Can anyone run slip angles past me again?
#1
Tried to explain understeer and oversteer (esp. the lift off type) in terms of slip angles and weight distribution to someone yesterday...
...ended up in a bit of a mess, myself So..
Is slip angle:
- angle (looking down) between rim (fore and aft) and contact patch on ground?
- angle (looking down) between rim and direction of travel?
Steady straight line .. slip angle = 0
Constant radius low speed 'steady' turn .. slip angle = 0 (?)
Understeering turn .. slip angle > 0 (?)
Oversteering turn .. slip angle < 0 (?)(where < and > can be interchanged depending on the point of reference)
etc etc ... or am I at entirely the wrong end of the stick?
Any good web sites or books that explain understeer/oversteer in pseudo-maths - don't want the whole nine yards, but want more than the 'understeer is where the car turns less than the steering input demanded' type qualitative statements
TIA
PS Sorry - I appreciate this is non-scooby related, but I thought I'd get the best replies in this forum (creep that I am)
PPS going off line for a few hours, so may not be able to respond quickly
...ended up in a bit of a mess, myself So..
Is slip angle:
- angle (looking down) between rim (fore and aft) and contact patch on ground?
- angle (looking down) between rim and direction of travel?
Steady straight line .. slip angle = 0
Constant radius low speed 'steady' turn .. slip angle = 0 (?)
Understeering turn .. slip angle > 0 (?)
Oversteering turn .. slip angle < 0 (?)(where < and > can be interchanged depending on the point of reference)
etc etc ... or am I at entirely the wrong end of the stick?
Any good web sites or books that explain understeer/oversteer in pseudo-maths - don't want the whole nine yards, but want more than the 'understeer is where the car turns less than the steering input demanded' type qualitative statements
TIA
PS Sorry - I appreciate this is non-scooby related, but I thought I'd get the best replies in this forum (creep that I am)
PPS going off line for a few hours, so may not be able to respond quickly
#2
Slip angle is (simply put) the angle between the rim/tire and the direction of travel.
If you make a turn at vitually zero speed the front tires/rims make an angle, this is your steering input. Because the speed and lateral forces are almost zero, the slip angles front and rear are also (almost zero).
If you increase you speed in the turn (not taking into account the forces on the tires for traction!) then your tires need to built up lateral forces. To built up lateral forces the tires need slip angles. The greater the slip angles, the greater the lateral forces. This relationship is not linear and depends on for example the tire pressures, the (dynamic) weight on the tire and the traction forces.
At a given speed in a turn your steering input for the front wheels is: steering input at (almost) zero speed + slip angle of front tires - slip angle at rear tires. If the slip angles of the front and rear tires are equal, your car (for this speed) is neutral. If the slip angles of the front tires are greater then your car is understeering. And course if the slip angles of the rear tires is greater then the car is oversteering.
The relationship between slip angle and lateral force is not linear and depends on for example the weight resting on the tire. So if your in a corner with increasing speed the weight transfer due to springs and anti roll bars shifts more to the front or the rear, your car can change from understeer to oversteer or the other way round.
Another situation when you are in a turn and lift off, you will have a weight transfer to the front. Your front tires will have it "easier" and can reduce the slip angle, but the rear tires will have to increase the slip angle. So the car will understeer lesser or even oversteer.
I hope I answered your questions without the mathematics.
If you make a turn at vitually zero speed the front tires/rims make an angle, this is your steering input. Because the speed and lateral forces are almost zero, the slip angles front and rear are also (almost zero).
If you increase you speed in the turn (not taking into account the forces on the tires for traction!) then your tires need to built up lateral forces. To built up lateral forces the tires need slip angles. The greater the slip angles, the greater the lateral forces. This relationship is not linear and depends on for example the tire pressures, the (dynamic) weight on the tire and the traction forces.
At a given speed in a turn your steering input for the front wheels is: steering input at (almost) zero speed + slip angle of front tires - slip angle at rear tires. If the slip angles of the front and rear tires are equal, your car (for this speed) is neutral. If the slip angles of the front tires are greater then your car is understeering. And course if the slip angles of the rear tires is greater then the car is oversteering.
The relationship between slip angle and lateral force is not linear and depends on for example the weight resting on the tire. So if your in a corner with increasing speed the weight transfer due to springs and anti roll bars shifts more to the front or the rear, your car can change from understeer to oversteer or the other way round.
Another situation when you are in a turn and lift off, you will have a weight transfer to the front. Your front tires will have it "easier" and can reduce the slip angle, but the rear tires will have to increase the slip angle. So the car will understeer lesser or even oversteer.
I hope I answered your questions without the mathematics.
#3
Rovo
I've read your words several times and....what a fantastic answer it is Thankyou very much.
I am getting more curious about this (yes, I'm sad) and would like to see some maths. I presume there is a mathematical analysis possible, and at varying degrees of complexity. It must be a standard university/college/design department treatise.
Anyone know of anything?
Thanks again Rovo
Martin
I've read your words several times and....what a fantastic answer it is Thankyou very much.
I am getting more curious about this (yes, I'm sad) and would like to see some maths. I presume there is a mathematical analysis possible, and at varying degrees of complexity. It must be a standard university/college/design department treatise.
Anyone know of anything?
Thanks again Rovo
Martin
#5
One of the hardest things to get your head around is that as soon as you start turning, there are slip angles. Even if you are driving steel wheels on a concrete floors, you can't change direction without slip angles.
Here's some further reading for you.
There are some nice diagrams here on the AutoZine Technical School.
Chapter 7 of the Stuttgart-West Physics of Racing is pretty good at explaining things, although there are a lot of formulae flying around.
Another place to read is the Physics of Motorsport.
I would also suggest taking a look at Driving Techniques.
Here's some further reading for you.
There are some nice diagrams here on the AutoZine Technical School.
Chapter 7 of the Stuttgart-West Physics of Racing is pretty good at explaining things, although there are a lot of formulae flying around.
Another place to read is the Physics of Motorsport.
I would also suggest taking a look at Driving Techniques.
#6
Martin,
I've got the technical (with formulas and pictures) explanation in English for the formula:
steering input for the front wheels = steering input at (almost) zero speed + slip angle of front tires - slip angle at rear tires
It's 6 pages A4-size. If your interested I can scan them and e-mail them to you?
Robert
I've got the technical (with formulas and pictures) explanation in English for the formula:
steering input for the front wheels = steering input at (almost) zero speed + slip angle of front tires - slip angle at rear tires
It's 6 pages A4-size. If your interested I can scan them and e-mail them to you?
Robert
#7
Robert, David
Thanks very much. I think that's all I need. It's enough to glaze everyone's eyes over in the pub, and enough to satisfy my slightly unnatural way of wanting to understand things
Cheers
Martin
Thanks very much. I think that's all I need. It's enough to glaze everyone's eyes over in the pub, and enough to satisfy my slightly unnatural way of wanting to understand things
Cheers
Martin
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