Path of a simple turning car?












3












$begingroup$


I am planning to build a small car that needs to travel through three specific points. Obviously, the car will need to travel in a partial circle to do this. In order to do this, I need to calculate the angle for the front wheels. I do not plan on using Ackermann steering, instead, the whole axle will turn slightly with the wheels attached. The angles will all be under ~15 degrees, and as such, both wheels will remain a relatively similar distance from the back axle < ~1-inch difference on a ~10-inch car.



Knowing the length of the car (I'd assume wheel center to wheel center would be needed), as well as the axles length (the front and back axle may or may not end up being different lengths, ie. tricycle vs car), and the angle of rotation of the wheels how can I determine an equation for the path the car will travel through? Additionally, would the fact that the entire axle is tilting slightly, meaning the wheels are not equidistance from the back be significant? If so, is it possible to determine the above calculations knowing that value as well?



Thank you!










share|cite|improve this question











$endgroup$

















    3












    $begingroup$


    I am planning to build a small car that needs to travel through three specific points. Obviously, the car will need to travel in a partial circle to do this. In order to do this, I need to calculate the angle for the front wheels. I do not plan on using Ackermann steering, instead, the whole axle will turn slightly with the wheels attached. The angles will all be under ~15 degrees, and as such, both wheels will remain a relatively similar distance from the back axle < ~1-inch difference on a ~10-inch car.



    Knowing the length of the car (I'd assume wheel center to wheel center would be needed), as well as the axles length (the front and back axle may or may not end up being different lengths, ie. tricycle vs car), and the angle of rotation of the wheels how can I determine an equation for the path the car will travel through? Additionally, would the fact that the entire axle is tilting slightly, meaning the wheels are not equidistance from the back be significant? If so, is it possible to determine the above calculations knowing that value as well?



    Thank you!










    share|cite|improve this question











    $endgroup$















      3












      3








      3





      $begingroup$


      I am planning to build a small car that needs to travel through three specific points. Obviously, the car will need to travel in a partial circle to do this. In order to do this, I need to calculate the angle for the front wheels. I do not plan on using Ackermann steering, instead, the whole axle will turn slightly with the wheels attached. The angles will all be under ~15 degrees, and as such, both wheels will remain a relatively similar distance from the back axle < ~1-inch difference on a ~10-inch car.



      Knowing the length of the car (I'd assume wheel center to wheel center would be needed), as well as the axles length (the front and back axle may or may not end up being different lengths, ie. tricycle vs car), and the angle of rotation of the wheels how can I determine an equation for the path the car will travel through? Additionally, would the fact that the entire axle is tilting slightly, meaning the wheels are not equidistance from the back be significant? If so, is it possible to determine the above calculations knowing that value as well?



      Thank you!










      share|cite|improve this question











      $endgroup$




      I am planning to build a small car that needs to travel through three specific points. Obviously, the car will need to travel in a partial circle to do this. In order to do this, I need to calculate the angle for the front wheels. I do not plan on using Ackermann steering, instead, the whole axle will turn slightly with the wheels attached. The angles will all be under ~15 degrees, and as such, both wheels will remain a relatively similar distance from the back axle < ~1-inch difference on a ~10-inch car.



      Knowing the length of the car (I'd assume wheel center to wheel center would be needed), as well as the axles length (the front and back axle may or may not end up being different lengths, ie. tricycle vs car), and the angle of rotation of the wheels how can I determine an equation for the path the car will travel through? Additionally, would the fact that the entire axle is tilting slightly, meaning the wheels are not equidistance from the back be significant? If so, is it possible to determine the above calculations knowing that value as well?



      Thank you!







      geometry trigonometry applications






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      edited Dec 28 '18 at 21:28







      Benjamin Zydney

















      asked Dec 28 '18 at 20:34









      Benjamin ZydneyBenjamin Zydney

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          1 Answer
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          $begingroup$

          If you prolong the axes of the rear and of the front wheels, the point where they meet is the center of rotation.



          Since the rear axle is fixed and steering is obtained by rotating the whole front axle, we get a scheme as the one shown.



          steer_car_1



          It is clear from the picture that, when the steering angle $beta$ has a large value, the rear and front turning radii
          are sensibly different: an effect that is well learnt by fresh drivers when parking backwards.



