- #1

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Px=C1(a2*C2+d4*S23)-d2*S1

Py=S1(a2*C2+d4*S23)+d2*C1

theta1=atan((Py*sqrt(Px^2+Py^2-d2^2)-d2*Px)/(Px*sqrt(Px^2+Py^2-d2^2)+d2*Py))

theta1=atan((-Py*sqrt(Px^2+Py^2-d2^2)-d2*Px)/(-Px*sqrt(Px^2+Py^2-d2^2)+d2*Py))

- Thread starter mikeley
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- #1

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Px=C1(a2*C2+d4*S23)-d2*S1

Py=S1(a2*C2+d4*S23)+d2*C1

theta1=atan((Py*sqrt(Px^2+Py^2-d2^2)-d2*Px)/(Px*sqrt(Px^2+Py^2-d2^2)+d2*Py))

theta1=atan((-Py*sqrt(Px^2+Py^2-d2^2)-d2*Px)/(-Px*sqrt(Px^2+Py^2-d2^2)+d2*Py))

- #2

AKG

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- #3

Fermat

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since S1 and C1 don't appear in the two final expressions, then perhaps S1 is sin(theta) and C1 is cos(theta) ??

Latex would be nicer.

Latex would be nicer.

Last edited:

- #4

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a2, d4, Px, Py are all constants

[tex]

Px=\cos(\theta1)*(a2*\cos(\theta2)+d4*\sin(\theta2+\theta3))-d2*\sin(\theta1) [/tex]

[tex]

Py=\sin(\theta1)*(a2*\cos(\theta2)+d4*\sin(\theta2+\theta3))+d2*\cos(\theta1) [/tex]

[tex]

\theta1=\arctan((Py*\sqrt(Px^2+Py^2-d2^2)-d2*Px)/(Px*\sqrt(Px^2+Py^2-d2^2)+d2*Py)) [/tex]

[tex]

\theta1=\arctan((-Py*\sqrt(Px^2+Py^2-d2^2)-d2*Px)/(-Px*\sqrt(Px^2+Py^2-d2^2)+d2*Py)) [/tex]

- #5

AKG

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- #6

AKG

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- #7

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Would you like some salt and pepper to go with thatAKG said:

in my equations I have only d4 not dA, and d2 is not d^2 and d4 is not d^4. d2 and d4 are two constants raised to the first degree. And yes, I checked my equations before posting, and they are ok. So some how in the result, t1=..., the expression containing d4 was canceled.

- #8

Fermat

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Multiply Px by sinθ and Py by cosθ. subtract Pxsinθ from Pycosθ to get rid of the bracketed term.

You should end up with Pycosθ - Pxsinθ = d2.

Divide both sides by cosθ.

Square both sides and get a quadratic in tanθ.

Solve ...

I ended up with a simpler form for tanθ.

[tex]tan\theta = \left(P_xP_y \pm d_2\sqrt{P_x^2 + P_y^2 - d_2^2}\right) / (P_x^2 - d_2^2)[/tex]

I manipulated your expression(s) for tanθ and ended up with my expression, so I suppose it could be done backwards - get your expression from mine - but it was a wee bit tedious.

Do you need to use the original form(s)?

- #9

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Thank you very much Fermat. No, I don't have to use the original form.

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