Designing an Ackermann steering mechanism

I'm designing a car with an Ackermann steering mechanism. According to everything I have read about Ackermann steering, if I set up my steering mechanism like this:

... then I should get behaviour like this:

Well, not according to my CAD. The front wheel axes crossing point actually traces out a path somewhat distant from the rear axle projection, like this:

Is this what I should expect? Or have the rules of geometry suddenly changed within my CAD package?

• turning should be that much so that the radius drawn from centre of tuning angle should coincide with and intersects at extended line of rear axle as shown in fig 1
– user18831
Jun 5, 2016 at 4:51

You need to make some modifications to your cad drawing so it is in line with what the Ackerman Theory is stating. I'm sure once you get things correct in your drawing, you'll find it will work just fine.

First, do you notice that on your example, you have the pivot point (turning point of your tire) right exactly at the edge of the tire. You'll see that in this image, the pivot point (red arrows marked A) is noticeably away from the tire a distance.

Second, the point at which the turning mechanism (red arrows marked B) are a come in board from the pivot points quite a bit. You have this in your drawing, but I'm pretty sure it's not enough. The point of where this point should be at is described as being thus: If you draw a line through the pivot point (A) to the center point of your rear axle (red arrow marked C), the arm pivot point (B) should be located on that line somewhere, but before the rear of the tire (I'm actually guessing on the length of the steering arm, but this length seems logical). It needs to be long enough to provide the difference, but not so long as to tie things up. If I were a betting man, I'd put it at ~70% of the tire radius (NOTE: I did say tire radius, not the swing arm radius). Nevertheless, the point of rotation on the steering arm needs to be located on this line.

So you aren't put out if it doesn't work exactly along the entire turning radius, it won't be. According to Carroll Smith, in Tune to Win he states (pg. 60):

No single intersection point will result in true Ackerman steering over the whole range, but by moving the intersect point in the longitudinal plane, you can come close in the normal range of steering angles.

Once you have these things corrected, I think you'll find your model to work much closer to what you expect.

As a side note, if you want to get technical about it, you could lay it out mathematically. Racetech.com.au spells it out (NOTE: They don't have a clear picture or I'd steal it and post it here. If I have time later, I will remake their diagram and edit this post.)

• Thanks for the answer. On point A. It makes no difference if you move the tyre along the axis. The point is that the projections of the axes should meet at the rear axle projection. Dec 10, 2014 at 20:30
• On Point B: That's exactly what I did, and what I tried to explain in my question. On my diagram, I haven't drawn the tyres, just the hubs, and my point B is about 70% along the tyre radius. Dec 10, 2014 at 20:33
• I realise that I won't get perfect Ackermann steering, but what I get even when I follow the advice exactly seems to be miles away from perfect, as shown by my series of red dots. If you have a CAD package, I urge you to try it too, and see if you can get it any closer. Dec 10, 2014 at 20:34
• @Rocketmagnet ... I'll leave you a note in the chat so we don't clutter things up here. I have some questions and requests of you where I can hopefully help you out. Dec 10, 2014 at 21:42

The Ackermann Theory states the meaning of your first drawing, ie that a line drawn through the center line of the track and the steering track rod end would pass through the centre of the rear axle. To achieve this with your with CAD programme you will need to include camber, caster and the suspensions included angle to facilitate this.

• Consider 'TOOT' in your calculations. Toe Out On Turns, ie the inner wheel travels a smaller diameter turning circle then the outside wheel does when turning off of a straight ahead direction. Dec 10, 2014 at 10:54

on first look, the reference image you used has a trapezoid formed of the points AABB, but your version seems to have something like a perfect 4 bar..a rectangle or parallelogram..try working on the angles..the hinge points basically.. hope that helps..

• You can see that it's not a parallelogram if you look at the blue linkage piece, which is clearly at an angle. Aug 27, 2016 at 14:59

Maybe the following link helps in understanding (notably figure 6)

https://www.quora.com/What-are-the-required-calculations-for-the-Anti-Ackermann-Steering-Mechanism

Actually, I have computed the common center point as well, in a spread sheet, and I find also that the center of turning is not on the rear axis.

• Could you please post the relevant parts of that here. Answers that are just links are not particularly helpful. Dec 1, 2016 at 15:27