          The (ideal) kynematics of the car obeys to the following set of equations
          $$ bbox[lightyellow] {
          left{ matrix{
          {bf u}_{,T} = left( {cos alpha ,sin alpha } right)quad {bf u}_{,H}
          = left( {cos left( {alpha + beta } right),sin left( {alpha + beta } right)} right) hfill cr
          {bf n}_{,T} = left( { - sin alpha ,cos alpha } right)quad {bf n}_{,H}
          = left( { - sin left( {alpha + beta } right),cos left( {alpha + beta } right)} right) hfill cr
          {bf H} = {bf T} + L,{bf u}_{,T} hfill cr
          0 = {bf n}_{,T} cdot ,{d over {dt}}{bf T} = {bf n}_{,H} cdot {d over {dt}}{bf H}
          = {bf n}_{,H} cdot left( {{d over {dt}}{bf T} + L,{d over {dt}}{bf u}_{,T} } right) hfill cr} right.
          }$$

          where $alpha$ and $beta$ and the other parameters are function of time.



          Now, if we assume that the car has a rear traction which provides, through a differential gear, a constant
          average speed $v$ at the center of the axle $T$, then
          $$
          {d over {dt}}{bf T} = v,{bf u}_{,T}
          $$

          and we get an equation linking $alpha$ and $beta$ as
          $$
          eqalign{
          & 0 = {bf n}_{,H} cdot left( {{d over {dt}}{bf T} + L,{d over {dt}}{bf u}_{,T} } right) = cr
          & = left( { - sin left( {alpha + beta } right),cos left( {alpha + beta } right)} right) cdot left( {v,left( {cos alpha ,sin alpha } right)
          + Lleft( { - sin alpha ,cos alpha } right){{dalpha } over {dt}}} right) = cr
          & = - vsin left( {alpha + beta } right)cos alpha + L{{dalpha } over {dt}}sin left( {alpha + beta } right)sin alpha + cr
          & + vcos left( {alpha + beta } right)sin alpha + L{{dalpha } over {dt}}cos left( {alpha + beta } right)cos alpha = cr
          & = - vsin beta + L{{dalpha } over {dt}}cos beta cr}
          $$

          i.e.:
          $$ bbox[lightyellow] {
          {{dalpha } over {dt}} = {v over L}tan beta
          }$$



          For a constant $beta$ and $alpha(0)=0$ we get
          $$ bbox[lightyellow] {
          left{ matrix{
          alpha (t) = left( {{v over L}tan beta } right)t = omega ,t hfill cr
          {d over {dt}}{bf T} = v,{bf u}_{,T} = v,left( {cos left( {omega ,t} right),sin left( {omega ,t} right)} right) hfill cr
          {bf T}(t) = {bf T}_{,0} + {v over omega }left( {sin left( {omega ,t} right),1 - cos left( {omega t} right)} right) hfill cr
          {bf H}(t) = {bf T}(t) + L,left( {cos left( {omega ,t} right),sin left( {omega ,t} right)} right) hfill cr} right.
          }$$



          The trajectories of $T$ and $H$ are concentric circles



          steer_car_2






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            +1: I would just add that as the control system should recalculate the angle for the steering axle at frequent intervals based on the actual position of the car, any errors in the approximate calculation should be self-correctding.
            $endgroup$
            – Rob Arthan
            Dec 28 '18 at 21:03












          • $begingroup$
            Those refinements are not primarily for the reasons you state -- they are geometrically necessary, because the front wheels turn in place, not on the end of a common axle. The inner wheel must turn more sharply then the outer wheel, so that their two normals intersect the line of the rear axle at the same place.
            $endgroup$
            – TonyK
            Dec 28 '18 at 21:40










          • $begingroup$
            Perhaps "The point where the line extending the front axle intersects with the line extending the rear axle, is the center of rotation."? (I'm not a native English speaker, and the "prolong" threw me off.)
            $endgroup$
            – Nominal Animal
            Dec 29 '18 at 10:24










          • $begingroup$
            @Nominal Animal : yes the meaning is that, I used the latin derivation of " extending in length"
            $endgroup$
            – G Cab
            Dec 29 '18 at 15:02












          • $begingroup$
            @TonyK: you are fully right: I kept it short!
            $endgroup$
            – G Cab
            Dec 29 '18 at 15:06











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          2












          $begingroup$

          If you prolong the axes of the rear and of the front wheels, the point where they meet is the center of rotation.



