Why does a car's steering wheel get lighter with increasing speed [closed] The Next CEO of Stack OverflowDoes ABS shorten stopping distance of a car?What is the direction of static friction?Accelerate the car by static frictionWhat causes a car's velocity to follow the front wheels' direction?A car with constant speed doing a turnWhy is it easier to steer a car at speed?Aircraft - static take off - how is this possible?slip definition that works in gameWhen a car's non-driving wheels are turned, what is the frictional force vector that actually causes the vehicle to turn in that direction?The physics of a cornering wheel and its centripetal force
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Why does a car's steering wheel get lighter with increasing speed [closed]
The Next CEO of Stack OverflowDoes ABS shorten stopping distance of a car?What is the direction of static friction?Accelerate the car by static frictionWhat causes a car's velocity to follow the front wheels' direction?A car with constant speed doing a turnWhy is it easier to steer a car at speed?Aircraft - static take off - how is this possible?slip definition that works in gameWhen a car's non-driving wheels are turned, what is the frictional force vector that actually causes the vehicle to turn in that direction?The physics of a cornering wheel and its centripetal force
$begingroup$
I've noticed it is difficult to turn the wheels of a car when the car is stationary, especially cars without power steering, which is why the power steering was invented. However, I've noticed it becomes feather light when traveling at speed (some models even stiffen the steering wheel electronically at speed). So, why does a car's steering wheel get lighter with increasing speed?
newtonian-mechanics everyday-life speed
edited Mar 8 at 19:30
jezzo
1594
asked Mar 8 at 14:09
securitydude5securitydude5
1875
$endgroup$
closed as off-topic by David Z♦ Mar 9 at 10:59
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question appears to be about engineering, which is the application of scientific knowledge to construct a solution to solve a specific problem. As such, it is off topic for this site, which deals with the science, whether theoretical or experimental, of how the natural world works. For more information, see this meta post." – David Z
add a comment |
$begingroup$
I've noticed it is difficult to turn the wheels of a car when the car is stationary, especially cars without power steering, which is why the power steering was invented. However, I've noticed it becomes feather light when traveling at speed (some models even stiffen the steering wheel electronically at speed). So, why does a car's steering wheel get lighter with increasing speed?
newtonian-mechanics everyday-life speed
edited Mar 8 at 19:30
jezzo
1594
asked Mar 8 at 14:09
securitydude5securitydude5
1875
$endgroup$
closed as off-topic by David Z♦ Mar 9 at 10:59
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question appears to be about engineering, which is the application of scientific knowledge to construct a solution to solve a specific problem. As such, it is off topic for this site, which deals with the science, whether theoretical or experimental, of how the natural world works. For more information, see this meta post." – David Z
add a comment |
$begingroup$
I've noticed it is difficult to turn the wheels of a car when the car is stationary, especially cars without power steering, which is why the power steering was invented. However, I've noticed it becomes feather light when traveling at speed (some models even stiffen the steering wheel electronically at speed). So, why does a car's steering wheel get lighter with increasing speed?
newtonian-mechanics everyday-life speed
edited Mar 8 at 19:30
jezzo
1594
asked Mar 8 at 14:09
securitydude5securitydude5
1875
$endgroup$
I've noticed it is difficult to turn the wheels of a car when the car is stationary, especially cars without power steering, which is why the power steering was invented. However, I've noticed it becomes feather light when traveling at speed (some models even stiffen the steering wheel electronically at speed). So, why does a car's steering wheel get lighter with increasing speed?
