Ackerman, Toe and a Few Other Random Thoughts«
The diagram below outlines the important geometry in determining the motions of the steer wheels in in a vehicle that uses Ackerman steering steering geometry. Ackerman is an interesting problem because it it is dynamic. That is to say that we have two components moving together ± the left left and right r ight steering knuckles, but the relationship between their motions changes as we move move them. This is is a real head scratcher. It¶s a bit like like having a bowling ball in a dark roo m and throwing other bowling balls in an attempt to locate it by listening for an impact. impact. Every time you you find the ball by crashing into into it, it moves and again don¶t know where it is.
e c n a t s i D r e t n e C o t r e t n e C n i P g n i K
D
C
T i e R o d L e n g t h
A
B
Ackerman Angle
R AA AA
Wheel Base
Luckily we have some mathematical voodoo voodoo that can help us figure it out. Let¶s look at the important important distances and angles. The two most most fundamental distances are the wheel base of the car and the kingpin kingpin center to center distance. If we draw two lines representing the wheelbase and the d istance from the car¶s center line to one of o f the king pins, we can make a triangle. triangle. By design, the line line that goes through the centers of the Ackerman arm forms forms the hypotenuse of this triangle. See below.
e c n a C t s i D r e t n e C o 2 t r e t n e C n i P g A n i K
Ackerman Angle
Wheelbase
B
Note that the angle with its vertex at A is 90 degrees by design, unless the vehicle has been crashed. If this angle experiences an unplanned adjustment due to an impact, the car will dog track. This can be checked using a tape measure and comparing distances from side to side. Also note that the line that forms the Ackerman angle with the hypotenuse is parallel with the thrust line, again by definition. Because of this, we can say that angle B and the Ackerman angle are similar, so if we know one, we know the other. But angle B isn¶t too hard to come up with. Recall that the tangent function gives the ratio between the opposite side and the adjacent side of the triangle. So« TAN Angle B = king pin center to center distance / 2 Wheelbase The problem is that we know the distances and are trying to find angle B. We need the inverse function ARCTAN. Rearranging, we get: ARCTAN king pin center to center distance / 2 = Angle B Wheelbase We can pick distances, turn the crank and find Angle B and by extension, the Ackerman Angle. For example, lets choose a wheel base of 72´ and a king pin to king pin distance of 36´. The formula would look as follows:
ARCTAN
36´ / 2 72´
= Angle B
Plugging the number into a calculator or Excel, or looking up in a table, ARCTAN (.25) = 14.036º So, the Ackerman Angle is 14.036 degrees. We can use this to find the length of the tie rod. e c n a t s i D r e t n e C o t r e t n e C n i P g n i K
D
C
T i e R o d L e n g t h
14.036º A
B
R AA
Wheel Base
To find the length of the tie rod, we can deco mpose the trapezoid ABCD into a rectangle and two triangles. D
C
B 14.036º
Y
A A
B
If you think logically about t he diagram above, the length of the tie rod (segment BC in the drawing) is equal to the king pin to king pin center distance minus distance Y on each side. So, what is distance Y? To find out, you have to pick an Ackerman Arm Radius. You may choose this by purchasing a standard Ackerman arm out of a catalog, or you may design your own. Either way, it is what we might refer to as a ³drawing board problem´, meaning that basically this is a parameter that the engineer chooses by his gut. Let¶s pick 6´ to make life easy. So, how long is distance Y? Well, recall that the SIN of an angle is the ratio between the side opposite the angle and the hypotenuse. In shorthand it looks as follows. SIN 14.036º = Y/6´ As you know, the name of the game in Algebra is getting the variable by itself, so« 6´ * SIN 14.036º = Y You can look SIN 14.036 up in a table, or punch it up on a calculator, giving you: 6´ * .243 = Y A few Dazzling Algebraic Contortions, And: 1.445´ = Y
So, the tie rod is 1.445´ inches shorter on the bottom and 1.445´ inches shorter on the top than the kingpin center to center distance. Expressed mathematically: LT = DKC ± 2*R AA*SIN Ackerman Angle Where: LT is the length of the tie rod DKC is the distance between king pins center to center R AA is the radius of the Ackerman Arm Plugging in our numbers« LT = 36´ ± 2*6´*SIN 14.036º Looking up the SIN value« LT = 36´ ± 2*6´*.243 Turning the crank« LT = 33.084´ So, for a car configured as this one is, the tie rod needs to be 33.084´ from the center of one rod end to the center of the other. We¶ve figured out all of the static values. Now the real fun begins. Let¶s contemplate a turn as diagrammed in red to the right. Suppose that D the Ackerman arm labeled AB steers 20 degrees to the left as shown. What C