          Since the rear axle is fixed and steering is obtained by rotating the whole front axle, we get a scheme as the one shown.



          steer_car_1



          It is clear from the picture that, when the steering angle $beta$ has a large value, the rear and front turning radii
          are sensibly different: an effect that is well learnt by fresh drivers when parking backwards.



          The (ideal) kynematics of the car obeys to the following set of equations
          $$ bbox[lightyellow] {
          left{ matrix{
          {bf u}_{,T} = left( {cos alpha ,sin alpha } right)quad {bf u}_{,H}
          = left( {cos left( {alpha + beta } right),sin left( {alpha + beta } right)} right) hfill cr
          {bf n}_{,T} = left( { - sin alpha ,cos alpha } right)quad {bf n}_{,H}
          = left( { - sin left( {alpha + beta } right),cos left( {alpha + beta } right)} right) hfill cr
          {bf H} = {bf T} + L,{bf u}_{,T} hfill cr
          0 = {bf n}_{,T} cdot ,{d over {dt}}{bf T} = {bf n}_{,H} cdot {d over {dt}}{bf H}
          = {bf n}_{,H} cdot left( {{d over {dt}}{bf T} + L,{d over {dt}}{bf u}_{,T} } right) hfill cr} right.
          }$$

          where $alpha$ and $beta$ and the other parameters are function of time.



          Now, if we assume that the car has a rear traction which provides, through a differential gear, a constant
          average speed $v$ at the center of the axle $T$, then
          $$
          {d over {dt}}{bf T} = v,{bf u}_{,T}
          $$

          and we get an equation linking $alpha$ and $beta$ as
          $$
          eqalign{
          & 0 = {bf n}_{,H} cdot left( {{d over {dt}}{bf T} + L,{d over {dt}}{bf u}_{,T} } right) = cr
          & = left( { - sin left( {alpha + beta } right),cos left( {alpha + beta } right)} right) cdot left( {v,left( {cos alpha ,sin alpha } right)
          + Lleft( { - sin alpha ,cos alpha } right){{dalpha } over {dt}}} right) = cr
          & = - vsin left( {alpha + beta } right)cos alpha + L{{dalpha } over {dt}}sin left( {alpha + beta } right)sin alpha + cr
          & + vcos left( {alpha + beta } right)sin alpha + L{{dalpha } over {dt}}cos left( {alpha + beta } right)cos alpha = cr
          & = - vsin beta + L{{dalpha } over {dt}}cos beta cr}
          $$

          i.e.:
          $$ bbox[lightyellow] {
          {{dalpha } over {dt}} = {v over L}tan beta
          }$$



          For a constant $beta$ and $alpha(0)=0$ we get
          $$ bbox[lightyellow] {
          left{ matrix{
          alpha (t) = left( {{v over L}tan beta } right)t = omega ,t hfill cr
          {d over {dt}}{bf T} = v,{bf u}_{,T} = v,left( {cos left( {omega ,t} right),sin left( {omega ,t} right)} right) hfill cr
          {bf T}(t) = {bf T}_{,0} + {v over omega }left( {sin left( {omega ,t} right),1 - cos left( {omega t} right)} right) hfill cr
          {bf H}(t) = {bf T}(t) + L,left( {cos left( {omega ,t} right),sin left( {omega ,t} right)} right) hfill cr} right.
          }$$



          The trajectories of $T$ and $H$ are concentric circles



          steer_car_2






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            +1: I would just add that as the control system should recalculate the angle for the steering axle at frequent intervals based on the actual position of the car, any errors in the approximate calculation should be self-correctding.
            $endgroup$
            – Rob Arthan
            Dec 28 '18 at 21:03












          • $begingroup$
            Those refinements are not primarily for the reasons you state -- they are geometrically necessary, because the front wheels turn in place, not on the end of a common axle. The inner wheel must turn more sharply then the outer wheel, so that their two normals intersect the line of the rear axle at the same place.
            $endgroup$
            – TonyK
            Dec 28 '18 at 21:40










          • $begingroup$
            Perhaps "The point where the line extending the front axle intersects with the line extending the rear axle, is the center of rotation."? (I'm not a native English speaker, and the "prolong" threw me off.)
            $endgroup$
            – Nominal Animal
            Dec 29 '18 at 10:24










          • $begingroup$
            @Nominal Animal : yes the meaning is that, I used the latin derivation of " extending in length"
            $endgroup$
            – G Cab
            Dec 29 '18 at 15:02












          • $begingroup$
            @TonyK: you are fully right: I kept it short!
            $endgroup$
            – G Cab
            Dec 29 '18 at 15:06
















          2












          $begingroup$

          If you prolong the axes of the rear and of the front wheels, the point where they meet is the center of rotation.



