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August 29th, 2016, 10:17 AM   #1
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Isolate 2 variables in 2 formulas

I am developing a robotic arm and I need to know the angles of 2 parts of this arm (bars) and I have 2 variables with 2 formulas and I tried my best and I still can't isolate those 2 variables.

In the formulas below, [A,B,C,D] are constants, and [x,y] are variables in degrees (not radians):

A⋅cos(x)−B=C⋅sin(y−90)
C⋅cos(y-90)−A⋅sin(x)=D

How can I isolate x and y?

I tried my best and I ended up with a big arc cos with sin inside which had also another cos inside which I could not work with it anymore and isolate x and y.

Knowing:

cos(y−90°)=sin(y)
sin(y−90°)=−cos(y)

I could simplify both equations tho the ones below but I still can't isolate x and y.

A⋅cos(x)−B=−C⋅cos(y)
C⋅sin(y)−A⋅sin(x)=D

Last edited by skipjack; August 30th, 2016 at 04:19 AM.
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August 29th, 2016, 10:52 AM   #2
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Quote:
Originally Posted by batata004 View Post
I am developing a robotic arm and I need to know the angles of 2 parts of this arm (bars) and I have 2 variables with 2 formulas and I tried my best and I still can't isolate those 2 variables.

In the formulas below, [A,B,C,D] are constants, and [x,y] are variables in degrees (not radians):

A⋅cos(x)−B=C⋅sin(y−90)
C⋅cos(y-90)−A⋅sin(x)=D

How can I isolate x and y?

I tried my best and I ended up with a big arc cos with sin inside which had also another cos inside which I could not work with it anymore and isolate x and y.

Knowing:

cos(y−90°)=sin(y)
sin(y−90°)=−cos(y)

I could simplify both equations tho the ones below but I still can't isolate x and y.

A⋅cos(x)−B=−C⋅cos(y)
C⋅sin(y)−A⋅sin(x)=D
Mathematica returns
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Last edited by skipjack; August 30th, 2016 at 04:19 AM.
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August 29th, 2016, 10:58 AM   #3
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Math Focus: real analysis, vector analysis, numerical analysis, discrete mathematics
If you wish to solve 'by hand', then proceed like this:

Solve $\displaystyle C\cos y$ and $\displaystyle C\sin y$, rise both sides of the two equations to square, add both sides of equations up and use the identity of $\displaystyle \sin ^2 + \cos ^2 = 1$. Now you have an equation containing only $\displaystyle x$. Can you solve that?

For $\displaystyle y$, once you've solved $\displaystyle x$, go back to the equations which you wrote for $\displaystyle C\cos y$ and $\displaystyle C\sin y$ and now divide the $\displaystyle \sin$-equation by $\displaystyle \cos$-equation and use the identity of $\displaystyle \frac{\sin}{\cos} = \tan$ and write the solution for $\displaystyle y$.
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August 29th, 2016, 12:28 PM   #4
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@romsek thank you so much! The equation got really big but it's not a problem to me! My microcontroller is gonna calculate everything fine with that formula. What is "conditionalExpression"?

@fysmat I would like to have a clean X and Y equations where X and Y are explicitly isolated. The way you told me I am not gonna end up with those variables isolated.
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August 29th, 2016, 01:31 PM   #5
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Quote:
Originally Posted by batata004 View Post
@romsek thank you so much! The equation got really big but it's not a problem to me! My microcontroller is gonna calculate everything fine with that formula. What is "conditionalExpression"?
basically that's in there to ensure the final answer lies between 0 and 2pi.

the C[1], C[2], just stand for constants and the conditional expression ensures those constants are integers. If you look carefully they are adding multiples of 2pi to the answer if necessary.

All those periodic solutions are valid but I specified that x and y lie within 0,2pi.
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August 30th, 2016, 07:29 PM   #6
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Acosx=B-Ccosy
Asinx=-D+Csiny
Square both sides and add to get an equation of the form:
Fcosy+Gsiny=H
Let
F=sint
G=cost, then t=arctan(F/G)
sin(t+y)=H
y=arcsinH-t
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