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This question is a duplicate of Does turning sharply on a bicycle conserve more energy than a wide turn?, but the accepted answer to that question is somewhat incorrect, so it's probably safe for now. In general, in the absence of friction, the force of turning applied to something like a bike curving through a turn is approximately perpendicular to the path, and hence does no work, and the magnitude of the velocity of the object will be unchanged.

However, realistically, friction losses will probably play a decent part in it. So, if I were you I'd try setting up the simulation so that the loss in speed is governed by $$\frac{d|v|}{dt}=-\alpha\left|v\frac{d\theta}{dt}\right|$$ or something of the sort, where $\alpha$ is a constant of proportionality which encodes the frictional energy loss associated with turning around the car and $\frac{d\theta}{dt}=\omega$ is the angular velocity of the car.

This is a model differential equation which states that the speed-damping coefficient is $\alpha\left|\omega(t)\right|$, ie, the sharper you turn, the more speed you lose due to friction. This way, when you take a sharp turn, you lose a lot of speed from inefficiency and friction, but when you take a slow, soft turn, you don't lose as much speed.

Try various values of $\alpha$, and if possible run tests. Which ones look realistic, and which ones look goofy? Testing and debugging to find something which looks "right" is probably the most important part.

This question is a duplicate of Does turning sharply on a bicycle conserve more energy than a wide turn?, but the accepted answer to that question is somewhat incorrect, so it's probably safe for now. In general, in the absence of friction, the force of turning applied to something like a bike curving through a turn is approximately perpendicular to the path, and hence does no work, and the magnitude of the velocity of the object will be unchanged.

However, realistically, friction losses will probably play a decent part in it. So, if I were you I'd try setting up the simulation so that the loss in speed is governed by $$\frac{d|v|}{dt}=-\alpha\left|v\frac{d\theta}{dt}\right|$$ or something of the sort, where $\alpha$ is a constant of proportionality which encodes the frictional energy loss associated with turning around the car and $\frac{d\theta}{dt}=\omega$ is the angular velocity of the car.

This is a model differential equation which states that the speed-damping coefficient is $\alpha\left|\omega(t)\right|$, ie, the sharper you turn, the more speed you lose due to friction. This way, when you take a sharp turn, you lose a lot of speed from inefficiency and friction, but when you take a slow, soft turn, you don't lose as much speed.

Try various values of $\alpha$, and if possible run tests. Which ones look realistic, and which ones look goofy? Testing and debugging to find something which looks "right" is probably the most important part.

This question is a duplicate of Does turning sharply on a bicycle conserve more energy than a wide turn?, but the accepted answer to that question is incorrect, so it's probably safe for now. In general, in the absence of friction, the force of turning applied to something like a bike curving through a turn is approximately perpendicular to the path, and hence does no work, and the magnitude of the velocity of the object will be unchanged.

However, realistically, friction losses will probably play a decent part in it. So, if I were you I'd try setting up the simulation so that the tangential loss component is $$\frac{dv}{dt}=-\alpha v \left|\frac{d\theta}{dt}\right|$$ or something of the sort, where $\alpha$ is a constant of proportionality which encodes the frictional energy loss associated with turning around the car and $\frac{d\theta}{dt}=\omega$ is the angular velocity of the car.

This is a model differential equation which states that the speed-damping coefficient is $\alpha\left|\omega(t)\right|$, ie, the sharper you turn, the more speed you lose due to friction. This way, when you take a sharp turn, you lose a lot of speed from inefficiency and friction, but when you take a slow, soft turn, you don't lose as much speed.

If you need this in 2D rectilinear coordinates, you can just use the fact that $\theta(t)=\mbox{ArcTan}[x'(t),y'(t)]$ in Mathematica notation, from which it follows that

$$x''(t)=-\alpha x'(t)\left|\frac{x'(t) y''(t)-x''(t) y'(t)}{x'(t)^2+y'(t)^2}\right|$$ and $$y''(t)=-\alpha y'(t)\left|\frac{x'(t) y''(t)-x''(t) y'(t)}{x'(t)^2+y'(t)^2}\right|.$$

However, if you're programming a car simulator then you're probably thinking more along the lines of "when I turn the joystick this far, the value of $d\theta/dt$ becomes this much", etc, so the equation I wrote at the beginning may be more useful for you than the rectilinear ones.

Try various values of $\alpha$, and if possible run tests. Which ones look realistic, and which ones look goofy? Testing and debugging to find something which looks "right" is probably the most important part.

This question is a duplicate of Does turning sharply on a bicycle conserve more energy than a wide turn?, but the accepted answer to that question is somewhat incorrect, so it's probably safe for now. In general, in the absence of friction, the force of turning applied to something like a bike curving through a turn is approximately perpendicular to the path, and hence does no work, and the magnitude of the velocity of the object will be unchanged.