newtonian-mechanics everyday-life speed
newtonian-mechanics everyday-life speed
edited Mar 8 at 19:30
jezzo
1594
asked Mar 8 at 14:09
securitydude5securitydude5
1875
edited Mar 8 at 19:30
jezzo
1594
asked Mar 8 at 14:09
securitydude5securitydude5
1875
edited Mar 8 at 19:30
jezzo
1594
edited Mar 8 at 19:30
jezzo
1594
edited Mar 8 at 19:30
jezzo
1594
1594
asked Mar 8 at 14:09
securitydude5securitydude5
1875
asked Mar 8 at 14:09
securitydude5securitydude5
1875
asked Mar 8 at 14:09
securitydude5securitydude5
1875
1875
closed as off-topic by David Z♦ Mar 9 at 10:59
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question appears to be about engineering, which is the application of scientific knowledge to construct a solution to solve a specific problem. As such, it is off topic for this site, which deals with the science, whether theoretical or experimental, of how the natural world works. For more information, see this meta post." – David Z
closed as off-topic by David Z♦ Mar 9 at 10:59
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question appears to be about engineering, which is the application of scientific knowledge to construct a solution to solve a specific problem. As such, it is off topic for this site, which deals with the science, whether theoretical or experimental, of how the natural world works. For more information, see this meta post." – David Z
add a comment |
add a comment |
4 Answers
4
active
oldest
votes
$begingroup$
Imagine the car stationary. The tire sits on the ground with the contact patch touching.
As you start turning the wheel, the linkage to the wheels starts to rotate the contact patch on the ground. (There are also more complex motions because of the non-zero caster angle of the front wheel).
This rotation is opposed by the static friction of the tire. As you continue turning, portions of the tread on the contact patch are pulled over and stressed.
Now imagine holding the steering wheel at that angle and allowing the car to roll forward a bit. The tread at the rear of the contact patch lifts away from the road and the stress in that portion of the tire is released. Meanwhile new tread rolls onto the contact patch in front, but at the correct angle. Once the contact patch is covered by new tread, the stress from the turn is gone and steering wheel is back to a near-neutral force (again, modulo several effects from the suspension angles).
The faster the car is moving forward, the faster it can put tread into the contact patch with no side stress. So the steering becomes easier.
answered Mar 8 at 16:53
BowlOfRedBowlOfRed
17.9k22745
$endgroup$
$begingroup$
And you can see the tire change shape when turning in this video of high speed cornering which can help you visualize what's happening where the rubber meets the road.
$endgroup$
– davidbak
Mar 8 at 21:53
add a comment |
$begingroup$
The work to turn the wheel is roughly proportional to how much you turn the wheel, and inversely proportional to the distance the car traveled. You feel less resistance at higher speed because the car moved farther for the same amount of steering wheel turn. This is because when the wheels on the ground are rolling, the distortion of the rubber, and the friction, that occurs when you turn the steering wheel is less than it is when those wheels are stationary to the ground.
answered Mar 8 at 14:26
DigiprocDigiproc
1,56349
$endgroup$
$begingroup$
I'd word this as "inversely proportional to the distance the car traveled"
$endgroup$
– Nayuki
Mar 8 at 22:17
$begingroup$
oops, you're right, I fixed it
$endgroup$
– Digiproc
Mar 8 at 22:57
add a comment |
$begingroup$
Even though a car is not a wing, most cars generate some lift as they travel through the air as well as a force moment which tends to torque the car down at the rear wheels and up at the front. In addition, the torque that the engine is applying to the driven wheels results in a countertorque on the body of the car that also tends to lift the front end of the car. All these effects tend to unload the front wheels, which lightens the steering forces.
answered Mar 8 at 18:48
niels nielsenniels nielsen
21k53062
$endgroup$
5
$begingroup$
This is probably less significant than the thing BowlOfRed suggested, but I don't doubt it has a measurable effect.
$endgroup$
– wizzwizz4
Mar 8 at 19:06
add a comment |
$begingroup$
As others have posted, the forward rotation of the wheel reduces the 'scrubbing', however, there is an opposing force which should be mentioned, the gyroscopic effect which would cause the steering to become more difficult to turn the faster the wheels rotate.
answered Mar 8 at 19:58
TopCatTopCat
311
$endgroup$
2
$begingroup$
Can you calculate the magnitude of the gyroscopic effect? My guess is that it hasn't been mentioned because we don't notice it at ordinary speeds, because it is negligible compared to static friction from 1000 kg on rubber-to-cement contact.
