Why is my car acceleration not matching the calculated force?

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Discussion Overview

The discussion revolves around the discrepancy between the calculated force required for a car's acceleration and the force derived from the car's torque specifications. Participants explore concepts of Newtonian physics, torque, gear ratios, and the implications of drivetrain mechanics on acceleration.

Discussion Character

  • Exploratory
  • Technical explanation
  • Debate/contested

Main Points Raised

  • Marius calculates the force needed for his car to accelerate and finds a significant difference between this force and the force derived from the car's torque.
  • Some participants suggest that gear ratios must be considered, as they can increase torque and affect acceleration.
  • Marius questions whether he should multiply the gear ratio with the final drive ratio to find the effective force.
  • Another participant explains that the gearbox allows for a higher overall gear ratio when starting, which is crucial for preventing engine stalling during acceleration.
  • There is a suggestion that the analysis presented by Marius may not be valid if acceleration occurs through multiple gears.

Areas of Agreement / Disagreement

Participants do not reach a consensus on the correct approach to calculating the effective force during acceleration, with multiple viewpoints on the role of gear ratios and the validity of Marius's initial calculations.

Contextual Notes

Participants note that Marius's calculations do not account for various factors such as gear ratios, which could significantly alter the effective torque and force during acceleration.

Jonsson
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Hello there,

Just got an old car, and I was looking at the specifications, and trying to teach myself Newtonian physics. But I cannot get it to work out correctly. Please tell me where I go wrong.

The car weighs 900 kg. The engine has 200nm of torque, it accelerates from still to 27.7 m/s in 8 seconds. The radius from the center of the drive axle to the outside of the tire is circa 0.3m.

[itex]F = m\,a[/itex]

Assuming constant acceleration:

[itex]F = m\frac{v}{t}[/itex]

[itex]F = 900\mathrm{ kg}\frac{27.7\mathrm{ m/s}}{8\mathrm{ s}}[/itex]

[itex]F = 3116\,\mathrm{ N}[/itex]

So, to go from zero to 100km/h in 8 seconds takes 3116 N. This is not taking aerodynamic drag and drivetrain friction into account.

This is where I go wrong I think. The car has 200 Nm of torque:

[itex]\tau = 200\,\mathrm{Nm}[/itex]

I have an arm of 0.3m (the distance from the centre of the wheel to the outside of the tyre)

[itex]F = \frac{\tau}{r}[/itex]

[itex]F = 666 N[/itex]

The problem is [itex]3116\,\mathrm{ N} >> 666\,\mathrm{ N}[/itex]

And this is without even taken aerodynamic drag and drivetrain losses/friction and all of that jazz into account.

How come my maths is off by so much?

Thank you for your time.

Kind regards,
Marius

NOTE: Looks like my numbers may be off by a factor of 2∏ ? Not sure where it should go though.
 
Last edited:
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You have to take the gear into account, it can increase the torque (and reduce the rotation frequency).

Quick check: a typical wheel has a diameter of ~50cm. At 30m/s, it rotates with 1200rpm. That is not the range where you operate the engine to accelerate.
 
These are the gear ratios of the car:


1st 3.80 : 1
2nd 2.06 : 1
3rd 1.26 : 1
4th .89 : 1

final drive ratio 4.125 : 1

Do you think I am right in multiplying the ratio of the gear with the differential final drive ratio?

F = 200Nm * gear ratio * final_drive_ratio/0.3m

Thank you for your time.

Kind regards,
Marius
 
That's what a gearbox is for. You have a higher overall gear ratio when starting in first gear to keep the engine from stalling. Once the car is moving, you change gears to reduce the overall ratio until you reach the point where the wheels are rotating at a speed which won't cause the engine to stall, and the engine is turning at sufficient RPM to keep the car moving at a constant travel speed.

When you are accelerating to speed, unless you do it in a single gear, the analysis in the OP is not valid.
 

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