However, realistically, friction losses will probably play a decent part in it. So, if I were you I'd try setting up the simulation so that the loss in speed is governed by $$\frac{d|v|}{dt}=-\alpha\left|v\frac{d\theta}{dt}\right|$$ or something of the sort, where $\alpha$ is a constant of proportionality which encodes the frictional energy loss associated with turning around the car and $\frac{d\theta}{dt}=\omega$ is the angular velocity of the car.

This is a model differential equation which states that the speed-damping coefficient is $\alpha\left|\omega(t)\right|$, ie, the sharper you turn, the more speed you lose due to friction. This way, when you take a sharp turn, you lose a lot of speed from inefficiency and friction, but when you take a slow, soft turn, you don't lose as much speed.

Try various values of $\alpha$, and if possible run tests. Which ones look realistic, and which ones look goofy? Testing and debugging to find something which looks "right" is probably the most important part.

This question is a duplicate of Does turning sharply on a bicycle conserve more energy than a wide turn?, but the accepted answer to that question is incorrect, so it's probably safe for now. In general, in the absence of friction, the force of turning applied to something like a bike curving through a turn is approximately perpendicular to the path, and hence does no work.

However, realistically, friction losses will probably play a decent part in it. So, if I were you I'd try setting up the simulation so that $$\frac{dv}{dt}=-\alpha v \left|\frac{d\theta}{dt}\right|$$ or something of the sort, where $\alpha$ is a constant of proportionality which encodes the frictional energy loss associated with turning around the car and $\frac{d\theta}{dt}=\omega$ is the angular velocity of the car.

This is a model differential equation which states that the speed-damping coefficient is $\alpha\left|\omega(t)\right|$, ie, the sharper you turn, the more speed you lose due to friction. This way, when you take a sharp turn, you lose a lot of speed from inefficiency and friction, but when you take a slow, soft turn, you don't lose as much speed.

If you need this in 2D rectilinear coordinates, you can just use the fact that $\theta(t)=\mbox{ArcTan}[x'(t),y'(t)]$ in Mathematica notation, from which it follows that

$$x''(t)=-\alpha x'(t)\left|\frac{x'(t) y''(t)-x''(t) y'(t)}{x'(t)^2+y'(t)^2}\right|$$ and $$y''(t)=-\alpha y'(t)\left|\frac{x'(t) y''(t)-x''(t) y'(t)}{x'(t)^2+y'(t)^2}\right|.$$

Try various values of $\alpha$, and if possible run tests. Which ones look realistic, and which ones look goofy? Testing and debugging to find something which looks "right" is probably the most important part.

This question is a duplicate of Does turning sharply on a bicycle conserve more energy than a wide turn?, but the accepted answer to that question is incorrect, so it's probably safe for now. In general, in the absence of friction, the force of turning applied to something like a bike curving through a turn is approximately perpendicular to the path, and hence does no work, and the magnitude of the velocity of the object will be unchanged.

However, realistically, friction losses will probably play a decent part in it. So, if I were you I'd try setting up the simulation so that the tangential loss component is $$\frac{dv}{dt}=-\alpha v \left|\frac{d\theta}{dt}\right|$$ or something of the sort, where $\alpha$ is a constant of proportionality which encodes the frictional energy loss associated with turning around the car and $\frac{d\theta}{dt}=\omega$ is the angular velocity of the car.

This is a model differential equation which states that the speed-damping coefficient is $\alpha\left|\omega(t)\right|$, ie, the sharper you turn, the more speed you lose due to friction. This way, when you take a sharp turn, you lose a lot of speed from inefficiency and friction, but when you take a slow, soft turn, you don't lose as much speed.

If you need this in 2D rectilinear coordinates, you can just use the fact that $\theta(t)=\mbox{ArcTan}[x'(t),y'(t)]$ in Mathematica notation, from which it follows that

$$x''(t)=-\alpha x'(t)\left|\frac{x'(t) y''(t)-x''(t) y'(t)}{x'(t)^2+y'(t)^2}\right|$$ and $$y''(t)=-\alpha y'(t)\left|\frac{x'(t) y''(t)-x''(t) y'(t)}{x'(t)^2+y'(t)^2}\right|.$$

However, if you're programming a car simulator then you're probably thinking more along the lines of "when I turn the joystick this far, the value of $d\theta/dt$ becomes this much", etc, so the equation I wrote at the beginning may be more useful for you than the rectilinear ones.

Try various values of $\alpha$, and if possible run tests. Which ones look realistic, and which ones look goofy? Testing and debugging to find something which looks "right" is probably the most important part.