physics question on tire spin...
Thread Starter
Pro
Joined: Mar 2008
Posts: 613
Likes: 1
From: Salt Lake City Utah
physics question on tire spin...
I know it is too late at night for a physics question, but here it is anyway. It seems as if when I am going up this particular steep on ramp on my way home and have the opportunity for some spirited driving the wheels tend to want to spin when shifting into 2nd and 3rd (don't go fast enough try spinning into 4th) more so than they want to when doing the same driving on a flat straight away.
The traction control comes on in a major way during shifting on the hill, much more so than flatter places. So the physics part is obvious... does driving up a hill create an easier opportunity for the wheels to break loose?
I know the car has to work harder to push itself up the hill, so is it because it's working harder that the light weight of the vehicle allows the tires to break loose because that is the easier for them to do than to push the car upward?
Just wondering
John
The traction control comes on in a major way during shifting on the hill, much more so than flatter places. So the physics part is obvious... does driving up a hill create an easier opportunity for the wheels to break loose?
I know the car has to work harder to push itself up the hill, so is it because it's working harder that the light weight of the vehicle allows the tires to break loose because that is the easier for them to do than to push the car upward?
Just wondering
John
Race Director
Joined: Mar 2007
Posts: 18,307
Likes: 21
From: Marlton. Increasing performance one speeding ticket at a time! NJ
Think of it this way...........you're a wise guy and don't heed a drawbridge barrier and make it to the top, only to find the bridge is opening up. You slam on your brakes, at the top wondering what's going to happen. As the bridge angles up, there comes a point where you no longer have friction between the tire and the bridge and you fall to your doom.
If you want a real scientific explanation, remember that the friction force is equal to the coefficient of friction (tire compound and road surface) multiplied by the force of mass and gravity. The mass doesn't change, but as the angle changes, the gravity vector changes with the cosine of the angle. So at 0° you have full friction force, and as you angle up, that friction is reduced until you reach 90°, where friction between your tire and road surface is impossible without other forces acting on your car.
Does that help?
If you want a real scientific explanation, remember that the friction force is equal to the coefficient of friction (tire compound and road surface) multiplied by the force of mass and gravity. The mass doesn't change, but as the angle changes, the gravity vector changes with the cosine of the angle. So at 0° you have full friction force, and as you angle up, that friction is reduced until you reach 90°, where friction between your tire and road surface is impossible without other forces acting on your car.
Does that help?
Le Mans Master
Joined: Mar 2002
Posts: 5,955
Likes: 161
From: PNW
Pretty much it. Find a hill steep enough and give it enough throttle and the tires will start to spin to the point where the car won't even move up the hill any more. Same thing happens on a slight hill with snow ... it's all a matter of the power to traction ratio and the slope of the hill.
Melting Slicks
Joined: Nov 2007
Posts: 2,336
Likes: 4
From: NW Arkansas
St. Jude Donor '08-'09
ascending an incline gravity starts working against you but at the same time weight transfer puts the center of gravity more toward the traction axle. hmmmm
gravity seems to have the upper hand
gravity seems to have the upper hand
Racer
Joined: May 2008
Posts: 307
Likes: 1
From: Saskatoon
Think of it this way...........you're a wise guy and don't heed a drawbridge barrier and make it to the top, only to find the bridge is opening up. You slam on your brakes, at the top wondering what's going to happen. As the bridge angles up, there comes a point where you no longer have friction between the tire and the bridge and you fall to your doom.
If you want a real scientific explanation, remember that the friction force is equal to the coefficient of friction (tire compound and road surface) multiplied by the force of mass and gravity. The mass doesn't change, but as the angle changes, the gravity vector changes with the cosine of the angle. So at 0° you have full friction force, and as you angle up, that friction is reduced until you reach 90°, where friction between your tire and road surface is impossible without other forces acting on your car.
Does that help?
If you want a real scientific explanation, remember that the friction force is equal to the coefficient of friction (tire compound and road surface) multiplied by the force of mass and gravity. The mass doesn't change, but as the angle changes, the gravity vector changes with the cosine of the angle. So at 0° you have full friction force, and as you angle up, that friction is reduced until you reach 90°, where friction between your tire and road surface is impossible without other forces acting on your car.
Does that help?
