# Draw Map Based on Speed and G Forces

1. Aug 2, 2011

### gggggggg

Hi All, I am trying to draw a map, the raw data I have is Speed and G_Force (Lat and Acce).

I have attached a XLS of the RAW data I have. Its from a race track, (I had to trim the file to 100kb, so it might or might not loop). I dont have a GPS, but I have seen software draw maps based on this data, so I know its possible, even though it might not be super accurate, it will help.

I have no idea where to start. Are you able to help please.

thanks
greg

File size:
76 KB
Views:
162
2. Aug 5, 2011

### gggggggg

I am not having much luck, though believe it can be done with just Lateral G force and Speed.

3. Aug 5, 2011

4. Aug 5, 2011

### xxChrisxx

This sort of thing isn't super accurate, as you have to assume that your driver is going flat out.

But it's like a skid pad. If you know the speed and the lateral g force, you can calcualte the radius of the corner and use speed to calculate the arc length (ie distance travelled).

5. Aug 5, 2011

### gggggggg

THat sounds like what I am after. NOt super accurate is not important. So the question is, how please?

6. Aug 5, 2011

### xxChrisxx

Doing it manually is going to be a pita, I'm not really sure how you could plot the answers graphically.

7. Aug 5, 2011

### gggggggg

I was thinking of something like.
1) If the course is a lap, then assume all Lateral G Force have to be equal to 360 degrees.
2) Break each tenth of a second down into little lenghts
3) THey each go off in the direction of the Lat G Force divide by the ratio of 1).
4) Use the distance from 2) as the hypotanuse and the angle from 3) to work out the oppersite and adjacent
5) Plot that point on a graph
6) Do the same for the next value, though making sure you are pointing in the right direction to start with

Maybe anyway, because I cant get it to look anything like the map I have.

8. Aug 5, 2011

### jack action

I would go this way (all in SI units):

find the radius of the curve:
R = V² / aL

find how many radians the car has turned:
dA = V / R dt

find the new direction the car is heading:
A = A0 + dA

find the X and Y components of the velocity:
Vx = V sin A0
Vy = V cos A0

find the displacements and final positions:
X = X0 + Vxdt
Y = Y0 + Vydt

Repeat with new values.

I did it for your data and it looks like a car spinning around in a 300mX300m parking lot. I don't know if it is what it's supposed to look like, but your data for acceleration don't seem to be very accurate as your «accel G» and velocity increase don't seem to agree (maybe I don't use the proper units?).

9. Aug 8, 2011

### gggggggg

THanks for this Jack, but it went right over my head. Can I get a copy of the XLS please? Or the above in PHP or Visual Basic, or even something a little more readable for my brain?

thanks
greg

10. Aug 9, 2011

### jack action

The next drawing shows the trajectory followed by the vehicle (in blue) within time dt, from position 1 to position 2 (click on the image to enlarge):

You can see the origin 0,0 which is the reference point for X,Y data. There is also the radius R and the velocity V, which is perpendicular to R. The radius (R) is related to the velocity (V) and the lateral acceleration (aL) by definition.

The angular displacement dA can be found with the angular velocity of the vehicle (w) with respect to the instantaneous center of rotation where w = dA/dt but also w = V/R.

The velocities of the vehicle are by definition Vx = (X-X0)/dt and Vy = (Y-Y0)/dt.

All of this leads to the previous equations.

As you can see in the drawing, the radius R changes from one position to another, but the instantaneous center of rotation too. So the longer is dt, the greater is the error.