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Topic: Ackerman Steering, Need Geometric Explanation
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Nehmo Sergheyev

Posts: 9
Registered: 12/13/04
Ackerman Steering, Need Geometric Explanation
Posted: Jan 28, 2012 8:59 PM
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On a 4 or 3 wheel vehicle with the front 2 doing the steering, both wheels need to track concentric arcs of different length radius.[img]http://img543.imageshack.us/img543/7991/ackermanprinciple.jpg[/img] [img]http://upload.wikimedia.org/wikipedia/commons/thumb/a/a0/Ackermann_turning.svg/488px-Ackermann_turning.svg.png[/img]
For a historical note, Ackerman didn't actually claim credit for the development, but he promoted it. There were a few people before him, Darwin (Erasmus Darwin, the grandfather of Charles) Jeantaud, Davis, and Lankensperger. This is a common theme in history. Edison didn't invent the bulb. Orville an Wilbur weren't first. Henry Ford followed several. Gates bought DOS (then called "Quick and Dirty Operating System", but Gates was too marketing-smart to stick with that). And (this one I just learned) Erasmus Darwin was really the one to get evolution down on paper. He wrote a poem in 1794, way before the Voyage of the Beagle published in 1839. Charles even gave him credit.

There is geometry proof on (page 190 of the whole pdf if you download it) page 21 of No. 11 December 13, 1899, of the Google Books pdf Horseless Age
The Ackerman steering geometry uses a deformable (in angles, the lengths of sides remain the same) isosceles US trapezoid (trapezium in English outside North America). The trapezoid is constructed so that in the straight-ahead position, the legs of the trapezoid extended will form an isosceles triangle which has a vertex at the center of the rear axle of the vehicle. [img]http://upload.wikimedia.org/wikipedia/commons/4/44/Ackermann.jpg[/img]
This is a proof of the Ackerman steering geometry which uses a deformable isosceles US trapezoid (trapezium in English outside North America)[img]http://books.google.com/books?id=MQsAAAAAMAAJ&pg=RA1-PA21&img=1&zoom=3&hl=en&sig=ACfU3U3yAbpnsbiR30L9E0NUl1YLnWEjnQ&ci=89%2C751%2C419%2C458&edge=0[/img][img]http://img810.imageshack.us/img810/1231/proofsteeringdiagram.gif[/img]

He is a paste of an OCR of the relevant text:

[The letter names of the points and lines may not have the correct subscripts because the OCR isn't accurate. For now, you'll have to figure it out.]

"Let ABC Fig 10 be the steering frame in its middle position and FG the axial line of the fore wheels at the same time
Bisect the angle BCG by the line CBo
and make BoCE a right angle the semi circle on EB0 will pass through A and C
and it is the locus of the vertices of all triangles having their base in the line EB0 and their vertical angle bisected by a line through B if one triangle on the same with its vertex on the semi circle has its vertical angle bisected by a line through BoEuclid VI 3and A At C make any angle BoCBi equal to BoCK and join ABi and AbJ By the theorem referred to the angle BiAB0 b ABo
The line Abi and Cbi are respectively the images of the actual axial lines Ab and Cb for the position of the steering vertex Bi The image point bi travels in the same straight line as Bi and therefore the actual point of intersection b travels in a straight line parallel to the path of Bj and at the same distance from FG which line ought therefore to be the axial line of the hind wheels "

The Ackerman angle is the one between a leg of the trapezoid and the center line of the vehicle in the straight forward position. [img]http://www.kartbuilding.net/PlAnS/ackermann_angle.gif[/img]

I once had a different geometric proof, but I forgot what it was. IICRC, it went something like...
[list=1]
[*]The trapezoid base also makes the base of a triangle.
[*]All tangles with that base and made with the arms of a quadrilateral with the same side-lengths as the original trapezoid, will have the same area. This is the step than needs to be proved. Once you have this, it's explained, at least to me.
[*]If all those triangles have the same area, and have the same base, then they have the same altitude.
[*]If they all have the same altitude, then all the vertices fall on a line parallel to the original front axle line, which will be the rear axle line.
[*]If we make lines at the Ackerman angle in opposite directions from both legs of the deformed trapezoid (which are the same as the sides of the triangle have a vertex on the rear axle line), then these lines will be tangents to concentric circles with a center on the rear axle line.[/list]

I have some questions on the proof on the first proof here, the one from Horseless Age . I need to draw a better diagram. Look for my next edit. If someone understands the proof as presented here, feel free to explain more clearly than what's above.
A trigonometry proof is in pdf Steering System Design
`~- Nehmo



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