Bond

08-10-2005, 09:21 AM

can anybody help me on what is the equation for moving my arm to a point(let say point A) with my two degree of rotational arm.there are 2 servo motor controlling two arm in horizontal position.anyone knows?

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Bond

08-10-2005, 09:21 AM

can anybody help me on what is the equation for moving my arm to a point(let say point A) with my two degree of rotational arm.there are 2 servo motor controlling two arm in horizontal position.anyone knows?

happy_99

08-11-2005, 12:11 AM

Give me a drawing of your mechanism.

Regds

happy_99

Regds

happy_99

Bond

08-11-2005, 03:58 PM

sorry,how am i going to post the drawing here?can give me ur email?or anything that i can post to let u see?thanx..[/list][/code][/quote]

happy_99

08-11-2005, 11:47 PM

Upload it to http://www.rapidshare.de/ so that other's can also offer their help.

Regds

happy_99

Regds

happy_99

Bond

08-12-2005, 10:15 PM

i have posted the picture of the arm and the 3rd movement is controlled by another DC motor and is not involved in this equation i suppose,rite?

any suggestion and can help me on the equation?

thx.all the brothers..

any suggestion and can help me on the equation?

thx.all the brothers..

08-12-2005, 11:14 PM

i have posted the picture of the arm and the 3rd movement is controlled by another DC motor and is not involved in this equation i suppose,rite?

Where's the picture? Give ur complete mechanism.

Bye

Where's the picture? Give ur complete mechanism.

Bye

Bond

08-13-2005, 11:32 AM

my picture being upload to http://www.rapidshare.de/ as asked by happy_99...can see my picture there?

i'm using two servo motor and both also spin according to the z-axis to the base.

which means my servo motor is attached to a base on the floor.there is an arm attached to my spining part of the motor.

another servo motor is attached to the 1st arm, and is also rotating according to he z-axis..and the second arm is attached to the spinning part of the motor.

get wat i mean?so.how can i attach my picture so that all can seen here?

help ya..thx.

i'm using two servo motor and both also spin according to the z-axis to the base.

which means my servo motor is attached to a base on the floor.there is an arm attached to my spining part of the motor.

another servo motor is attached to the 1st arm, and is also rotating according to he z-axis..and the second arm is attached to the spinning part of the motor.

get wat i mean?so.how can i attach my picture so that all can seen here?

help ya..thx.

happy_99

08-14-2005, 12:53 PM

Bond,

You should get a download link after you upload files to www.rapidshare.de

Eg. http://rapidshare.de/files/1965406/179_1_.part1.rar.html

Post that link here. Then only we can download your files.

Regds

happy_99

You should get a download link after you upload files to www.rapidshare.de

Eg. http://rapidshare.de/files/1965406/179_1_.part1.rar.html

Post that link here. Then only we can download your files.

Regds

happy_99

Bond

08-14-2005, 06:56 PM

http://rapidshare.de/files/3913555/pic33147.jpg.html

ok..here is my picture...pls help.

another thing i wanted to know is wat is the standard program for the IR protocol that used for IR?how to get the it?anyone knows?

the standard can be used in any of the IR devices..which means standard that can be used mostly in industrial..help again..thz...

ok..here is my picture...pls help.

another thing i wanted to know is wat is the standard program for the IR protocol that used for IR?how to get the it?anyone knows?

the standard can be used in any of the IR devices..which means standard that can be used mostly in industrial..help again..thz...

happy_99

08-17-2005, 12:02 AM

http://www.sendmepic.com//out.php/i3746_mechanism.png

Overall transformation from O(0,0,0) to P(x,y,z) is given by:

Rot(z,theta1)*Tran(A,0,0)*Rot(z,theta2)*Trans(B,0,-C)

Solving the matrices the final equations for P(x,y,z) are:

X = B(C1C2-S1S2) + AC1

Y = B(S1C2+C1S2) + AS1

Z = (-C)

or

X = BC12 + AC1

Y = BS12 + AS1

Z = (-C)

Finding the inverse I got:

Theta2 = CosInv{[X^2+Y^2-(A^2+B^2)]/2*A*B}

Theta1 = CosInv{[X(BC2+A)+Y(BS2)]/(A^2+B^2+2ABC2)}

Where

A = Length of first arm

B = Length of second arm

C = Lenght of third arm

Theta1 = Angle betn. base and the first arm

Theta2 = Angle betn. first arm and second arm

S1=Sin(theta1)

C1=Cos(theta1)

S2=Sin(theta2)

C2=Cos(theta2)

S12=Sin(theta1+theta2)

C12=Cos(theta1+theta2)

Note:

-I've not considered the thickness of the arms

-U need to find the signs of angles (correct quadrants)

-For the inverse there are multiple solutions, but i've considered only positive values

There's a VB program to solve the equations. Forward equations are working well. Inverse equations are ok, but u need to fix the proper quadrants.

Download it from here http://rapidshare.de/files/4013817/KinSolver.zip.html

Regds

happy_99

Overall transformation from O(0,0,0) to P(x,y,z) is given by:

Rot(z,theta1)*Tran(A,0,0)*Rot(z,theta2)*Trans(B,0,-C)

Solving the matrices the final equations for P(x,y,z) are:

X = B(C1C2-S1S2) + AC1

Y = B(S1C2+C1S2) + AS1

Z = (-C)

or

X = BC12 + AC1

Y = BS12 + AS1

Z = (-C)

Finding the inverse I got:

Theta2 = CosInv{[X^2+Y^2-(A^2+B^2)]/2*A*B}

Theta1 = CosInv{[X(BC2+A)+Y(BS2)]/(A^2+B^2+2ABC2)}

Where

A = Length of first arm

B = Length of second arm

C = Lenght of third arm

Theta1 = Angle betn. base and the first arm

Theta2 = Angle betn. first arm and second arm

S1=Sin(theta1)

C1=Cos(theta1)

S2=Sin(theta2)

C2=Cos(theta2)

S12=Sin(theta1+theta2)

C12=Cos(theta1+theta2)

Note:

-I've not considered the thickness of the arms

-U need to find the signs of angles (correct quadrants)

-For the inverse there are multiple solutions, but i've considered only positive values

There's a VB program to solve the equations. Forward equations are working well. Inverse equations are ok, but u need to fix the proper quadrants.

Download it from here http://rapidshare.de/files/4013817/KinSolver.zip.html

Regds

happy_99

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