analysis of wishbones

[SNasello]

Cam,

don't abandon it, your almost there!

http://en.wikipedia.org/wiki/Cross_product

Basically the tension in the link, the moment arm, and the resultant moment are all vectors. They have magnitude and direction. The moment is a result of the cross product of the force with the moment arm vector.

The error is in how you built up your A Matrix.

[Z]

Originally posted by CamMoreRon:
Sorry to have to beat this horse to death, but I really hate to abandon something I'm struggling with!

Cam,

Perserverance is a good thing. Stick with it!

I have had a closer look at the matrix equations, and, as I suspected, they are too good to be true. Namely, too simple, and hence wrong. (Too much "cogitatio caeca" = "thinking blind", which I will have to rant about sometime...)

The top three rows of AX=B, which sum forces in X, Y, and Z directions, are ok. The bottom three rows, which sum the moments, are the problem. As suggested by Stefan, the entries in A have to be fixed to reflect the "cross" (= "vector") product nature of moments. Incidentally, the link in Stefan's post above is one of the better examples of a Wiki page.
~~~o0o~~~

Anyway, here is how I would write out row 4 of the equations in full (sort of). Note I use "F1" for "force" in link 1 (save "Ts" for torques/couples), "dx1" for x-component of unit direction vector in link 1 (ie. = entry at A(1,1)), "Rx1" for x-coordinate of outer-end of link 1 (from R for radius), "*" is usual arithmetic multiply, and so on...

[dz1*Ry1 - dy1*Rz1]*F1 + A(4,2)*F2 + A(4,3)*F3 + A(4,4)*F4 + A(4,5)*F5 + [dz6*Ry6 - dy6*Rz6]*F6 = Mx.

So, entry of A at row 4, column 1 = A(4,1) = [dz1*Ry1 - dy1*Rz1], and similarly across the row.

As before, the RHS, namely B(4), is,
Mx = Fz.wp*Ry.wp - Fy.wp*Rz.wp,
with ".wp" = the force/position at the wheelprint.

You can do the entries of rows 5 and 6 of A similarly to above (with reference to my previous post for My and Mz).
~~~o0o~~~

Hope I didn't get too many typos in there...

Let us know if it works...

Z

[CamMoreRon]

Gentlemen, thank you so much!

Using cross products for the values in rows 4-6 of Matrix A has solved all my problems. I am now getting sensible answers and most importantly they stay the same no matter where I put my origin!

I'll admit I was relying on the method described initially a little too much, and I think that was definitely confusing things as it made me think I could run before I could walk, but all is good now. Gears have crunched into mesh and I definitely understand this a lot better, so I can't thank you enough for the perseverance.

[Z]

