Calculating max acceleration of a Tesla car

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barryj
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Homework Statement
Given:
Tesla car weight = 2083 kg
max torque = 660 n-m
how long to accelerate to 27 m/sec (60 mph)
Relevant Equations
F = M x A
T = F x R
V = A X T
[Mentor Note: Two threads with similar titles merged]

First calculate the equivalent force on the car from the max torque.
An average tire is 15 inch diameter = about .2 meter radius

Since T = F X R , F = T/R = 660/.2 = 3300 Newtons.
since F = M X A, A = F/M = 3300/2083 = 1.6 m/sec^2
since V = A X T, T = V/A = 27/1.6 = 16.8 seconds.

However, Tesla claims that their car goes from 0 - 27 m/s (6o mph) in 2.4 seconds.

Where is the error?
 
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Calculate the force on the car based on a 15 inch (.4 meter) diameter tire, r = about .2 meters
since T = f x r, f = T / r = 660/.2 = 3300 Newtons

since a = f/m, a = 3300/2083 = 1.58 m/sec^2

60 mph = about 27 m/sec
Since V = a x t, t = V/a = 27/1.58 = 17 seconds.

However, Tesla claims their car will accelerate 0 - 60 mph in 2.4 seconds.
17 vs 2.4 is a big difference.
Where is my error?
 
Attached is the Tesla spec sheet. I used 660 n-m (488 ft-lb) not a big difference to explain the discrepency.
 

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barryj said:
since T = f x r, f = T / r = 660/.2 = 3300 Newtons
.
Where is my error?
Here's a (only semi-educated) guess.

Although electric cars don't have a conventional gearbox, there is some gearing to reduce the high rotational speeds of the electric motor to the lower rotational speeds of the road-wheels.

If the given torque (660Nm) is the motor's torque, then (because of the gearing) the torque delivered to the road-wheels is bigger. You need to include the gearing ratio into the calculation.

And note that using a single torque value isn't very accurate. Torque varies with engine speed to some degree, e.g. https://images.theconversation.com/...0&q=45&auto=format&w=600&h=360&fit=crop&dpr=1

An accurate calculation would take this into account.
 
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BvU said:
Which one has 2.4 s and 660 N.m ?
How many wheels are driven in your Tesla ?

Is that the rim or the tire ?
I am assuming a 15 inch diameter tire. This is a .381 meter diameter and I use a radius of .2 meters.
 
barryj said:
Attached is the Tesla spec sheet. I used 660 n-m (488 ft-lb) not a big difference to explain the discrepency.
There is no way around the second law of Newton.
In order to accelerate a mass of 2083 kg at a rate of 11.175 m/s for each of the 2.4 seconds that the test last, a force of 23278 Newtons needs to be applied.
For a 4-wheel drive car, with assumed 35 front/65 rear weight distribution under acceleration, that means tangential forces of 4074 Newtons for each front tire and 7565 Newtons for each rear tire.

If we consider the around 9:1 torque multiplication of the motor's gears for the model S, we get close to those numbers.

Copied from:
https://www.quora.com/What-sort-of-...curve-Is-the-magic-in-the-controller-or-motor

"Tesla Model S Rear Drive Unit (eBay) The 3-phase 4-pole induction motor (upper right), inverter drive electronics (left), 9.73:1 reduction gearing and rear differential (center) are assembled in a single oil-filled rear Drive Unit. The rear wheels are driven directly by this unit, there’s no clutch or transmission (no shifting gears or “Neutral”, run the motor “forward” for Drive and “backwards” for Reverse). Power (~400Vdc) flows from the battery pack via the two heavy orange cables to the inverter where it’s converted to 3-phase ac.

Model S All-wheel Drive versions feature a similar Front Drive Unit with a second induction motor and 8.28:1 reduction gearing that directly drives the front wheels. The lower reduction ratio (compared to the rear’s 9.73:1 ratio) reportedly helps overall vehicle acceleration/power/efficiency/range."
 
Lnewqban said:
There is no way around the second law of Newton.
In order to accelerate a mass of 2083 kg at a rate of 11.175 m/s for each of the 2.4 seconds that the test last, a force of 23278 Newtons needs to be applied.
For a 4-wheel drive car, with assumed 35 front/65 rear weight distribution under acceleration, that means tangential forces of 4074 Newtons for each front tire and 7565 Newtons for each rear tire.

If we consider the around 9:1 torque multiplication of the motor's gears for the model S, we get close to those numbers.

Copied from:
https://www.quora.com/What-sort-of-...curve-Is-the-magic-in-the-controller-or-motor

"Tesla Model S Rear Drive Unit (eBay) The 3-phase 4-pole induction motor (upper right), inverter drive electronics (left), 9.73:1 reduction gearing and rear differential (center) are assembled in a single oil-filled rear Drive Unit. The rear wheels are driven directly by this unit, there’s no clutch or transmission (no shifting gears or “Neutral”, run the motor “forward” for Drive and “backwards” for Reverse). Power (~400Vdc) flows from the battery pack via the two heavy orange cables to the inverter where it’s converted to 3-phase ac.

Model S All-wheel Drive versions feature a similar Front Drive Unit with a second induction motor and 8.28:1 reduction gearing that directly drives the front wheels. The lower reduction ratio (compared to the rear’s 9.73:1 ratio) reportedly helps overall vehicle acceleration/power/efficiency/range."
Ah, a very good explanation.
I thought the torque was at the wheels.

Thanks
 
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