Steering Wheel Angle & Radius of Curvature

[Dorje]

Is there a formula which relates the angle of a steering wheel from normal position (theta) and the radius of curvature (r) of the arc that an average vehicle moves through? (assuming power steering, since that's the most common)

Such a formula would be useful in calculating centripetal forces acting on vehicles as a function of how sharp the steering wheel is turned (in general physics classes).

 

[Averagesupernova]

Depends on many variables. Wheel base for one. Age of the vehicle for another. Older vehicles even with power steering were not as quick steering as the newer ones.

 

[brewnog]

The relationship between the amount you turn the steering wheel and the amount the road wheels turn is simply called the Steering Ratio (just divide one number into the other). Sports cars have a much lower steering ratio than big trucks, for example. Some newer BMWs incorporate a variable steering ratio, - when the car is going slowly (parking, for instance) a low ratio is used, this decreases with increasing vehicle speed to improve driver response.

As for the relationship between the amount of lock and the curvature of radius, well, get a pen and paper! Draw a plan view of the car with the front wheels steering to some angle, draw a line along the rear axle of the car, and a line along each of the front axles. The points of intersection give the centre of the circle swept. Pen in some angles, do some trig, and you should get a rough idea.

 

[Dissident Dan]

As for the relationship between the amount of lock and the curvature of radius, well, get a pen and paper! Draw a plan view of the car with the front wheels steering to some angle, draw a line along the rear axle of the car, and a line along each of the front axles. The points of intersection give the centre of the circle swept. Pen in some angles, do some trig, and you should get a rough idea.

I would think that it would be more complicated than that. Your method seems to assume that the rear wheels will follow where the front wheels were, but in reality, they are closer to the center of the circle by the time they reach where the front tires were at the point in time the image represents.

 

[brewnog]

Well it's obvious that the rear wheels follow a different path to the front wheels, but I think the centre of the circle swept must be common to all wheels (this gives rise to the Ackerman principle) so radii for each wheel can be calculated this way.

Do you still think I'm wrong?

 

[Dorje]

Based on brewnog's idea, I used some geometry and the law of cosines to derive an *estimate* of the radius of curvature as a function of the steering wheel angle and wheel base.

s = wheel base
a = steering wheel angle
n = steering ratio (e.g. for 16:1, n = 16)
r = radius of curvature, in the same units as the wheel base

So:

r = s / (sqrt(2 - 2 * cos(2*a/n))

For an angle of zero degrees, the radius of curvature is infinite, which is expected. There would be a maximum value for "a" and thus a minimum value for "r", which would be equal to the turn radius.

I used Dodge Neon values (my car) for a sample:

s = 8.75ft
a = 45 deg, 90 deg, 135 deg
n = 16
r = ?

Using the formula:

r = 89.2 ft for 45 deg
r = 44.6 ft for 90 deg
r = 29.8 ft for 135 deg

The turn radius for a Dodge Neon is 17.9 ft. When the formula is solved backwards for the unknown steering wheel angle, I get a value of 226 deg, which seems reasonable, considering that the steering wheel cannot be turned the entire way around. I'll have to do some measurements on the vehicle to see how accurate the formula may be.

 

[arolin]

Howdy folks I know this post is a bit old but here is an answer that I've used in a couple projects now and it works well.

>r = s / (sqrt(2 - 2 * cos(2*a/n))

I could be wrong but I think this is wrong

I've always used 1/r= sin(a/n)/s

this is a measure of curvature which relates nicely to integration in that driving along a path length * curvature gets you dTheta...

dTheta/dt=speed*sin(Theta)/wheelbase

hope this helps someone.
cheers

 

[mark77]

How accurate is this when going above 30mph though? I know there is deformation in the tire causing a slip angle and thus the car is not going where the wheel is aimed. Check slip angle on wikepedia for a better explenation. I am guessing this formula only works with low speeds...?

 

[Ranger Mike]

Listen to Brewnog..one wise teacher and savvy guy..and

I suggest you read Race car suspension Class in this forum to gain insight on what happens when you turn left

 

[Oldfyze]

I’m trying to design a land sailer, and I wanted to make sure I had the right formula.