$endgroup$
– Sam
Mar 8 at 20:10
add a comment |
4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Imagine the car stationary. The tire sits on the ground with the contact patch touching.
As you start turning the wheel, the linkage to the wheels starts to rotate the contact patch on the ground. (There are also more complex motions because of the non-zero caster angle of the front wheel).
This rotation is opposed by the static friction of the tire. As you continue turning, portions of the tread on the contact patch are pulled over and stressed.
Now imagine holding the steering wheel at that angle and allowing the car to roll forward a bit. The tread at the rear of the contact patch lifts away from the road and the stress in that portion of the tire is released. Meanwhile new tread rolls onto the contact patch in front, but at the correct angle. Once the contact patch is covered by new tread, the stress from the turn is gone and steering wheel is back to a near-neutral force (again, modulo several effects from the suspension angles).
The faster the car is moving forward, the faster it can put tread into the contact patch with no side stress. So the steering becomes easier.
answered Mar 8 at 16:53
BowlOfRedBowlOfRed
17.9k22745
$endgroup$
$begingroup$
And you can see the tire change shape when turning in this video of high speed cornering which can help you visualize what's happening where the rubber meets the road.
$endgroup$
– davidbak
Mar 8 at 21:53
add a comment |
$begingroup$
Imagine the car stationary. The tire sits on the ground with the contact patch touching.
As you start turning the wheel, the linkage to the wheels starts to rotate the contact patch on the ground. (There are also more complex motions because of the non-zero caster angle of the front wheel).
This rotation is opposed by the static friction of the tire. As you continue turning, portions of the tread on the contact patch are pulled over and stressed.
Now imagine holding the steering wheel at that angle and allowing the car to roll forward a bit. The tread at the rear of the contact patch lifts away from the road and the stress in that portion of the tire is released. Meanwhile new tread rolls onto the contact patch in front, but at the correct angle. Once the contact patch is covered by new tread, the stress from the turn is gone and steering wheel is back to a near-neutral force (again, modulo several effects from the suspension angles).
The faster the car is moving forward, the faster it can put tread into the contact patch with no side stress. So the steering becomes easier.
answered Mar 8 at 16:53
BowlOfRedBowlOfRed
17.9k22745
$endgroup$
$begingroup$
And you can see the tire change shape when turning in this video of high speed cornering which can help you visualize what's happening where the rubber meets the road.
$endgroup$
– davidbak
Mar 8 at 21:53
add a comment |
$begingroup$
Imagine the car stationary. The tire sits on the ground with the contact patch touching.
As you start turning the wheel, the linkage to the wheels starts to rotate the contact patch on the ground. (There are also more complex motions because of the non-zero caster angle of the front wheel).
This rotation is opposed by the static friction of the tire. As you continue turning, portions of the tread on the contact patch are pulled over and stressed.
Now imagine holding the steering wheel at that angle and allowing the car to roll forward a bit. The tread at the rear of the contact patch lifts away from the road and the stress in that portion of the tire is released. Meanwhile new tread rolls onto the contact patch in front, but at the correct angle. Once the contact patch is covered by new tread, the stress from the turn is gone and steering wheel is back to a near-neutral force (again, modulo several effects from the suspension angles).
The faster the car is moving forward, the faster it can put tread into the contact patch with no side stress. So the steering becomes easier.
answered Mar 8 at 16:53
BowlOfRedBowlOfRed
17.9k22745
$endgroup$
Imagine the car stationary. The tire sits on the ground with the contact patch touching.
As you start turning the wheel, the linkage to the wheels starts to rotate the contact patch on the ground. (There are also more complex motions because of the non-zero caster angle of the front wheel).
This rotation is opposed by the static friction of the tire. As you continue turning, portions of the tread on the contact patch are pulled over and stressed.
