How Do You Calculate Movements of a Robotic Arm?

[supermiedos]

Homework Statement

Given the following robotic arm:
http://i.imgur.com/TIb1vOk.png

and knowing that [itex]x_{c}[/itex] = Lcosθ and [itex]y_{c}[/itex] = Lsinθ, find expressions for
[itex]x_{m}[/itex] and [itex]y_{m}[/itex] and show that:
d[itex]x_{m}[/itex] = -[itex]y_{m}[/itex]dθ + ([itex]y_{c}[/itex]-[itex]y_{m}[/itex])d[itex]\phi[/itex]

Homework Equations

(Shown on picture below)

The Attempt at a Solution

Here it is my attempt at a solution:
http://i.imgur.com/Ee7iZWk.png

First I made a traslation to the origin and then I found the angles. Then, I got that triangle (4th pic) and found the relations between the angles and the sides. Next, I simplified both expressions using trigonometric identities and finally got d[itex]x_{m}[/itex].

But I don't get the book's answer! Could you help me please?

 

[mighty2000]

Dear fellow scientist,

Your attempt at solving this problem is commendable. However, I believe there may be a small error in your calculations. In the first step, you translated the coordinates to the origin, but in the next step, you used the coordinates from the original position of the robotic arm. This could be the reason for the discrepancy in your answer.

Here is my attempt at solving the problem:

First, let's define the coordinates of the end effector as x_{e} and y_{e} in the original position of the robotic arm. From the given information, we know that x_{c} = Lcosθ and y_{c} = Lsinθ. We also know that x_{m} = x_{e} - x_{c} and y_{m} = y_{e} - y_{c}.

Now, using the Pythagorean theorem, we can express x_{e} and y_{e} in terms of L and θ:

x_{e} = Lcosθ + Lcos\phi

y_{e} = Lsinθ + Lsin\phi

Substituting these values in the expressions for x_{m} and y_{m}, we get:

x_{m} = (Lcosθ + Lcos\phi) - Lcosθ

y_{m} = (Lsinθ + Lsin\phi) - Lsinθ

Simplifying, we get:

x_{m} = Lcos\phi

y_{m} = Lsin\phi

Now, to show that dx_{m} = -y_{m}dθ + (y_{c}-y_{m})d\phi, we first need to find the values of y_{c} and y_{m} in terms of L and θ.

From the given information, we know that y_{c} = Lsinθ and y_{m} = Lsin\phi. Substituting these values, we get:

dx_{m} = -Lsin\phi dθ + (Lsinθ - Lsin\phi)d\phi

Simplifying, we get:

dx_{m} = -y_{m}dθ + (y_{c}-y_{m})d\phi

Hence, we have shown that dx_{m} = -y_{m}dθ + (y_{c}-y_{m})d