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Heavy Mathematical magic - Calculating Acceleration from Torque

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Okay, so I have a wheel of radius r (Meters), attached to a vehicle of weight M (Kilograms). If I apply a constant torque T (Newton-Meters), how much acceleration do I get?

My shaky math comes out to:

Inertia I=W*r²
Acceleration = T/I (In Radians/sec)
Therefore Acceleration = (T/I)*r (In Meters/second²)

Right?
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Okay, so I have a wheel of radius r (Meters), attached to a vehicle of weight M (Kilograms). If I apply a constant torque T (Newton-Meters), how much acceleration do I get?

My shaky math comes out to:

Inertia I=W*r²
Acceleration = T/I (In Radians/sec)
Therefore Acceleration = (T/I)*r (In Meters/second²)

Right?

there's lots more involved.... real wheels are not solid discs which means that the 'r' for moment of inertia is not the rim. ;) and then you are assuming that there is nothing attached to the wheel? you'd have to figure ALL the rotating weight you are accelerating AND the F=ma for accelerating the non-rotating weight thru space.
 

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Discussion Starter · #3 ·
I was trying to simplify things, and just assuming you had to move what was effectively a M-massed weight on the end of a r-length arm (Effectively, the force necessary to move the wheel against the ground). Essentially I need to calculate the force created by the motion of the wheel against the ground, and use that as the F in an F=Ma (Rearranged to solve for a, so F/M=a). Yay, more equations!

Could I use the size of the wheel as a ersatz "Gear", and if so, how would I calculate it's ratio?
 

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The wheel torque is equal to the cross product of the force the wheels apply to the road (tangential to the wheel contact area) and the wheel radius, here simply equal the product r*F since the angle between r and F is 90 degree. The force F = m*a = T/r where m is the mass of the vehicle and a is its acceleration, so a = T/r*m. The product of this force and the distance the vehicle moves is the work done by the motor to move the vehicle against the drag and rolling resistance forces. It does not include the work done to increase the energy of rotating parts, or losses in the motor, controller, batteries, and friction in the drive train. The wheel torque equals the product of motor torque and overall gear ratio (trans plus rear end).
 

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The wheel torque is equal to the cross product of the force the wheels apply to the road (tangential to the wheel contact area) and the wheel radius, here simply equal the product r*F since the angle between r and F is 90 degree. The force F = m*a = T/r where m is the mass of the vehicle and a is its acceleration, so a = T/r*m. The product of this force and the distance the vehicle moves is the work done by the motor to move the vehicle against the drag and rolling resistance forces. It does not include the work done to increase the energy of rotating parts, or losses in the motor, controller, batteries, and friction in the drive train. The wheel torque equals the product of motor torque and overall gear ratio (trans plus rear end).

this all handles the non-rotating parts.... I was just trying to point out that to be accurate you would need to find moments of inertia for all the rotating parts too to predict the acceleration. Not insignificant with the wheels plus brake rotors, all the gears/clutch/flywheel in the drivetrain, and the guts of the motor itself.
 

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Discussion Starter · #6 ·
The wheel torque is equal to the cross product of the force the wheels apply to the road (tangential to the wheel contact area) and the wheel radius, here simply equal the product r*F since the angle between r and F is 90 degree. The force F = m*a = T/r where m is the mass of the vehicle and a is its acceleration, so a = T/r*m. The product of this force and the distance the vehicle moves is the work done by the motor to move the vehicle against the drag and rolling resistance forces. It does not include the work done to increase the energy of rotating parts, or losses in the motor, controller, batteries, and friction in the drive train. The wheel torque equals the product of motor torque and overall gear ratio (trans plus rear end).
Okay, let's throw some numbers in here.

Example car: 2010 Chevy Malibu, stock weight, with a Warp9 in direct-drive through a 4:1 differential.

Vehicle weight: 1556KG
Wheel radius: 0.3345m
Motor torque: 321N-m

So, Final torque at-the-wheel is 1284N-m (4x mechanical advantage from the differential, 1/4 the speed. Right?)
acceleration = 1284/(0.3345 * 1556) = 2.4669. If that's in Radians/sec, then acceleration = 2.4669...*0.3345 = 0.82519 m/s^2

Either way, that's not much acceleration at all. 0.25G? Definitely not going to be winning any drag races with that. So where's my math wrong?
 

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That's 2.47 N/kg or meters/second squared, which is 5.52 mph/second, or a bit under 11 seconds 0 to 60 mph - if you maintain that value of torque, which of course is not constant over the entire rpm range for either an ICE or electric motor/controller. The torque falls off at higher rpm for the electric motor/controller, that be the problem.
 

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Discussion Starter · #9 ·

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Well, here's the end result: http://evthoughts.blogspot.com/2011/10/vehicle-speed-calculator-mark-2now-with.html

Do let me know if you spot any problems. I'm sorry about the Javascript being all on one line, but apparently Blogger loves to insert br tags everywhere. :/
nice little thing : )
just add fields filling exaples, note or comment in rpm field: torque up to given rpm - constant, also may be a title - calculated theoretical constant acceleration; constant torque model as opposite to contant power (ideal CVT model), where torque and acceleration decreases wile speed increases

may be specify (to be clear) - accel calc. w/o taking into account resistance (wind, rolling)
 

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That's 2.47 N/kg or meters/second squared, which is 5.52 mph/second, or a bit under 11 seconds 0 to 60 mph - if you maintain that value of torque, which of course is not constant over the entire rpm range for either an ICE or electric motor/controller. The torque falls off at higher rpm for the electric motor/controller, that be the problem.
Dear Sir,

but isnt this equation bit deceptive. It depicts that if one increases the wheel torque to a very large number by using high motor torque and gear ratio, the acceleration would be enormous. Practically such a high acc would cause tyres to skid. Is there way to calculate max allowable acceleration considering the point after which the tyre skids?
 

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wow, resurrected after 6 years :)

I think people are a little more conscious of the limits of tire traction these days, hopefully. It is damn nice to have limits when you are designing a system too!
 
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