Hello!
I have a real tyre model and all it's Pacejka's MF5.2 coefficients, and I am trying to put this model into rFactor1.
I've found a lot of information about the parameters contained in the TBC file but unfortunately 90% is either inaccurate or useless information. What I've found so far:
And as I understand, combining the SlipCurve with the variables DryLatLong, LatPeak, LongPeak and the GripMultiplier curve (which has as input the variable LoadSens), one should arrive on the same curves from a real tyre model.
But I can't find any information about how this is done.
For example, the variables LatPeak and LongPeak I've compiled the following explanations:
Slip range where lateral/longitudinal peak force occurs depending on load. Defines a curve where the load on the tire is along the x-axis and the y-axis is the sine of the slip angle where peak forces occur.
Since long goes by slip percent not angle this is simple: 0.015 = 1.5 percent slip and 0.035 = 3.5 percent.
Example1: (0.0817, 0.233, 13825.0)
At zero load, the sine of this slip angle is 0.0817, so the slip angle at which maximum lateral force is generated is asin(0.0817) = 4.7 deg. At a load on the tire of 13825 N, the sine of the slip angle is then the second value, or 0.233, so the angle is 13.5 deg.
Well, the example leads to two points (x y): (0 0.0817) and (0.233 13825). With two points one can make a line, not a curve. But anyway, no idea what I can do with this afterwards.
Example2: (0.1,0.3,15000)
The peak of the slip curve at 0 load will now be at 0.1, or asin(0.1) = 5.74 deg. The entire slip curve will be moved according to this value. If peak grip is acheived at 0.18 (10.37 degs) according to the slip curve, this will mean the entire slip curve slip step is multiplied by 0.1/0.18 = 0.555 rad. You will now be able to slide approximately, 0.555x as much as the slip curve would normally allow.
Now this example suggests the following calculation, that uses only the first value of three from the variable. So still not clear.
SlipCurve = SlipCurve * (LatPeak(1stValue) / (SlipCurve's SlipAngle for maximum value))
Does anyone have a guide of how to make this calculations and make a model, with units and real equations?
Thanks!
I have a real tyre model and all it's Pacejka's MF5.2 coefficients, and I am trying to put this model into rFactor1.
I've found a lot of information about the parameters contained in the TBC file but unfortunately 90% is either inaccurate or useless information. What I've found so far:
- Bristow's TBC File Bible;
- Kangaloosh CarFactory help file;
- http://yoss.free.fr/rfactornfr/Modding_Tutorial_Website/PhysicsGlossary.shtml#Tire
- https://jdmfactor.boards.net/thread/158/rfactor-physics-tutorial-danger-walls\
And as I understand, combining the SlipCurve with the variables DryLatLong, LatPeak, LongPeak and the GripMultiplier curve (which has as input the variable LoadSens), one should arrive on the same curves from a real tyre model.
But I can't find any information about how this is done.
For example, the variables LatPeak and LongPeak I've compiled the following explanations:
Slip range where lateral/longitudinal peak force occurs depending on load. Defines a curve where the load on the tire is along the x-axis and the y-axis is the sine of the slip angle where peak forces occur.
Since long goes by slip percent not angle this is simple: 0.015 = 1.5 percent slip and 0.035 = 3.5 percent.
Example1: (0.0817, 0.233, 13825.0)
At zero load, the sine of this slip angle is 0.0817, so the slip angle at which maximum lateral force is generated is asin(0.0817) = 4.7 deg. At a load on the tire of 13825 N, the sine of the slip angle is then the second value, or 0.233, so the angle is 13.5 deg.
Well, the example leads to two points (x y): (0 0.0817) and (0.233 13825). With two points one can make a line, not a curve. But anyway, no idea what I can do with this afterwards.
Example2: (0.1,0.3,15000)
The peak of the slip curve at 0 load will now be at 0.1, or asin(0.1) = 5.74 deg. The entire slip curve will be moved according to this value. If peak grip is acheived at 0.18 (10.37 degs) according to the slip curve, this will mean the entire slip curve slip step is multiplied by 0.1/0.18 = 0.555 rad. You will now be able to slide approximately, 0.555x as much as the slip curve would normally allow.
Now this example suggests the following calculation, that uses only the first value of three from the variable. So still not clear.
SlipCurve = SlipCurve * (LatPeak(1stValue) / (SlipCurve's SlipAngle for maximum value))
Does anyone have a guide of how to make this calculations and make a model, with units and real equations?
Thanks!