Now imagine holding the steering wheel at that angle and allowing the car to roll forward a bit. The tread at the rear of the contact patch lifts away from the road and the stress in that portion of the tire is released. Meanwhile new tread rolls onto the contact patch in front, but at the correct angle. Once the contact patch is covered by new tread, the stress from the turn is gone and steering wheel is back to a near-neutral force (again, modulo several effects from the suspension angles).
The faster the car is moving forward, the faster it can put tread into the contact patch with no side stress. So the steering becomes easier.
answered Mar 8 at 16:53
BowlOfRedBowlOfRed
17.9k22745
edited Mar 8 at 23:36
answered Mar 8 at 16:53
BowlOfRedBowlOfRed
17.9k22745
answered Mar 8 at 16:53
BowlOfRedBowlOfRed
17.9k22745
answered Mar 8 at 16:53
BowlOfRedBowlOfRed
17.9k22745
17.9k22745
$begingroup$
And you can see the tire change shape when turning in this video of high speed cornering which can help you visualize what's happening where the rubber meets the road.
$endgroup$
– davidbak
Mar 8 at 21:53
add a comment |
$begingroup$
And you can see the tire change shape when turning in this video of high speed cornering which can help you visualize what's happening where the rubber meets the road.
$endgroup$
– davidbak
Mar 8 at 21:53
$begingroup$
And you can see the tire change shape when turning in this video of high speed cornering which can help you visualize what's happening where the rubber meets the road.
$endgroup$
– davidbak
Mar 8 at 21:53
$begingroup$
And you can see the tire change shape when turning in this video of high speed cornering which can help you visualize what's happening where the rubber meets the road.
$endgroup$
– davidbak
Mar 8 at 21:53
add a comment |
$begingroup$
The work to turn the wheel is roughly proportional to how much you turn the wheel, and inversely proportional to the distance the car traveled. You feel less resistance at higher speed because the car moved farther for the same amount of steering wheel turn. This is because when the wheels on the ground are rolling, the distortion of the rubber, and the friction, that occurs when you turn the steering wheel is less than it is when those wheels are stationary to the ground.
answered Mar 8 at 14:26
DigiprocDigiproc
1,56349
$endgroup$
$begingroup$
I'd word this as "inversely proportional to the distance the car traveled"
$endgroup$
– Nayuki
Mar 8 at 22:17
$begingroup$
oops, you're right, I fixed it
$endgroup$
– Digiproc
Mar 8 at 22:57
add a comment |
$begingroup$
The work to turn the wheel is roughly proportional to how much you turn the wheel, and inversely proportional to the distance the car traveled. You feel less resistance at higher speed because the car moved farther for the same amount of steering wheel turn. This is because when the wheels on the ground are rolling, the distortion of the rubber, and the friction, that occurs when you turn the steering wheel is less than it is when those wheels are stationary to the ground.
answered Mar 8 at 14:26
DigiprocDigiproc
1,56349
$endgroup$
$begingroup$
I'd word this as "inversely proportional to the distance the car traveled"
$endgroup$
– Nayuki
Mar 8 at 22:17
$begingroup$
oops, you're right, I fixed it
$endgroup$
– Digiproc
Mar 8 at 22:57
add a comment |
$begingroup$
The work to turn the wheel is roughly proportional to how much you turn the wheel, and inversely proportional to the distance the car traveled. You feel less resistance at higher speed because the car moved farther for the same amount of steering wheel turn. This is because when the wheels on the ground are rolling, the distortion of the rubber, and the friction, that occurs when you turn the steering wheel is less than it is when those wheels are stationary to the ground.
answered Mar 8 at 14:26
DigiprocDigiproc
1,56349
$endgroup$
The work to turn the wheel is roughly proportional to how much you turn the wheel, and inversely proportional to the distance the car traveled. You feel less resistance at higher speed because the car moved farther for the same amount of steering wheel turn. This is because when the wheels on the ground are rolling, the distortion of the rubber, and the friction, that occurs when you turn the steering wheel is less than it is when those wheels are stationary to the ground.