angle does the other Ackerman arm transect? You might think 20 degrees, Something but this would result in the steering wheels being p arallel in a turn, which less than 20º would be unsatisfactory as we discussed in the previous packet. In reality, because the car pictured is turning to the left, the right Ackerman ar m (CD) needs to steer something less than 20 degrees. But how much less? This becomes a moving target. I tried a lot of high flutin¶ mathematical tricks until I discovered a rather straightforward way to attack this. Let us D AA C consider a line drawn diagonally from point D to B. This creates three angles that add together to give the B angle of the wheel that pivots at point D. We¶ll call A the first angle K, the second angle (pronounced k gamma), and the third angle is of course, the 20º Ackerman angle. Now we can set to work on determining each. If you think about angle k, we can determine it because for any steer angle, we know the positions of the ends of the diagonal line. If we assigned point A the coordinate of 0,0 then point D would have the coordinates Kingpin Center to Center Distance,0. In the B A
our case specifically point D¶s coordinates would be 36,0. Point B¶s coordinates take a little bit more elbow grease to find. We can calculate it¶s locations with the following formulae: Point B¶s X coordinate = R AA * COS(AA + SAL) Point B¶s Y coordinate = R AA * SIN(AA + SAL) Where: R AA is the Ackerman Arm Radius AA is the Ackerman Angle SAL is the steering angle of the left wheel. Zero degrees is straight ahead. Positive values are a left turn, negative values are a right turn. D
Plugging in our numbers for a 20º left turn« Point B¶s X coordinate = 6´ * COS(14.036º + 20º) Point B¶s Y coordinate = 6´ * SIN(14.036º + 20º) Jumping through a few hoops« Point B¶s X coordinate = 4.972 Point B¶s Y coordinate = 3.358
AA D
C
k
So, the coordinates of Point B at a 20º left turn are 4.972,3.358. We can project straight to the left of point B and straight up from point A to create a new point called point E. Because we projected straight left and straight up, the angle at E is by definition 90º. Also, because point E falls on segment AD, we can calculate distance DE with the formula: DE = AD ± AE Plugging in our numbers DE = 36´ ± 3.358´
4.972 8 5 3 . 3
E
A
B
Crunching the numbers« DE = 32.642´
Now that we know EB and ED, we can find the length of BD because it is a hypotenuse of the triangle formed. Using Pythagorean Theorem:
BD =
2
(EB + (DE)
2
Plugging our numbers in« BD = (32.642´)2 + (4.972´)2 BD = 33.019´ Furthermore, because we know the sides of the triangle we can determine ang le k in the following manner: TAN k = EB/ED Of course, we¶re trying to find k, so let¶s get that by itself by taking the ARCTAN ARCTAN (EB/ED) = k Plugging in our numbers« ARCTAN (4.972´/32.642´) = k Some mathematical acrobatics and« 8.661º = k So now that we know angle k and the Ackerman angle, the problem is two thirds licked. All we have left is to find angle (pronounced gamma). Note that is included in triangle BDC. Let¶s think about what we know about this triangle. We know that side DC is the length of the Ackerman arm, which we chose to be 6´. We know that side CB is the length of the tie rod, which we calculated earlier to be 33.084´. Finally, we know the distance BD, which we determined using Pathagorean Theorem to be 33.019´. So we have a triangle and we know the lengths of each o f the three sides. Luckily, there is a somewhat abstract relationship between t he sides of non-right triangles called law of cosines. It can be expressed a number of ways, but we will use the permutation shown below. 2
2
2
COS = A + B ± C 2AB
Per usual, we are trying to get the thing we don¶t know by itself, so we¶ll need to beat this up a little bit to make it useful. Rearranging gives:
AA C
D
k
4.972 8 5 3 . 3
E
A
B
2
2
2
ARCCOS A + B ± C = 2AB Plugging in our numbers« 2
2
2
ARCCOS (33.019) + (6) ± (33.084) = 2(33.019)(6) Crunching the numbers« 85.411° = Now if we add up angle k, and the Ackerman angle, we¶ll have the tire¶s steer angle from the line that connects the two kingpins. To get the steer angle, we have to subtract 90°. The formula is: Steer Angle = k + + Ackerman Angle - 90° Plugging our numbers in« Steer Angle = 8.661º + 85.411° + 14.036° - 90° Turning the crank: Steer Angle = 18.108° As with any engineering math, we must ask if this is a reasonable number. Let¶s think about it. The car is executing a left turn. The left front wheel is steered 20° to the left. The right wheel is tracing a larger arc, and therefore should have a lesser steer angle. In short the left side steering angle is 20° and the right side should be something less than 20°. We¶ve passed that test. Additionally, experience on real cars on the alignment rack indicates that these numbers are reasonable. Attached you will find an Excel file that will do the heavy lifting for you. Good luck!
