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#1
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Slip Ratio and Tration Force
Greetings,
Searching for a slip ratio equation and its relation to the traction force, I found two seemingly different formulas: slip_ratio = (angular_velocity * object_radius - longtitudinal_velocity) / longtitudinal_velocity F = slip_ratio * traction_coefficient And, slip_ratio = (angular_velocity * object_radius - longtitudinal_velocity) / (angular_velocity * object_radius) F = engine_torque * wheel_radius * slip_ratio First question that comes to mind is which one is right? Second, I've used the first equation, however I didn't know what would the slip ratio be if the velocity is 0.0 (Starting up). If it would be zero then the car would never start in the first place. Also what value does the traction coefficient usually have? Now, if the second formula was the right one then the wheel would never stop spinning, because when there's no engine torque it would mean that the traction force would be zero, so the acceleration would never go negative (When going forward). So how can I solve this problem? As a side question, is there any other torque (that I need to worry about) exerted on the wheel besides traction, engine and brakes? Thanks, Abdo Haji-Ali Programmer In|Framez |
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#2
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Slip Ratio and Tration Force
> F = engine_torque * wheel_radius * slip_ratio
This one can't be right. Force would be relative to rear wheel torque divided by tire radius. > I didn't know what would the slip > ratio be if the velocity is 0.0 (Starting up). You need a different formula. Some formulas treat a tire as if it were a spring. Note that these are "idealized" formulas to make modeling a car simpler. The actual real world formulas for a tire are more complex. > Also what value does the traction coefficient usually have? For a peformance tire, around 1.0. For a road racing slick, 1.5 or more. For a drag racing slick, over 4.0. You can't make tires out of table tennis rubber (way too soft), but it has a coefficient around 7 or 8. > So how can I solve this problem? As mentioned, treat the tire as a spring. > As a side question, is there any other torque (that I need to worry > about) exerted on the wheel besides traction, engine and brakes? Traction just puts a limit of the torque. The engine/gearing, and brakes generate a torque. If the car is on an incline, then gravity applies a force to the car and tires. |
#3
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Slip Ratio and Tration Force
The complication of traction, the loading does make a difference, especially for
smaller objects (smooth surfaces tend to attract each other, which is probably the main reason that the smaller block seems to have much more grip). Look at the latter part of clip # 2: http://www.gyroscopes.org/1974lecture.asp The relationship between slip angle and side force is not linear with the load factor, look at the graph of force versus slip angle and note that the curves differ depending on the load (this is slip angle, not slip ratio): http://www.smithees-racetech.com.au/ackerman.html |
#4
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Slip Ratio and Tration Force
The first equation for slip ratio is actually almost correct. The SAE
definition of slip ratio is slip_ratio = (angular_speed_tire/ABS(speed_over_ground)) - 1. If you recall from math class you can convert angular velocities into linear velocities with the equation V= R*w where R is the radius or the wheel/tire and w is the angular velocity. The only flaw in the equation is that you need to use the absolute value of the denominator or you will not maintain the sign when going in reverse. I would stick to using angular velocities, there is really no reason to convert then linear for the calculation. As a side note, remember that the slip_ratio is a percentage. But a percentage of what? I'll tackle that next.... The equations for force are lacking. The maximum tractive force, in very simple terms, is the traction coefficient of the tire times the vertical load on the tire. What the slip ratio is, is a percentage of this total possible force. So the equation for force would be F = slip_ratio * vertical_load * traction_coeff. Of course the actual force generated by the tire is much more complex. Once the tire has passed 100% of it's possible force it enters a state of sliding and the force will fall off. To get accurate forces you will need to implement an equation such as Pacejka's "magic formula". To complicate things further, you will also have to take into account the lateral forces at the same time as the longitudinal forces (the tire only has so much traction to go around). As pointed out by Jeff in another post, you will need to handle the low speed case in a different manner (especially at 0 velocity where there is a singularity). The spring method he mentioned will work just fine. It is also interesting to point out that you need to update the tire forces at a fairly high frequency or you will get jerky response. Also, the engine torque does not have any direct influence on the tire force other than to increase the angular velocity of the wheel which then affects the slip_ratio calculation. I hope this ramble helps. -Brent Abdo Haji-Ali wrote: > Greetings, > > Searching for a slip ratio equation and its relation to the traction > force, I found two seemingly different formulas: > > slip_ratio = (angular_velocity * object_radius - > longtitudinal_velocity) / longtitudinal_velocity > F = slip_ratio * traction_coefficient > > And, > > slip_ratio = (angular_velocity * object_radius - > longtitudinal_velocity) / (angular_velocity * object_radius) > F = engine_torque * wheel_radius * slip_ratio > > First question that comes to mind is which one is right? Second, I've > used the first equation, however I didn't know what would the slip > ratio be if the velocity is 0.0 (Starting up). If it would be zero then > the car would never start in the first place. Also what value does the > traction coefficient usually have? > Now, if the second formula was the right one then the wheel would never > stop spinning, because when there's no engine torque it would mean that > the traction force would be zero, so the acceleration would never go > negative (When going forward). So how can I solve this problem? > > As a side question, is there any other torque (that I need to worry > about) exerted on the wheel besides traction, engine and brakes? > > Thanks, > Abdo Haji-Ali > Programmer > In|Framez |
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