answered Mar 8 at 14:26
DigiprocDigiproc
1,56349
edited Mar 8 at 22:56
answered Mar 8 at 14:26
DigiprocDigiproc
1,56349
answered Mar 8 at 14:26
DigiprocDigiproc
1,56349
answered Mar 8 at 14:26
DigiprocDigiproc
1,56349
1,56349
$begingroup$
I'd word this as "inversely proportional to the distance the car traveled"
$endgroup$
– Nayuki
Mar 8 at 22:17
$begingroup$
oops, you're right, I fixed it
$endgroup$
– Digiproc
Mar 8 at 22:57
add a comment |
$begingroup$
I'd word this as "inversely proportional to the distance the car traveled"
$endgroup$
– Nayuki
Mar 8 at 22:17
$begingroup$
oops, you're right, I fixed it
$endgroup$
– Digiproc
Mar 8 at 22:57
$begingroup$
I'd word this as "inversely proportional to the distance the car traveled"
$endgroup$
– Nayuki
Mar 8 at 22:17
$begingroup$
I'd word this as "inversely proportional to the distance the car traveled"
$endgroup$
– Nayuki
Mar 8 at 22:17
$begingroup$
oops, you're right, I fixed it
$endgroup$
– Digiproc
Mar 8 at 22:57
$begingroup$
oops, you're right, I fixed it
$endgroup$
– Digiproc
Mar 8 at 22:57
add a comment |
$begingroup$
Even though a car is not a wing, most cars generate some lift as they travel through the air as well as a force moment which tends to torque the car down at the rear wheels and up at the front. In addition, the torque that the engine is applying to the driven wheels results in a countertorque on the body of the car that also tends to lift the front end of the car. All these effects tend to unload the front wheels, which lightens the steering forces.
answered Mar 8 at 18:48
niels nielsenniels nielsen
21k53062
$endgroup$
5
$begingroup$
This is probably less significant than the thing BowlOfRed suggested, but I don't doubt it has a measurable effect.
$endgroup$
– wizzwizz4
Mar 8 at 19:06
add a comment |
$begingroup$
Even though a car is not a wing, most cars generate some lift as they travel through the air as well as a force moment which tends to torque the car down at the rear wheels and up at the front. In addition, the torque that the engine is applying to the driven wheels results in a countertorque on the body of the car that also tends to lift the front end of the car. All these effects tend to unload the front wheels, which lightens the steering forces.
answered Mar 8 at 18:48
niels nielsenniels nielsen
21k53062
$endgroup$
5
$begingroup$
This is probably less significant than the thing BowlOfRed suggested, but I don't doubt it has a measurable effect.
$endgroup$
– wizzwizz4
Mar 8 at 19:06
add a comment |
$begingroup$
Even though a car is not a wing, most cars generate some lift as they travel through the air as well as a force moment which tends to torque the car down at the rear wheels and up at the front. In addition, the torque that the engine is applying to the driven wheels results in a countertorque on the body of the car that also tends to lift the front end of the car. All these effects tend to unload the front wheels, which lightens the steering forces.
answered Mar 8 at 18:48
niels nielsenniels nielsen
21k53062
$endgroup$
Even though a car is not a wing, most cars generate some lift as they travel through the air as well as a force moment which tends to torque the car down at the rear wheels and up at the front. In addition, the torque that the engine is applying to the driven wheels results in a countertorque on the body of the car that also tends to lift the front end of the car. All these effects tend to unload the front wheels, which lightens the steering forces.
answered Mar 8 at 18:48
niels nielsenniels nielsen
21k53062
answered Mar 8 at 18:48
niels nielsenniels nielsen
21k53062
answered Mar 8 at 18:48
niels nielsenniels nielsen
21k53062
answered Mar 8 at 18:48
niels nielsenniels nielsen
21k53062
21k53062
5
$begingroup$
This is probably less significant than the thing BowlOfRed suggested, but I don't doubt it has a measurable effect.
