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Drive train calculations.
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Daniel
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Joined: 30 Aug 2005
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Drive train calculations.

I've been thinking about how to find out the best drive train setup for a robot. I was reading on the american forums about how they calculate the best gear ratio for certain motors to get the best performance for the robot. Basically they don't want to gear the motors too hot. They don't list the calculations so I went and figured out what they were talking about.

Basically what they want is for the robot the start moving without the wheels skidding. This happens when you loose traction with the floor. If you put too much torque through the wheels they will loose grip with the ground and your robot won't go anywhere. This point is easy to find, you just need to find the firctional coefficent between your robots wheels and the floor. I guessed this to be 0.7, but I am most likely wrong.

So here is the calculations for a 100W scooter motors driven robot.

The max force the robot can exert before the wheels begin to slip is:
Force = (Mass of the robot x Gravity) x Frictional coefficient between wheels and floor
F = 13.6 x 9.81 x 0.7 = 93.3912N

Scooter come with 150mm wheels, so I'll work with that.
So the maximum torque you can output before the wheels slip is:
Max Torque = Radius of wheel (m) x Force
T = 0.075 x 93.3912 = 7Nm

Because we know the rated power of the motors we can find the velocity at this point. (this isn't acurate because the power curve of a motor isn't constant, but I can't be bothered with that detail)
Power = Torque x Radial Velocity (radians/sec)
Velocity = 200 (2 motors, 2 wheel drive) / 7 = 28.57 radians/sec

rpm = (rad/sec x 60)/(2 x pi)
28.57rad/sec = 272.83rpm

The 100W scooter motors put out 2300 rpm, but not when they are at 100W. I'll do the same as my axe calculations and halve that and say 1150rpm.
So the best gear ration for 100W scooter motors is:
1150rpm / 272.83rpm = 4.2:1

How does that sound?

272.83rpm on 150mm wheels = velocity?
m/sec = (pi x Diameter) x (rpm/60)
(pi x 0.15) x (272.83/60) = 2.14m/sec
= 7.7km/h
(now you think I'm wrong Razz )

Acceleration of the robot:
Acceleration = Force / mass
93.3912 / 13.6 = 6.687m/s^2

The distance it would take to get to top speed:
Velocity squared = 2 x acceleration x distance
V^2 / (2 x a) = S
(2.14^2) / (2 x 6.687) = 0.34m

Post Thu Feb 09, 2006 8:36 am 
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DumHed
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don't forget each wheel only has 1/4 of the bot's weight on it for 4 wheel drive or 1/2 if it's 2 wheel drive.
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Post Thu Feb 09, 2006 8:46 am 
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Daniel
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Ok, I took it one set further and went backward. Now I can find out what would be the maximum power you can have in a light weight drive train. Very Happy

Basically you are limited by the friction between the wheels and the ground and how far you have to get to your top speed. There is no point in getting the wheels to spin or to slam into the oppisite wall of the arena before getting to full speed.

The calcuations start off the same:


quote:
The max force the robot can exert before the wheels begin to slip is:
Force = (Mass of the robot x Gravity) x Frictional coefficient between wheels and floor
F = 13.6 x 9.81 x 0.7 = 93.3912N

Scooter come with 150mm wheels, so I'll work with that.
So the maximum torque you can output before the wheels slip is:
Max Torque = Radius of wheel (m) x Force
T = 0.075 x 93.3912 = 7Nm


Then we cut to the acceleration part:


quote:
Acceleration of the robot:
Acceleration = Force / mass
93.3912 / 13.6 = 6.687m/s^2


We can find the maxium velocity a robot would hit after traveling the full distance of the arena. Lets say an arena is 6m across.

Velocity squared = 2 x acceleration x distance
V^2 = 2 x 6.687 x 6
Velocity = 8.96m/s or 32.3km/h

We can then find the speed of the wheels:
pi x Diameter = 0.47m
Rev/sec = 8.96m/sec / 0.47m = 19.06 rev/sec or 1143.8rpm
19.06 rev/sec = 119.76 red/sec

So the maximum power you can have in a feather drive train in a 6m arena with 0.7 friction is:
Power = angular velocity x torque
119.76 rad/sec x 7Nm = 838.3 W

So you can have 2 300W scooter motors, but not the 400W motors. Very Happy

Post Thu Feb 09, 2006 8:49 am 
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Daniel
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quote:
Originally posted by DumHed:
don't forget each wheel only has 1/4 of the bot's weight on it for 4 wheel drive or 1/2 if it's 2 wheel drive.


That won't effect the friction usless 2 of the 4 wheels arn't powered. This is all refering to rammers which rely of getting the most power to the ground.

Weight distrabution won't effect it if all the wheels are powered.

Post Thu Feb 09, 2006 8:51 am 
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Woody



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I've no idea if these links have been posted before .....so.... apologies in advance if they have.

http://www.architeuthis-dux.net/torquecalc.asp

http://www.4qd.co.uk/faq/current.html

Regards Woody

Post Thu Feb 09, 2006 9:02 am 
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DumHed
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[quote="Daniel"]

quote:
Originally posted by DumHed:
That won't effect the friction usless 2 of the 4 wheels arn't powered. This is all refering to rammers which rely of getting the most power to the ground.
Weight distrabution won't effect it if all the wheels are powered.


No, but usually each wheel is powered by its own motor, so the power output per wheel depends on how many wheels you have, and how much of the weight is on them.

