Equation to describe vehicle position versus time

[BronYrAur]

Anyone know if there is a theoretical equation to explain a vehicle's position versus time at full throttle? I'm guessing there are too many variables with tires, wind, etc.

I have plotted several drag race data points on a position versus time graph and curve fit the data. A 5th or 6th degree polynomial seems to be working the best to match the empirical data. The problem is that the typical 1/4 mile drag race only reports time at 5 locations (60', 330', 660', 1000', and 1320'). It also reports velocity at 2 locations (660' and 1320'). When I take the first derivative of the position equation in orderto get a velocity equation, I can't get the reported velocities to match up exactly. I am usually 1 or 2 MPH off. I am trying to use this equation to determine the 0-60mph time.

Now I realize that the instruments may not be calibrated correctly and I also know that with only 5 position data points, my curve fitted equation has some error. So I'm wondering if there is a theoretical equation that is supposed to explain position vs time. If so, I would like to see how well it stacks up to my empirical data.

I am specifically interested in a Dodge Magnum.

 

[IRstuff]

What do the times look like? Your polynomial order is way to large, as this will cause the curve to pass directly through each data point regardless of any noise in the data.

You might consider doing a spline fit over the segment where you hit 60 mph and interpolate.

TTFN

 

[MintJulep]

This should get you started.

http://www.uwm.edu/~horowitz/PropulsionResistance.html

 

[patprimmer]

Gear changes might upset the calculation also.

Regards

eng-tips, by professional engineers for professional engineers
Please see FAQ731-376 for tips on how to make the best use of Eng-Tips Fora.

 

[Greenlight]

Regarding the two MPH data points associated with a typical 1/4 mile timing system, please remember that the velocity timers for the 1/8 mile are placed 66' before the 1/8 mile marker and at the 1/8 mile marker. The MPH is determined NOT at the 1/8 mile lights, but averaged from the two readings. The same at the 1/4 mile. So, a reasonable assumption would be that the MPH shown on the time slip is really the MPH 33' prior to that particular area of the track (i.e. @627' and 1287').

 

[BronYrAur]

I had specifically chosen such large order polynominals because they matched my data directly. Maybe that wasn't such a good idea. Here is a typical run:

60' 2.004 sec
330' 5.531 sec
660' 8.383 sec 86.13 MPH
1000' 10.834 sec
1320' 12.908 sec 108.55 MPH

 

[BronYrAur]

That last set of data I posted seemed to fit best to a 6th deg poly. The equation I came up with is:

6th Degree Polynomial Fit: y=a+bx+cx^2+dx^3...
Coefficient Data:
a = 6.7290643e-013
b = 11.171725
c = 9.2796363
d = 0.16608009
e = -0.071281314
f = 0.0051347964
g = -0.00012727154

The positions matched exactly. Taking the 1st derivative, the velocities are as follows:

8.838 sec 87.94 MPH versus the reported 86.13 MPH
12.908 sec 108.84 MPH versus the reported 108.55 MPH
The 0-60MPH time then comes out to be 4.506 sec.

Greenlight's comments may actually bring the MPH’s a little more in line. Otherwise, I think this is a pretty good fit considering that I only have 5 data points to start with. Any other ideas?

 

[MacGyverS2000]

...and if you had 50 data points, a 49th-order equation would fit it best, but it wouldn't tell you anything useful about the data itself.

Neglecting shift points, acceleration between shift points should be (roughly) increasing monotonically. A 2nd-order should fit those data points nicely. When you include the shift points, acceleration will drop off sharply for a short time, so you'll have some discontinuities that a 2nd-order won't be able to handle with much grace.


Dan - Owner

http://www.Hi-TecDesigns.com

 

[IRstuff]

Is your equation time as a function of velocity of position or position as a function of time?

In any case, the prediction is poor because the discontinuities are not captured in the data. You have a linear term with a relatively large coefficient. This represents an offset in the derivative, which can't possible exist in reality, since you start at t=0 and v=0

I've played around a bit with your data and there doesn't seem to be any way around that minor difficulty, so there isn't any simple polynomial that can accurately capture what you want with so few datapoints.

I would suggest using a spline interpolation and leave it at that. Without using some other instrument with more resolution, you can't get there from here.

TTFN

 

[MechanicalChef]

This sounds bench racing for smart-people.