$endgroup$
– wizzwizz4
Mar 8 at 19:06
add a comment |
5
$begingroup$
This is probably less significant than the thing BowlOfRed suggested, but I don't doubt it has a measurable effect.
$endgroup$
– wizzwizz4
Mar 8 at 19:06
5
5
$begingroup$
This is probably less significant than the thing BowlOfRed suggested, but I don't doubt it has a measurable effect.
$endgroup$
– wizzwizz4
Mar 8 at 19:06
$begingroup$
This is probably less significant than the thing BowlOfRed suggested, but I don't doubt it has a measurable effect.
$endgroup$
– wizzwizz4
Mar 8 at 19:06
add a comment |
$begingroup$
As others have posted, the forward rotation of the wheel reduces the 'scrubbing', however, there is an opposing force which should be mentioned, the gyroscopic effect which would cause the steering to become more difficult to turn the faster the wheels rotate.
answered Mar 8 at 19:58
TopCatTopCat
311
$endgroup$
2
$begingroup$
Can you calculate the magnitude of the gyroscopic effect? My guess is that it hasn't been mentioned because we don't notice it at ordinary speeds, because it is negligible compared to static friction from 1000 kg on rubber-to-cement contact.
$endgroup$
– Sam
Mar 8 at 20:10
add a comment |
$begingroup$
As others have posted, the forward rotation of the wheel reduces the 'scrubbing', however, there is an opposing force which should be mentioned, the gyroscopic effect which would cause the steering to become more difficult to turn the faster the wheels rotate.
answered Mar 8 at 19:58
TopCatTopCat
311
$endgroup$
2
$begingroup$
Can you calculate the magnitude of the gyroscopic effect? My guess is that it hasn't been mentioned because we don't notice it at ordinary speeds, because it is negligible compared to static friction from 1000 kg on rubber-to-cement contact.
$endgroup$
– Sam
Mar 8 at 20:10
add a comment |
$begingroup$
As others have posted, the forward rotation of the wheel reduces the 'scrubbing', however, there is an opposing force which should be mentioned, the gyroscopic effect which would cause the steering to become more difficult to turn the faster the wheels rotate.
answered Mar 8 at 19:58
TopCatTopCat
311
$endgroup$
As others have posted, the forward rotation of the wheel reduces the 'scrubbing', however, there is an opposing force which should be mentioned, the gyroscopic effect which would cause the steering to become more difficult to turn the faster the wheels rotate.
answered Mar 8 at 19:58
TopCatTopCat
311
answered Mar 8 at 19:58
TopCatTopCat
311
answered Mar 8 at 19:58
TopCatTopCat
311
answered Mar 8 at 19:58
TopCatTopCat
311
311
2
$begingroup$
Can you calculate the magnitude of the gyroscopic effect? My guess is that it hasn't been mentioned because we don't notice it at ordinary speeds, because it is negligible compared to static friction from 1000 kg on rubber-to-cement contact.
$endgroup$
– Sam
Mar 8 at 20:10
add a comment |
2
$begingroup$
Can you calculate the magnitude of the gyroscopic effect? My guess is that it hasn't been mentioned because we don't notice it at ordinary speeds, because it is negligible compared to static friction from 1000 kg on rubber-to-cement contact.
$endgroup$
– Sam
Mar 8 at 20:10
2
2
$begingroup$
Can you calculate the magnitude of the gyroscopic effect? My guess is that it hasn't been mentioned because we don't notice it at ordinary speeds, because it is negligible compared to static friction from 1000 kg on rubber-to-cement contact.
$endgroup$
– Sam
Mar 8 at 20:10
$begingroup$
Can you calculate the magnitude of the gyroscopic effect? My guess is that it hasn't been mentioned because we don't notice it at ordinary speeds, because it is negligible compared to static friction from 1000 kg on rubber-to-cement contact.
$endgroup$
– Sam
Mar 8 at 20:10
add a comment |