A 4WD bot under acceleration will have more weight on the rear wheels than the front, so it will be able to put more power to the ground at the back wheels than the front wheels.
This is assuming that the center of thrust is below the center of mass (which is is unless you have somehow made a bot that drives under the floor!)
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Post Thu Feb 09, 2006 11:28 am 
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Daniel
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The force acting down on the robot is always the same, no matter how it is orrientated or accelerating. If it is taking off and there is more force acting down on the rear wheels (assuming this is a 4 wheel drive, 4 motor robot) then the rear wheels will have more load. They motors will be working harder then the front motors and would therefore be out putting more torque and less speed then the front motors. However the average down force on all four wheels is still the same as when the robot is at rest and the average output onto the ground from all four wheels, as asumed in the calculations, will be the same as when the intial take off when all four wheel will have the same down force.
Wheel slipage does not occour when the robot is moving. It is when the load is first applied to the wheels. Or when the robot is changing acceleration, eg forward to reverse. At this point the load of the wheels is the same and all 4 wheel will slip.
As with all calculations we are asuming ideal cercumstances, the load is distrabuted evenly. The calculation are also taken from a the robot moving from a standing start, hence the no initial velocity in the distance equations.

Post Thu Feb 09, 2006 12:19 pm 
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DumHed
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I think there are a few too many variables involved (motor efficiency / torque curve, different surfaces, etc)

Either way, if you have a 4WD bot, and give each motor the maximum amount of power it can put to the ground without wheel spin, the front wheels will spin, as weight is tranferrered to the rear wheels instantly.
Both motors will be making the same power, as they're starting from stall and are given the same input wattage.
The front motors will reduce torque once they speed up, but the wheels will already be spinning.

In cars it's referred to as geometric weight transfer, because it's a byproduct of the car's geometry.
This is why a car can be balanced mid corner by keeping the steering in the same place and varying the throttle input.
Changing the torque going to the wheels instantly changes the amount of load each tyre has on it, and hence its grip level.

Also don't forget that most rammer / wedge bots will also need to push other bots around, so designing around the maximum power that can be put to the ground under its own weight could leave you with an underpowered bot when you have another one sitting on top of it.
You'd be better off have a dual preset system for power, so you can drive around with power limited normally, and then go to maximum power when you need it (or can use it)
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Post Thu Feb 09, 2006 12:34 pm 
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Valen
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like say a *ram button* muwhahahaha ;->
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Post Thu Feb 09, 2006 4:13 pm 
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DumHed
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you weren't getting enough power to spin the wheels anyway Razz
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Post Thu Feb 09, 2006 4:40 pm 
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kkeerroo
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quote:
Originally posted by DumHed:
In cars it's referred to as geometric weight transfer, because it's a byproduct of the car's geometry.


Cars have suspension and are therfor a "live" mass (the center of gravity moves as the force acting on the body changes) while a robot general has no suspension and therfor is a "dead weight". So my thinking is that the center of gravity on a robot will no move over the rear wheels on initial acceleration of a 4wd robot.
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Post Thu Feb 09, 2006 4:45 pm 
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DumHed
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cars have weight transfer due to the suspension movement, but they also have the geometric weight transfer.

Put it this way:

Assume a square bot with 4 wheels, one at each corner, and only the rear wheels driven.
If it has enough grip and power on takeoff, it will do a "wheelie" due to geometric weight transfer - or in other words, the lever action between the axle line and the plane of thrust (being the floor).
Having the front wheels driven doesn't change the weight transfer - they'll still be in the air if the rears have enough grip, and all weight will be transferred to the rear.
This obviously means the rears get more traction, and can put more power to the ground, but only under the right conditions.
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Post Thu Feb 09, 2006 4:54 pm 
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kkeerroo
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quote:
Originally posted by DumHed:
but only under the right conditions.


Exactly. Only after the initial acceleration which is what this topic is about.


Boy, am I in a bad mood today.
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Post Thu Feb 09, 2006 5:05 pm 
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DumHed
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I mean the right conditions as in floor, wheel type, power available.

The rear will always get more grip than the front under acceleration.

What do you define as "initial" acceleration?
As soon as *any* torque (for a forward direction) is applied by the wheels, there is a weight shift to the rear.
It's instantaneous, and so are the effects.

Even if it's front wheel drive only, the lever action still works in the same way, but is more noticable as weight coming off the front wheels rather than weight going on the back wheels.

Drag cars have basically no suspenion, but front wheel drives are always slower than rear wheel drives - even though they tend to have more weight over the front wheels.
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Post Thu Feb 09, 2006 5:14 pm 
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Spockie-Tech
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Actually, most Pro Drag cars have *very* sophisticated suspension. Its just not designed for going over bumps comfortably like "normal" suspension is.

4-Links, Ladder-Bars and quite a few other types of precisely engineered linkages are used in the rear end of a drag car to turn the rotational torque reaction along the rear axle axis into downforce to press the tyres against the track.

Differential-Rate shock absorbers allow the front of the car to rear up easily, and then settle back slowly assist in weight transfer to the rear of the car (where the power is being applied), the "crinkle" effect of the drag-slicks tyre sidewalls is carefully considered and a lot of other factors I probably dont know about.

I have seen some of the simulation software available for the many configurations of drag-car suspension. It is very involved and takes into account linkage lengths, unsprung mass vs sprung mass, polar moments of inertia for the rotating masses and a million other factors. Your average 14-second street drag car might not worry too much about this sort of thing, but once you get up into the sub-9 second brackets, it really turns into rocket science.

Anyway, thats all about *acceleration* which is different to static-power output that mr Kero is discussing, but dont get the idea that putting 1000+ horpsepower to the ground in 8 seconds is simple just because the cars go in a straight line.. Wink

Have you tinkered with EDTSim to see what factors it takes into account in this area ?
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Last edited by Spockie-Tech on Thu Feb 09, 2006 7:29 pm; edited 1 time in total

Post Thu Feb 09, 2006 7:01 pm 
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