          Since the rear axle is fixed and steering is obtained by rotating the whole front axle, we get a scheme as the one shown.



          steer_car_1



          It is clear from the picture that, when the steering angle $beta$ has a large value, the rear and front turning radii
          are sensibly different: an effect that is well learnt by fresh drivers when parking backwards.



          The (ideal) kynematics of the car obeys to the following set of equations
          $$ bbox[lightyellow] {
          left{ matrix{
          {bf u}_{,T} = left( {cos alpha ,sin alpha } right)quad {bf u}_{,H}
          = left( {cos left( {alpha + beta } right),sin left( {alpha + beta } right)} right) hfill cr
          {bf n}_{,T} = left( { - sin alpha ,cos alpha } right)quad {bf n}_{,H}
          = left( { - sin left( {alpha + beta } right),cos left( {alpha + beta } right)} right) hfill cr
          {bf H} = {bf T} + L,{bf u}_{,T} hfill cr
          0 = {bf n}_{,T} cdot ,{d over {dt}}{bf T} = {bf n}_{,H} cdot {d over {dt}}{bf H}
          = {bf n}_{,H} cdot left( {{d over {dt}}{bf T} + L,{d over {dt}}{bf u}_{,T} } right) hfill cr} right.
          }$$

          where $alpha$ and $beta$ and the other parameters are function of time.



          Now, if we assume that the car has a rear traction which provides, through a differential gear, a constant
          average speed $v$ at the center of the axle $T$, then
          $$
          {d over {dt}}{bf T} = v,{bf u}_{,T}
          $$

          and we get an equation linking $alpha$ and $beta$ as
          $$
          eqalign{
          & 0 = {bf n}_{,H} cdot left( {{d over {dt}}{bf T} + L,{d over {dt}}{bf u}_{,T} } right) = cr
          & = left( { - sin left( {alpha + beta } right),cos left( {alpha + beta } right)} right) cdot left( {v,left( {cos alpha ,sin alpha } right)
          + Lleft( { - sin alpha ,cos alpha } right){{dalpha } over {dt}}} right) = cr
          & = - vsin left( {alpha + beta } right)cos alpha + L{{dalpha } over {dt}}sin left( {alpha + beta } right)sin alpha + cr
          & + vcos left( {alpha + beta } right)sin alpha + L{{dalpha } over {dt}}cos left( {alpha + beta } right)cos alpha = cr
          & = - vsin beta + L{{dalpha } over {dt}}cos beta cr}
          $$

          i.e.:
          $$ bbox[lightyellow] {
          {{dalpha } over {dt}} = {v over L}tan beta
          }$$



          For a constant $beta$ and $alpha(0)=0$ we get
          $$ bbox[lightyellow] {
          left{ matrix{
          alpha (t) = left( {{v over L}tan beta } right)t = omega ,t hfill cr
          {d over {dt}}{bf T} = v,{bf u}_{,T} = v,left( {cos left( {omega ,t} right),sin left( {omega ,t} right)} right) hfill cr
          {bf T}(t) = {bf T}_{,0} + {v over omega }left( {sin left( {omega ,t} right),1 - cos left( {omega t} right)} right) hfill cr
          {bf H}(t) = {bf T}(t) + L,left( {cos left( {omega ,t} right),sin left( {omega ,t} right)} right) hfill cr} right.
          }$$



          The trajectories of $T$ and $H$ are concentric circles



          steer_car_2






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            +1: I would just add that as the control system should recalculate the angle for the steering axle at frequent intervals based on the actual position of the car, any errors in the approximate calculation should be self-correctding.
            $endgroup$
            – Rob Arthan
            Dec 28 '18 at 21:03