Cam,

Good to see that your perseverance has paid off! (And btw thanks for the spell check... )
~~~o0o~~~

In case anyone still doesn't understand the alphabet soup of the "cross product" equations, here is a brief translation.

"Mx = Fz.Ry - Fy.Rz" simply means that the moment about the longitudinal X-axis (Mx) is caused by the vertical component of force (Fz) acting along a LoA offset in the lateral direction (Ry), together with the lateral component of force (Fy) acting along a LoA offset in the vertical direction Rz. So if X is on the LHS of the equation, then only Ys and Zs on the RHS. Similarly for My and Mz.

The only tricky bit is working out the signs (+/-) on the RHS. I draw a small 3-D sketch and use the "Right Hand Rule" - right thumb points in +ve direction of axis, and if right fingers curl in same direction as the force, then +ve, else -ve.

Z

[Ceboe]

Hi, I was guided here by the other topic.

I did everything that is written here and I understand why. But my forces are still pretty high, in region of 20000N, witch is impossible.

I did the A-matrix as said above, my first three rows are like the A-matrix on the previous page. Then I calculated all the cross products. But I think there was a problem here.

For example for the A[4,1] I did:

[dZ1*Ry1 - dy1*Rz1]
witch would be concrete dZ1 = -0.058.
The value in A[1;3]. Then Ry1 I took the value of the Lower WB Outer Ball Joint witch is at -12 685 150. So Ry1 = 685 but I divided this value by 1000 to have the same units everywhere.

This gives me for A[4,1] = 0.105
Thus: (-0.058*0.685)-(-0.966*0.150)= 0.150

To expand this for the fifth row I did the following: [dx1*Rz1 - dz1*Rx1]

Witch gives the following: [(-0.254*0.150)-(-0.058*0.012)]
gives -0.039

I Think this is the correct way of doing it? No?

But when I have a 1000N of force in the X direction I yield a force of more than 16000N in the rear lower A-arm

The thing I might consider to be wrong is the R..1 value. This should maybe be the difference between the pivot point and the outer ball joint?

[SNasello]

Ceboe, why don't you post the equations that you used to derive your A matrix. Then we might be able to help better determine where you went wrong. The magnitude of your result looks similar to what was posted earlier on the previous page.

[Ceboe]

SO I used the points of the topic starter, so I can check the things I did along the way. I am making a pull-rod suspension, but that should not matter, I would just put in my points and the outcome should be correct.

So:

X Y Z
1 Lower WB front pivot -100 350 130
2 Lower WB rear pivot 250 350 150
3 Lower WB outer ball joint -12 685 150
4 Upper WB front pivot -80 450 362
5 Upper WB rear pivot 200 450 362
6 Upper WB outer ball joint 12 660 410

7 Pull Rod wishbone end -5 645 180
8 Pull Rod rocker end -5 485 430

9 Outer Track Rod Ball Joint -120 668 200
10 Inner Track Rod Ball joint -125 370 176

11 Tire Contact Patch 0 750 0

Are the points I used

This gives me three unit vectors:

X Y Z
3to1 -0,254 -0,966 -0,058
3to2 0,616 -0,788 0,000

6to4 -0,393 -0,897 -0,205
6to5 0,658 -0,734 -0,168

7to8 0,000 -0,539 0,842


9to10 -0,017 -0,997 -0,080

This seems to be correct, considering the answers given above.
So this gives me following matrix A. Where the first 3 rows are filled with X Y Z, X for row 1, Y for row 2 and Z for row 3. And then every column representing a member.
3to1, with this I mean, force from point 3 to point 1, so the front lower A-arm.



This is the matrix I become.
So I think the problem is in the last 3 rows.
The 0.105 of row 1 column 1, I became by solving [dz1*Ry1 - dy1*Rz1] so [(-0.058*685/1000)-(-0.966*150/1000)] witch gives 0.105
so dz1 is the value of the unit vector witch is on the third row at column 1, so -0.058. Ry1 is the Y value of the outer ball joint of the lower A-arm so 685, but this I divide by 1000 so everything is in meters. I do the same thing for dy1, witch is then -0.966 (the Y value of the unit vector) and 150/1000 with 150 being the Z value of the outer ball joint of the lower A-arm.

For the fifth row I used [dx1*Rz1 - dz1*Rx1] with dx1 = -0.254, Rz1 = 150/1000, dz1 = -0.058 and Rx1 = -12/1000, this gives a value of -0.039
And the sixt row [dy1*Rx1 - dx1*Ry1] with dy1 = -0.966, Rx1 = -12/1000, dx1 = -0.254 and Ry1 = 685/1000, witch gives a value of 0.185

I did the same thing for every column. So for example for the last column I did following:
For row 1: [dz6*Ry6 - dy6*Rz6], [(0.842*645/1000)-(-0.539*180/1000)] gives 0.640
where Ry6 is the Y value of the point where the push rod is attached to the wishbone, so the point closest to the wheel. Just like I took the point of the outer ball joint at the A arms.
But I think here might be a mistake?

[SNasello]

Ceboe,

I think you have the same problem as the previous poster. The way that you have built this up is using the coordinate of the mounting point of the link to the upright, which is essentially the moment arm about the origin.

The question is, what does the equation for your Force/moment input vector look like?

[Ceboe]

Originally posted by SNasello:
Ceboe,

I think you have the same problem as the previous poster. The way that you have built this up is using the coordinate of the mounting point of the link to the upright, which is essentially the moment arm about the origin.

The question is, what does the equation for your Force/moment input vector look like?

With the force/moment input vector, I think you mean the values of matrix B. I made this that I can just fill in the 3 forces on the tire contact patch. From these forces Mx, My, Mz will be filled in. So Mx = Fz*dz - Fy*dy where Fz and Fy are the forces on the third and second row. And dz is are the points of the contact patch. So here dz = 0 and dy = 750. So if I fill in Fx = 1000 I get a moment of -750 for Mz. Same way, a force of 1000N in the Z direction gives me an Mx of 750

But when I fill 1000N in for Fx, Fy and Fz. Then I will have a moment of 750Nm for Mx, a moment of -750Nm of Mz. My is zero.

[Ceboe]

I think I found my error. I rechecked all my values and found a couple wrong in my excel file. Some values of the upper wishbone were linked to the outer point of the lower wishbone...

Now I get some values that I think could be true.

When I enter 1000N for Fx, Fy and Fz I get following vector B:

x 1000
y -1000
z 1000
moment x 750
moment y 0
moment z -750

This then gives me following values for vector X:

Front Lower Wishbone 846 N
Rear Lower Wishbone 3058 N
Front Upper Wishbone 871 N
Rear Upper Wishbone -578 N
Steering Rod -3155 N
Pull Rod 1041 N


where a minus is pressure and a positive force is a pulling force.

Can somebody check this please?
I think this could be correct!

But a strange thing, the rear upper wishbone has a pressure force, but this is under braking and in a corner, a situation where I don't think there could be a pressure force on that member..