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










19












$begingroup$


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?










share|cite|improve this question











$endgroup$



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
If this question can be reworded to fit the rules in the help center, please edit the question.




















    19












    $begingroup$


    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?










    share|cite|improve this question











    $endgroup$



    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
    If this question can be reworded to fit the rules in the help center, please edit the question.


















      19












      19








      19


      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?










      share|cite|improve this question











      $endgroup$




      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






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Mar 8 at 19:30









      jezzo

      1594




      1594










      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
      If this question can be reworded to fit the rules in the help center, please edit the question.







      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
      If this question can be reworded to fit the rules in the help center, please edit the question.




















          4 Answers
          4






          active

          oldest

          votes


















          39












          $begingroup$

          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











          $endgroup$












          • $begingroup$
            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



















          10












          $begingroup$

          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











          $endgroup$












          • $begingroup$
            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


















          6












          $begingroup$

          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









          $endgroup$








          • 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


















          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









          $endgroup$








          • 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

















          4 Answers
          4






          active

          oldest

          votes








          4 Answers
          4






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes









          39












          $begingroup$

          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











          $endgroup$












          • $begingroup$
            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
















          39












          $begingroup$

          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











          $endgroup$












          • $begingroup$
            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














          39












          39








          39





          $begingroup$

          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











          $endgroup$



          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.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Mar 8 at 23:36

























          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

















          • $begingroup$
            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












          10












          $begingroup$

          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











          $endgroup$












          • $begingroup$
            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















          10












          $begingroup$

          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











          $endgroup$












          • $begingroup$
            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













          10












          10








          10





          $begingroup$

          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











          $endgroup$



          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.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Mar 8 at 22:56

























          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
















          • $begingroup$
            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











          6












          $begingroup$

          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









          $endgroup$








          • 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















          6












          $begingroup$

          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









          $endgroup$








          • 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













          6












          6








          6





          $begingroup$

          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









          $endgroup$



          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.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          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












          • 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











          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









          $endgroup$








          • 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















          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









          $endgroup$








          • 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













          1












          1








          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









          $endgroup$



          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.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          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












          • 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



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