          • $begingroup$
            Those refinements are not primarily for the reasons you state -- they are geometrically necessary, because the front wheels turn in place, not on the end of a common axle. The inner wheel must turn more sharply then the outer wheel, so that their two normals intersect the line of the rear axle at the same place.
            $endgroup$
            – TonyK
            Dec 28 '18 at 21:40










          • $begingroup$
            Perhaps "The point where the line extending the front axle intersects with the line extending the rear axle, is the center of rotation."? (I'm not a native English speaker, and the "prolong" threw me off.)
            $endgroup$
            – Nominal Animal
            Dec 29 '18 at 10:24










          • $begingroup$
            @Nominal Animal : yes the meaning is that, I used the latin derivation of " extending in length"
            $endgroup$
            – G Cab
            Dec 29 '18 at 15:02












          • $begingroup$
            @TonyK: you are fully right: I kept it short!
            $endgroup$
            – G Cab
            Dec 29 '18 at 15:06














          2












          2








          2





          $begingroup$

          If you prolong the axes of the rear and of the front wheels, the point where they meet is the center of rotation.



          Since the rear axle is fixed and steering is obtained by rotating the whole front axle, we get a scheme as the one shown.



          steer_car_1



          It is clear from the picture that, when the steering angle $beta$ has a large value, the rear and front turning radii
          are sensibly different: an effect that is well learnt by fresh drivers when parking backwards.



          The (ideal) kynematics of the car obeys to the following set of equations
          $$ bbox[lightyellow] {
          left{ matrix{
          {bf u}_{,T} = left( {cos alpha ,sin alpha } right)quad {bf u}_{,H}
          = left( {cos left( {alpha + beta } right),sin left( {alpha + beta } right)} right) hfill cr
          {bf n}_{,T} = left( { - sin alpha ,cos alpha } right)quad {bf n}_{,H}
          = left( { - sin left( {alpha + beta } right),cos left( {alpha + beta } right)} right) hfill cr
          {bf H} = {bf T} + L,{bf u}_{,T} hfill cr
          0 = {bf n}_{,T} cdot ,{d over {dt}}{bf T} = {bf n}_{,H} cdot {d over {dt}}{bf H}
          = {bf n}_{,H} cdot left( {{d over {dt}}{bf T} + L,{d over {dt}}{bf u}_{,T} } right) hfill cr} right.
          }$$

          where $alpha$ and $beta$ and the other parameters are function of time.



          Now, if we assume that the car has a rear traction which provides, through a differential gear, a constant
          average speed $v$ at the center of the axle $T$, then
          $$
          {d over {dt}}{bf T} = v,{bf u}_{,T}
          $$

          and we get an equation linking $alpha$ and $beta$ as
          $$
          eqalign{
          & 0 = {bf n}_{,H} cdot left( {{d over {dt}}{bf T} + L,{d over {dt}}{bf u}_{,T} } right) = cr
          & = left( { - sin left( {alpha + beta } right),cos left( {alpha + beta } right)} right) cdot left( {v,left( {cos alpha ,sin alpha } right)
          + Lleft( { - sin alpha ,cos alpha } right){{dalpha } over {dt}}} right) = cr
          & = - vsin left( {alpha + beta } right)cos alpha + L{{dalpha } over {dt}}sin left( {alpha + beta } right)sin alpha + cr
          & + vcos left( {alpha + beta } right)sin alpha + L{{dalpha } over {dt}}cos left( {alpha + beta } right)cos alpha = cr
          & = - vsin beta + L{{dalpha } over {dt}}cos beta cr}
          $$

          i.e.:
          $$ bbox[lightyellow] {
          {{dalpha } over {dt}} = {v over L}tan beta
          }$$



          For a constant $beta$ and $alpha(0)=0$ we get
          $$ bbox[lightyellow] {
          left{ matrix{
          alpha (t) = left( {{v over L}tan beta } right)t = omega ,t hfill cr
          {d over {dt}}{bf T} = v,{bf u}_{,T} = v,left( {cos left( {omega ,t} right),sin left( {omega ,t} right)} right) hfill cr
          {bf T}(t) = {bf T}_{,0} + {v over omega }left( {sin left( {omega ,t} right),1 - cos left( {omega t} right)} right) hfill cr
          {bf H}(t) = {bf T}(t) + L,left( {cos left( {omega ,t} right),sin left( {omega ,t} right)} right) hfill cr} right.
          }$$



          The trajectories of $T$ and $H$ are concentric circles



          steer_car_2






          share|cite|improve this answer











          $endgroup$



          If you prolong the axes of the rear and of the front wheels, the point where they meet is the center of rotation.



