Finding velocity/time graph for a car?

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Discussion Overview

The discussion revolves around deriving a velocity versus time graph for a car under ideal conditions, taking into account factors such as air resistance and rolling resistance. Participants explore the relationship between power, force, and velocity, and how these concepts apply to the dynamics of the car being fixed.

Discussion Character

  • Technical explanation
  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • One participant describes their approach to finding a velocity versus time graph using a power versus velocity graph derived from engine data, incorporating gear ratios and tire radius.
  • Another participant corrects the initial claim that the force from the engine is the derivative of power with respect to velocity, stating instead that it is power divided by velocity.
  • A participant questions the relationship between force and power, seeking clarification on why the force of an engine would be expressed as power over velocity.
  • Several participants discuss the implications of varying force, suggesting methods to account for this in the equations.
  • One participant expresses confusion over a calculated force value, questioning its reasonableness given the car's specifications.
  • Another participant notes that 160 horsepower can indeed produce the calculated force, but only at a low speed of 14 mph.

Areas of Agreement / Disagreement

Participants do not reach a consensus on the correct interpretation of the relationship between power and force, and there are varying views on how to account for non-constant forces in the equations. The discussion remains unresolved regarding the calculations and interpretations presented.

Contextual Notes

There are limitations in the assumptions made regarding the constancy of forces and the specific conditions under which the equations apply. The discussion also highlights the complexity of deriving accurate models for vehicle dynamics.

andrewr0x
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I have been recently playing around with some figures in an attempt to find a velocity versus time graph under ideal conditions including air resistance and rolling resistance for a car that I am in the process of fixing. What better incentive to get it running than knowing (at least a rough estimate) of how fast it will be? However, I have run into some problems.

I have Approximated a Power versus velocity graph based on a graph that I got from finding a cubic regression from a dynamometer chart of the engine in the car. I then altered tha graph so that instead of having power versus rpm of the flywheel to power versus velocity given the gear ratios and radius of the tire.
v=(2pi*x/60)/(Ratio of gear * Ratio of Differential)*Radius of Tire
where v is velocity and x is the RPM of the engine.

Once this has been found, I determined that the derivative of P(v) would equal F(v) due to the engine. I also looked up the formula for F(v) of air resistance to be (1/2)*Coefficient of drag*Frontal area*Air density*velocity^2. Rolling resistance's would be Coefficient of rolling friction*Normal.

From this I get a differential equation that m(dv/dt)=Fengine(v)-Fair(v)-Ffriction(v).

However, upon solving this, I get a result that does not make sense at all. Could someone verify that I am doing this correctly, and if not, give me some instructions on how to do this the correct way? Any and all help is appreciated.
 
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The force given by engine is not the derivative dP/dv. It is P/v.
 
krab said:
The force given by engine is not the derivative dP/dv. It is P/v.
But I thought that the integral of Force versus Velocity gives you Power?

Can you explain why the force of an engine would be power/velocity?
 
[tex]P = \frac{dW}{dt}[/tex]

[tex]F = \frac{dW}{dx}[/tex] so [tex]dW = F dx[/tex]

then

[tex]P = \frac{F dx}{dt}[/tex]

If F is assumed to be constant it simplifies to the very nice

[tex]P = F \frac{dx}{dt} = Fv[/tex]
 
The thing about this is that the force of my engine is not constant. So I need a way to allow for this in my equations
 
If the force is varying with time, just differentiate the second last expression, if it is varying with some other variable, then use the chain rule to find the net rate of change WRT to time.
 
whozum said:
[tex]P = \frac{dW}{dt}[/tex]

[tex]F = \frac{dW}{dx}[/tex] so [tex]dW = F dx[/tex]

then

[tex]P = \frac{F dx}{dt}[/tex]

If F is assumed to be constant it simplifies to the very nice

[tex]P = F \frac{dx}{dt} = Fv[/tex]
That's all true except the last sentence. F needn't be constant.
[tex]P = \frac{F dx}{dt} = F \frac{dx}{dt} = Fv[/tex]
 
Thank you guys, I will try and play around with that formula. I was playing around in Derive (very nice once you get the hang of it) and I got a velocity curve that made no sense at all. After thinking about that I remember reviewing that Power = Force * Average Velocity in my Physics C class.
 
  • #10
I am getting a force of 19k Newtons. This seems unreasonably high to be pulling about 1.4 G's in a 3000 pound car with only 160 hp to the wheels. What would cause this?
 
  • #11
160hp can get you this force, but at only 14 mph.
 

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