OK, I'm stuck.. see attached pic

Given r, v, h, a, d and u, I want to know phi

so far I have got:

Sin phi = x/u

x + y = sqrt(a^2 + d^2)
x = sqrt(a^2 + d^2) - y

Sin phi = (sqrt(a^2 + d^2) - y) / u

y = z - h
Sin phi = (sqrt(a^2 + d^2) - (z - h)) / u
Sin phi = (sqrt(a^2 + d^2) - z + h) / u

r^2 = t^2 + z^2
z = sqrt(r^2 - t^2)

Sin phi = (sqrt(a^2 + d^2) - (sqrt(r^2 - t^2)) + h) / u

and now I'm stuck. I need to loose t from the equation and get v in there.

grrrrr!!

Is the angle between x and w a right angle?
If so break down U to get x, sin phi is x/u

[Edited on 19/1/08 by emsfactory]

I'll assume that what you're trying to model here is a double wishbone suspension where u is the upright and r and a are the wishbones. The generalised solution to this problem is found in the subject of kinematics - in this case, 4-link kinematics (the chassis is considered a fixed link). Here's a brief summary I wrote on the Locost_Theory mailing list way back in December 1999.

Here's a potted summary (with a very brief introduction to planar kinematics).

<---h--->
A o---------o B
|
a | b
|
O |<-----g-----> C
__o______________o__
////////////////////

The above diagram is a 4-element planar linkage (basically, all of the
hinge axes are parallel). The 4 joints are labeled O, A, B & C. The
lengths of the 4 elements are a, h, b & g (clockwise from the left).
Element OC is considered the "fixed" element for these discussions. In a
suspension design problem this would be the chassis and the diagram would
need to be rotated 90 degrees one way or the other. If you were modeling
the motion of a suspension upright you would represent the upper and lower
wishbones with (say) OA and CB; AB would be the upright and O & C the
chassis mounting points for the wishbones.

Lets call the inside angle between OA and OC the input angle and call it
theta or T (hmmm, email program can't do greek characters - what a surprise
. The output angle (call it phi or P) is between OC and CB but on the
*outside* of the link assembly (ie roughly where the letter 'C' is in the
diagram). .

The formula to solve for P(T) (that's phi as a function of theta) is pretty
simple.

A(T) = 2ab.cos(T) - 2gb
B(T) = 2ab.sin(T)
C(T) = g^2 + b^2 + a^2 - h^2 - 2ag.cos(T)

P(T) = arctan(B/A) + arccos(C/sqrt(A^2 + B^2))
P(T) = arctan(B/A) - arccos(C/sqrt(A^2 + B^2))

Note that there are two solutions as the output link may be able to sit in
two different positions for the same input angle. Also, if

A^2 + B^2 - C^2 < 0

then the linkage can't be assembled in that position.


The diagram is a little scrambled fom cutting and pasting etc but I assume you can work out the principles. If not, search for "Four Link Kinematics" and you should find plenty of reference material.

Dominic

quote:

---

Originally posted by emsfactory
Is the angle between x and w a right angle?
If so break down U to get x, sin phi is x/u

[Edited on 19/1/08 by emsfactory]

---

quote:

---

Originally posted by TheGecko
I'll assume that what you're trying to model here is a double wishbone suspension where u is the upright and r and a are the wishbones. The generalised solution to this problem is found in the subject of kinematics - in this case, 4-link kinematics (the chassis is considered a fixed link). Here's a brief summary I wrote on the Locost_Theory mailing list way back in December 1999.

Here's a potted summary (with a very brief introduction to planar kinematics).

<---h--->
A o---------o B
|
a | b
|
O |<-----g-----> C
__o______________o__
////////////////////

The above diagram is a 4-element planar linkage (basically, all of the
hinge axes are parallel). The 4 joints are labeled O, A, B & C. The
lengths of the 4 elements are a, h, b & g (clockwise from the left).
Element OC is considered the "fixed" element for these discussions. In a
suspension design problem this would be the chassis and the diagram would
need to be rotated 90 degrees one way or the other. If you were modeling
the motion of a suspension upright you would represent the upper and lower
wishbones with (say) OA and CB; AB would be the upright and O & C the
chassis mounting points for the wishbones.

Lets call the inside angle between OA and OC the input angle and call it
theta or T (hmmm, email program can't do greek characters - what a surprise
. The output angle (call it phi or P) is between OC and CB but on the
*outside* of the link assembly (ie roughly where the letter 'C' is in the
diagram). .

The formula to solve for P(T) (that's phi as a function of theta) is pretty
simple.

A(T) = 2ab.cos(T) - 2gb
B(T) = 2ab.sin(T)
C(T) = g^2 + b^2 + a^2 - h^2 - 2ag.cos(T)

P(T) = arctan(B/A) + arccos(C/sqrt(A^2 + B^2))
P(T) = arctan(B/A) - arccos(C/sqrt(A^2 + B^2))

Note that there are two solutions as the output link may be able to sit in
two different positions for the same input angle. Also, if

A^2 + B^2 - C^2 < 0

then the linkage can't be assembled in that position.


The diagram is a little scrambled fom cutting and pasting etc but I assume you can work out the principles. If not, search for "Four Link Kinematics" and you should find plenty of reference material.

Dominic

---

Could someone check this for me, please.

T = 180 - Q - S
Sin S = d/a
Tan Q = v/h

T = 180 - ATan(v/h) - ASin(d/a)

A(T) = 2cu.Cos(T) - 2au
B(T) = 2cu.Sin(T)
C(T) = a^2 + u^2 + c^2 -r^2 - 2ca.Cos(T)

c = sqrt (v^2 + h^2)

P(T) = ATan(B/A) +or- ACos(C/sqrt(A^2+B^2))


KPI = 180 - P - V
Cos V = d/a
V = ACos (d/s)

KPI = 180 - P - ACos(d/a)

[Edited on 19/1/08 by Jesus-Ninja]