Using the PDF diagram attached, I was able to derive Arolin’s formula from geometry. But I think it’s only an approximation, which works best when the steering angle (angle of the wheels, not the steering wheel) is 90 degrees from straight ahead.

In the diagram, the “car” is initially the line AC, with one fixed rear wheel at A and a steerable front wheel at C. As it moves forward an infinitesimal amount dy (the distance between A and B), its front end describes an arc from C to E, determined by the steering angle θ between CD and CE, which we assume is finite and constant. The length of the “car” remains constant at s, but it moves infinitesimally to position BE.

As this happens, the front end of the “car” moves along the arc CE. By the definition of sine, the linear distance CE relates to the “tangential” distance DE (in red) that the front end of the “car” moves perpendicularly to the “straight ahead” direction AD by the sine of the steering angle θ, thus:

DE/CE = sin θ or CE = DE/sin θ

But DE is also related to the infinitesimal turning angle dψ as follows:

DE= AE sin dψ. So CE = (AE sin dψ)/sin θ

The calculations get complicated if you take the actual length of AE according to Pythagoras’ theorem. But since the changes are infinitesimal, to first order of approximation AE ≅ s and, by first-order Taylor expansion sin dψ ≅ dψ (in radians).

With these approximations, the length traced by the front end of the “car” is

CE ≈ s dψ/sin θ

Since s and θ are constants, it’s a trivial matter to integrate the length of the arc CE over the angle dψ until the finite arc equals the steering radius AC, which (by definition of radian) occurs when ψ = 1:

Steering radius = ∫ (CE)dψ = s/sin θ ∫ dψ = s/sin θ, where the integration is from 0 to 1.

This is the same as Arolin’s formula, with his a = my θ, the usual designation of angle, and his n = 1 (because we’re using the steering angle, i.e., the angle of the wheels themselves relative to straight ahead, not the angle of the steering wheel).

The formula is exact for θ = 90 degrees because then AE actually equals s. Of course, when θ = 90 degrees you burn up your tires trying to move the car forward.

I’m not sure how the formula should vary as θ → 0. At the limit it seems to work, because the steering radius becomes infinite, as it ought, when the car moves straight ahead. But the formula might not work well for small steering angles.

Anyway, wish me luck with the cactus.

 

[sketch]

I take it you have already planned around the much altered centre of gravity? I wouldn't want too small a radius of steering on a land sailer...

 

[Oldfyze]

Altered from what?

I plan to use 10" diameter wheels, so with a 5" seat cushion, riders' bodies will be approximately 10" off the ground. And my design will allow for widening the wheel base (within limits) to accommodate strong winds in our mountain valley.

But I'll be using (I think) a mast and sail assembly from a commercial dinghy. For reasons of safety, the boom will have to clear riders' heads. So the sail forces might be a little top-heavy.

I don't think I'll have to worry about too tight a steering radius, as I plan to use a tiller-like arrangement, rather than a steering wheel, which will probably limit the wheel angle to about 30 degrees. The problem won't be so much tipping as turning around, which is why I want to be able to calculate the steering radius accurately.

If you have any specific suggestions or other things to consider, I'd like to see them.

 

[sketch]

Altered from what?

I plan to use 10" diameter wheels, so with a 5" seat cushion, riders' bodies will be approximately 10" off the ground. And my design will allow for widening the wheel base (within limits) to accommodate strong winds in our mountain valley.

But I'll be using (I think) a mast and sail assembly from a commercial dinghy. For reasons of safety, the boom will have to clear riders' heads. So the sail forces might be a little top-heavy.

I don't think I'll have to worry about too tight a steering radius, as I plan to use a tiller-like arrangement, rather than a steering wheel, which will probably limit the wheel angle to about 30 degrees. The problem won't be so much tipping as turning around, which is why I want to be able to calculate the steering radius accurately.

If you have any specific suggestions or other things to consider, I'd like to see them.

It's the extra weight and stresses caused by the long pole with a sail attached that I was concerned about but it looks like you've thought that through. I guess you could use a smaller sail if it's very windy.

I'd probably want to make the way the tiller affects the angle of the steering wheels adjustable to some extent to allow for different conditions.