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Why does a car's steering wheel get lighter with increasing speed

Does 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

15

2

$begingroup$

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

share|cite|improve this question

edited 10 hours ago

jezzo

434

asked 15 hours ago

securitydude5securitydude5

1654

$endgroup$

add a comment | 

15

2

$begingroup$

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

share|cite|improve this question

edited 10 hours ago

jezzo

434

asked 15 hours ago

securitydude5securitydude5

1654

$endgroup$

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

share|cite|improve this question

edited 10 hours ago

jezzo

434

asked 15 hours ago

securitydude5securitydude5

1654

$endgroup$

share|cite|improve this question

edited 10 hours ago

jezzo

434

asked 15 hours ago

securitydude5securitydude5

1654

share|cite|improve this question

edited 10 hours ago

jezzo

434

asked 15 hours ago

securitydude5securitydude5

1654

edited 10 hours ago

jezzo

434

edited 10 hours ago

jezzo

434

asked 15 hours ago

securitydude5securitydude5

1654

asked 15 hours ago

securitydude5securitydude5

1654

4 Answers
4

active

oldest

votes

33

$begingroup$

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.

share|cite|improve this answer

edited 6 hours ago

answered 13 hours ago

BowlOfRedBowlOfRed

17.4k22743

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  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
  8 hours ago
  
  
  

  
  
  
  

add a comment | 

9

$begingroup$

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.

share|cite|improve this answer

edited 7 hours ago

answered 15 hours ago

DigiprocDigiproc

1,50848

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I'd word this as "inversely proportional to the distance the car traveled"
  $endgroup$
  – Nayuki
  7 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  oops, you're right, I fixed it
  $endgroup$
  – Digiproc
  7 hours ago
  

  
  
  
  

add a comment | 

5

$begingroup$

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.

share|cite|improve this answer

answered 11 hours ago

niels nielsenniels nielsen

20.3k53061

$endgroup$

• 
  
  
  

  
  4
  

  
  
  
  
  
  

  
  $begingroup$
  This is probably less significant than the thing BowlOfRed suggested, but I don't doubt it has a measurable effect.
  $endgroup$
  – wizzwizz4
  10 hours ago
  

  
  
  
  

add a comment | 

1

$begingroup$

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.

share|cite|improve this answer

answered 10 hours ago

TopCatTopCat

111

New contributor

TopCat is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

• 
  
  
  

  
  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
  9 hours ago
  

  
  
  
  

add a comment | 

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33

$begingroup$

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.

share|cite|improve this answer

edited 6 hours ago

answered 13 hours ago

BowlOfRedBowlOfRed

17.4k22743

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  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
  8 hours ago
  
  
  

  
  
  
  

add a comment | 

33

$begingroup$

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.

share|cite|improve this answer

edited 6 hours ago

answered 13 hours ago

BowlOfRedBowlOfRed

17.4k22743

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  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
  8 hours ago
  
  
  

  
  
  
  

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.

share|cite|improve this answer

edited 6 hours ago

answered 13 hours ago

BowlOfRedBowlOfRed

17.4k22743

$endgroup$

share|cite|improve this answer

edited 6 hours ago

answered 13 hours ago

BowlOfRedBowlOfRed

17.4k22743

answered 13 hours ago

BowlOfRedBowlOfRed

17.4k22743

answered 13 hours ago

BowlOfRedBowlOfRed

17.4k22743

9

$begingroup$

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.

share|cite|improve this answer

edited 7 hours ago

answered 15 hours ago

DigiprocDigiproc

1,50848

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I'd word this as "inversely proportional to the distance the car traveled"
  $endgroup$
  – Nayuki
  7 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  oops, you're right, I fixed it
  $endgroup$
  – Digiproc
  7 hours ago
  

  
  
  
  

add a comment | 

9

$begingroup$

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.

share|cite|improve this answer

edited 7 hours ago

answered 15 hours ago

DigiprocDigiproc

1,50848

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I'd word this as "inversely proportional to the distance the car traveled"
  $endgroup$
  – Nayuki
  7 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  oops, you're right, I fixed it
  $endgroup$
  – Digiproc
  7 hours ago
  

  
  
  
  

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.

share|cite|improve this answer

edited 7 hours ago

answered 15 hours ago

DigiprocDigiproc

1,50848

$endgroup$

share|cite|improve this answer

edited 7 hours ago

answered 15 hours ago

DigiprocDigiproc

1,50848

answered 15 hours ago

DigiprocDigiproc

1,50848

answered 15 hours ago

DigiprocDigiproc

1,50848

5

$begingroup$

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.

share|cite|improve this answer

answered 11 hours ago

niels nielsenniels nielsen

20.3k53061

$endgroup$

• 
  
  
  

  
  4
  

  
  
  
  
  
  

  
  $begingroup$
  This is probably less significant than the thing BowlOfRed suggested, but I don't doubt it has a measurable effect.
  $endgroup$
  – wizzwizz4
  10 hours ago
  

  
  
  
  

add a comment | 

5

$begingroup$

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.

share|cite|improve this answer

answered 11 hours ago

niels nielsenniels nielsen

20.3k53061

$endgroup$

• 
  
  
  

  
  4
  

  
  
  
  
  
  

  
  $begingroup$
  This is probably less significant than the thing BowlOfRed suggested, but I don't doubt it has a measurable effect.
  $endgroup$
  – wizzwizz4
  10 hours ago
  

  
  
  
  

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.

share|cite|improve this answer

answered 11 hours ago

niels nielsenniels nielsen

20.3k53061

$endgroup$

share|cite|improve this answer

answered 11 hours ago

niels nielsenniels nielsen

20.3k53061

answered 11 hours ago

niels nielsenniels nielsen

20.3k53061

answered 11 hours ago

niels nielsenniels nielsen

20.3k53061

1

$begingroup$

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.

share|cite|improve this answer

answered 10 hours ago

TopCatTopCat

111

New contributor

TopCat is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

• 
  
  
  

  
  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
  9 hours ago
  

  
  
  
  

add a comment | 

1

$begingroup$

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.

share|cite|improve this answer

answered 10 hours ago

TopCatTopCat

111

New contributor

TopCat is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

• 
  
  
  

  
  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
  9 hours ago
  

  
  
  
  

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.

share|cite|improve this answer

answered 10 hours ago

TopCatTopCat

111

New contributor

TopCat is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

share|cite|improve this answer

answered 10 hours ago

TopCatTopCat

111

New contributor

TopCat is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

answered 10 hours ago

TopCatTopCat

111

New contributor

TopCat is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

answered 10 hours ago

TopCatTopCat

111