Why does a car's steering wheel get lighter with increasing speed [on hold]












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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?










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put on hold as off-topic by David Z 2 days ago


This question appears to be off-topic. The users who voted to close gave this specific reason:


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    17












    $begingroup$


    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











    $endgroup$



    put on hold as off-topic by David Z 2 days ago


    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.



















      17












      17








      17


      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






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      edited Mar 8 at 19:30









      jezzo

      434




      434










      asked Mar 8 at 14:09









      securitydude5securitydude5

      1754




      1754




      put on hold as off-topic by David Z 2 days ago


      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.







      put on hold as off-topic by David Z 2 days ago


      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


















          36












          $begingroup$

          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



















          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









          $endgroup$









          • 4




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








          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
            Mar 8 at 20:10


















          4 Answers
          4






          active

          oldest

          votes








          4 Answers
          4






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes









          36












          $begingroup$

          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


















          36












          $begingroup$

          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
















          36












          36








          36





          $begingroup$

          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

          17.7k22744




          17.7k22744












          • $begingroup$
            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

          1,51849




          1,51849












          • $begingroup$
            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











          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









          $endgroup$









          • 4




            $begingroup$
            This is probably less significant than the thing BowlOfRed suggested, but I don't doubt it has a measurable effect.
            $endgroup$
            – wizzwizz4
            Mar 8 at 19:06
















          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









          $endgroup$









          • 4




            $begingroup$
            This is probably less significant than the thing BowlOfRed suggested, but I don't doubt it has a measurable effect.
            $endgroup$
            – wizzwizz4
            Mar 8 at 19:06














          5












          5








          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









          $endgroup$



          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

          20.5k53061




          20.5k53061








          • 4




            $begingroup$
            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














          • 4




            $begingroup$
            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








          4




          4




          $begingroup$
          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








          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
            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








          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
            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








          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.






          $endgroup$



          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








          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.









          share|cite|improve this answer



          share|cite|improve this answer






          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 Mar 8 at 19:58









          TopCatTopCat

          111




          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.





          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.






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








          • 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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