I'm just kidding. I liked that expaination and thought I would be a jerk.
Racer
Joined: Jan 2007
Posts: 293
Likes: 0
From: Maryville IL
Not quite right. The weight of the car does head to the rear axle, but it is not pushing straight down as it would be if the car were flat. This reduces the amount of friction between the tires and the road.
Le Mans Master
Joined: Mar 2002
Posts: 5,955
Likes: 161
From: PNW
... extreme example would be a car trying to climb an incline that is at say 95%. Not much weight on the wheels at all.
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Pro
Joined: Nov 2002
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From: Toronto Ontario
Think of it this way...........you're a wise guy and don't heed a drawbridge barrier and make it to the top, only to find the bridge is opening up. You slam on your brakes, at the top wondering what's going to happen. As the bridge angles up, there comes a point where you no longer have friction between the tire and the bridge and you fall to your doom.
If you want a real scientific explanation, remember that the friction force is equal to the coefficient of friction (tire compound and road surface) multiplied by the force of mass and gravity. The mass doesn't change, but as the angle changes, the gravity vector changes with the cosine of the angle. So at 0° you have full friction force, and as you angle up, that friction is reduced until you reach 90°, where friction between your tire and road surface is impossible without other forces acting on your car.
Does that help?
If you want a real scientific explanation, remember that the friction force is equal to the coefficient of friction (tire compound and road surface) multiplied by the force of mass and gravity. The mass doesn't change, but as the angle changes, the gravity vector changes with the cosine of the angle. So at 0° you have full friction force, and as you angle up, that friction is reduced until you reach 90°, where friction between your tire and road surface is impossible without other forces acting on your car.
Does that help?
Great explaination
Melting Slicks
Joined: Nov 2007
Posts: 2,336
Likes: 4
From: NW Arkansas
St. Jude Donor '08-'09
and exactly where does the weight of the car go?
Le Mans Master
Joined: Mar 2002
Posts: 5,955
Likes: 161
From: PNW
The weight of the car doesn't change, but the weight (ie, force) of the tires on the ground decreases as the angle of the car increases ... that's what I mean. Look at the force vector of the tires on the ground as a function of car angle. The force vector is a function of the gravity vector, and as the car's angle increases the perpendicular force vector of the tires to the ground decreases as the cosine of the inclination angle. Cosine of 90 degrees is zero.
Melting Slicks
Joined: Nov 2007
Posts: 2,336
Likes: 4
From: NW Arkansas
St. Jude Donor '08-'09
The weight of the car doesn't change, but the weight (ie, force) of the tires on the ground decreases as the angle of the car increases ... that's what I mean. Look at the force vector of the tires on the ground as a function of car angle. The force vector is a function of the gravity vector, and as the car's angle increases the perpendicular force vector of the tires to the ground decreases as the cosine of the inclination angle. Cosine of 90 degrees is zero.
dude...... this simple ol' country only recently figured out how to set the clock on my Sony Beta Max!!
I was pulling your chain a little.
Le Mans Master
Joined: Sep 2007
Posts: 7,604
Likes: 8
From: Belleville Il
Sounds like the road surface might not have optimal traction as well. Years ago our main street had well worn asphalt that took on a sheen. That road surface would let you break them loose even on the level quite easily!! This could also be a factor!! By the way, for a little more fun,try COMPETITION DRIVING mode!! Wheel spin with stability control still active! Of course tire purchase factor may go up
!!
!!
Instructor
Joined: Nov 2006
Posts: 233
Likes: 2
I wonder if there is a transition point where the angle of the slope is actually beneficial. Obviously, you're going to lose traction on a 95% slope, but I wonder if a 5% slope actually allows for more traction. If it does, then the next question is going to be: Is the weight you now have to more counterproductive to acceleration than the extra traction provided?
Thanks,
Chris
Thanks,
Chris
Safety Car
Joined: Dec 2006
Posts: 3,623
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From: Sparks, Nevada
"so, all those years at the drag strip when we used 90/10 shocks on the front to raise the front and transfer weight was just a waste of time?"
Only when we were drag racing up hill. Don't forget the tilt fiberglass frontends and the straight axle either. 
Them were my good old days.
Only when we were drag racing up hill.
Don't forget the tilt fiberglass frontends and the straight axle either. Them were my good old days.