          Since the rear axle is fixed and steering is obtained by rotating the whole front axle, we get a scheme as the one shown.



          steer_car_1



          It is clear from the picture that, when the steering angle $beta$ has a large value, the rear and front turning radii
          are sensibly different: an effect that is well learnt by fresh drivers when parking backwards.



          The (ideal) kynematics of the car obeys to the following set of equations
          $$ bbox[lightyellow] {
          left{ matrix{
          {bf u}_{,T} = left( {cos alpha ,sin alpha } right)quad {bf u}_{,H}
          = left( {cos left( {alpha + beta } right),sin left( {alpha + beta } right)} right) hfill cr
          {bf n}_{,T} = left( { - sin alpha ,cos alpha } right)quad {bf n}_{,H}
          = left( { - sin left( {alpha + beta } right),cos left( {alpha + beta } right)} right) hfill cr
          {bf H} = {bf T} + L,{bf u}_{,T} hfill cr
          0 = {bf n}_{,T} cdot ,{d over {dt}}{bf T} = {bf n}_{,H} cdot {d over {dt}}{bf H}
          = {bf n}_{,H} cdot left( {{d over {dt}}{bf T} + L,{d over {dt}}{bf u}_{,T} } right) hfill cr} right.
          }$$

          where $alpha$ and $beta$ and the other parameters are function of time.



          Now, if we assume that the car has a rear traction which provides, through a differential gear, a constant
          average speed $v$ at the center of the axle $T$, then
          $$
          {d over {dt}}{bf T} = v,{bf u}_{,T}
          $$

          and we get an equation linking $alpha$ and $beta$ as
          $$
          eqalign{
          & 0 = {bf n}_{,H} cdot left( {{d over {dt}}{bf T} + L,{d over {dt}}{bf u}_{,T} } right) = cr
          & = left( { - sin left( {alpha + beta } right),cos left( {alpha + beta } right)} right) cdot left( {v,left( {cos alpha ,sin alpha } right)
          + Lleft( { - sin alpha ,cos alpha } right){{dalpha } over {dt}}} right) = cr
          & = - vsin left( {alpha + beta } right)cos alpha + L{{dalpha } over {dt}}sin left( {alpha + beta } right)sin alpha + cr
          & + vcos left( {alpha + beta } right)sin alpha + L{{dalpha } over {dt}}cos left( {alpha + beta } right)cos alpha = cr
          & = - vsin beta + L{{dalpha } over {dt}}cos beta cr}
          $$

          i.e.:
          $$ bbox[lightyellow] {
          {{dalpha } over {dt}} = {v over L}tan beta
          }$$



          For a constant $beta$ and $alpha(0)=0$ we get
          $$ bbox[lightyellow] {
          left{ matrix{
          alpha (t) = left( {{v over L}tan beta } right)t = omega ,t hfill cr
          {d over {dt}}{bf T} = v,{bf u}_{,T} = v,left( {cos left( {omega ,t} right),sin left( {omega ,t} right)} right) hfill cr
          {bf T}(t) = {bf T}_{,0} + {v over omega }left( {sin left( {omega ,t} right),1 - cos left( {omega t} right)} right) hfill cr
          {bf H}(t) = {bf T}(t) + L,left( {cos left( {omega ,t} right),sin left( {omega ,t} right)} right) hfill cr} right.
          }$$



          The trajectories of $T$ and $H$ are concentric circles



          steer_car_2







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Jan 1 at 15:07

























          answered Dec 28 '18 at 20:52









          G CabG Cab

          19.7k31339




          19.7k31339












          • $begingroup$
            +1: I would just add that as the control system should recalculate the angle for the steering axle at frequent intervals based on the actual position of the car, any errors in the approximate calculation should be self-correctding.
            $endgroup$
            – Rob Arthan
            Dec 28 '18 at 21:03












          • $begingroup$
            Those refinements are not primarily for the reasons you state -- they are geometrically necessary, because the front wheels turn in place, not on the end of a common axle. The inner wheel must turn more sharply then the outer wheel, so that their two normals intersect the line of the rear axle at the same place.
            $endgroup$
            – TonyK
            Dec 28 '18 at 21:40










