurr durr highschool physics you are stupid lrn2math omg" faggots.

Anonymous 2014-10-15 01:47:47 No.711082
There is a reason for this thread. Motion. keep it going.

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