          • $begingroup$
            Perhaps "The point where the line extending the front axle intersects with the line extending the rear axle, is the center of rotation."? (I'm not a native English speaker, and the "prolong" threw me off.)
            $endgroup$
            – Nominal Animal
            Dec 29 '18 at 10:24










          • $begingroup$
            @Nominal Animal : yes the meaning is that, I used the latin derivation of " extending in length"
            $endgroup$
            – G Cab
            Dec 29 '18 at 15:02












          • $begingroup$
            @TonyK: you are fully right: I kept it short!
            $endgroup$
            – G Cab
            Dec 29 '18 at 15:06


















          • $begingroup$
            +1: I would just add that as the control system should recalculate the angle for the steering axle at frequent intervals based on the actual position of the car, any errors in the approximate calculation should be self-correctding.
            $endgroup$
            – Rob Arthan
            Dec 28 '18 at 21:03












          • $begingroup$
            Those refinements are not primarily for the reasons you state -- they are geometrically necessary, because the front wheels turn in place, not on the end of a common axle. The inner wheel must turn more sharply then the outer wheel, so that their two normals intersect the line of the rear axle at the same place.
            $endgroup$
            – TonyK
            Dec 28 '18 at 21:40










          • $begingroup$
            Perhaps "The point where the line extending the front axle intersects with the line extending the rear axle, is the center of rotation."? (I'm not a native English speaker, and the "prolong" threw me off.)
            $endgroup$
            – Nominal Animal
            Dec 29 '18 at 10:24










          • $begingroup$
            @Nominal Animal : yes the meaning is that, I used the latin derivation of " extending in length"
            $endgroup$
            – G Cab
            Dec 29 '18 at 15:02












          • $begingroup$
            @TonyK: you are fully right: I kept it short!
            $endgroup$
            – G Cab
            Dec 29 '18 at 15:06
















          $begingroup$
          +1: I would just add that as the control system should recalculate the angle for the steering axle at frequent intervals based on the actual position of the car, any errors in the approximate calculation should be self-correctding.
          $endgroup$
          – Rob Arthan
          Dec 28 '18 at 21:03






          $begingroup$
          +1: I would just add that as the control system should recalculate the angle for the steering axle at frequent intervals based on the actual position of the car, any errors in the approximate calculation should be self-correctding.
          $endgroup$
          – Rob Arthan
          Dec 28 '18 at 21:03














          $begingroup$
          Those refinements are not primarily for the reasons you state -- they are geometrically necessary, because the front wheels turn in place, not on the end of a common axle. The inner wheel must turn more sharply then the outer wheel, so that their two normals intersect the line of the rear axle at the same place.
          $endgroup$
          – TonyK
          Dec 28 '18 at 21:40




          $begingroup$
          Those refinements are not primarily for the reasons you state -- they are geometrically necessary, because the front wheels turn in place, not on the end of a common axle. The inner wheel must turn more sharply then the outer wheel, so that their two normals intersect the line of the rear axle at the same place.
          $endgroup$
          – TonyK
          Dec 28 '18 at 21:40












          $begingroup$
          Perhaps "The point where the line extending the front axle intersects with the line extending the rear axle, is the center of rotation."? (I'm not a native English speaker, and the "prolong" threw me off.)
          $endgroup$
          – Nominal Animal
          Dec 29 '18 at 10:24




          $begingroup$
          Perhaps "The point where the line extending the front axle intersects with the line extending the rear axle, is the center of rotation."? (I'm not a native English speaker, and the "prolong" threw me off.)
          $endgroup$
          – Nominal Animal
          Dec 29 '18 at 10:24












          $begingroup$
          @Nominal Animal : yes the meaning is that, I used the latin derivation of " extending in length"
          $endgroup$
          – G Cab
          Dec 29 '18 at 15:02






          $begingroup$
          @Nominal Animal : yes the meaning is that, I used the latin derivation of " extending in length"
          $endgroup$
          – G Cab
          Dec 29 '18 at 15:02














          $begingroup$
          @TonyK: you are fully right: I kept it short!
          $endgroup$
          – G Cab
          Dec 29 '18 at 15:06




          $begingroup$
          @TonyK: you are fully right: I kept it short!
          $endgroup$
          – G Cab
          Dec 29 '18 at 15:06


















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