I want to draw the workspace of a gripper finger

Greetings, friends.

My goal is to draw the workspace of the gripper finger with two joints.

It is common knowledge that the workspace is the set of points that can be reached with the fingertip. Therefore, I have the necessary math equations to draw the points and visualize the area that the finger can reach.

It is possible for me to draw all the points using the equations I derived from the mechanism, but I cannot rotate the points to simulate the rotation of the finger. You can see the red-marked area that needs to be rotated.

And this is the code used to draw the figure above.

clc;

oe=48.218;

j4p=166.696;

pb=86.488;

theta1=28.219;

j3j4=42.219;

j2j3=42;

hold on

for theta3=30:0.5:120;

for pb=110:-1:20

rdash=pb;

% rotation of the finger calculation

theta7=((90-theta3)*3.14159265359)/180; %theta 7 is radian

x=rdash*cos(theta7);

y=rdash*sin(theta7);

tb=y;

pt=pb-y;

theta5=atan(pt/x); %*180)/3.14159265359; theta 5 is radian

thetapp=theta7+theta5;

theta3rad=(theta3*3.14159265359)/180; %theta 3 is radian

% thetapprad=(thetapp*3.14159265359)/180;

pp=(pb*sin(theta3rad))/sin(thetapp);

pa=pb-oe;

j4a=sqrt(j4p^2-pa^2);

phi1=atan(pa/j4a)*180/3.14159265359;

%let px+x be p_x

p_x=j4a+24.658;

op=sqrt(pb^2+p_x^2);

phi2=atan(pb/p_x)*180/3.14159265359;

theta2=theta1-phi1;

theta2rad=(theta2*3.14159265359)/180;

j3d=j3j4*sin(theta2rad);

j4d=j3j4*cos(theta2rad);

j3g=j4d+24.668;

j3h=oe-j3d;

j2h=sqrt(j2j3^2-j3h^2);

j3f=p_x-j3g;

px=j3f+j2h;

x=p_x-px;

grid on

plot3(pp,p_x,pb,'b.');

end

end

Answer 1

[William Rose]

@Ahmed, your code follows:

clc;
oe=48.218;
j4p=166.696;
pb=86.488;
theta1=28.219; 
j3j4=42.219;
j2j3=42;
hold on
for theta3=30:0.5:120;
    for pb=110:-1:20
        rdash=pb;
        % rotation of the finger calculation
        theta7=((90-theta3)*3.14159265359)/180; %theta 7 is radian
        x=rdash*cos(theta7);
        y=rdash*sin(theta7);
        tb=y;
        pt=pb-y;
        theta5=atan(pt/x); %*180)/3.14159265359; theta 5 is radian
        thetapp=theta7+theta5;
        theta3rad=(theta3*3.14159265359)/180;  %theta 3 is radian
    %     thetapprad=(thetapp*3.14159265359)/180;
        pp=(pb*sin(theta3rad))/sin(thetapp);
        pa=pb-oe;
        j4a=sqrt(j4p^2-pa^2);
        phi1=atan(pa/j4a)*180/3.14159265359;
        %let px+x be p_x
        p_x=j4a+24.658;
        op=sqrt(pb^2+p_x^2);
        phi2=atan(pb/p_x)*180/3.14159265359;
        theta2=theta1-phi1;
        theta2rad=(theta2*3.14159265359)/180;
        j3d=j3j4*sin(theta2rad); 
        j4d=j3j4*cos(theta2rad);
        j3g=j4d+24.668;
        j3h=oe-j3d;
        j2h=sqrt(j2j3^2-j3h^2);
        j3f=p_x-j3g;
        px=j3f+j2h;
        x=p_x-px;
        grid on
        plot3(pp,p_x,pb,'b.');
    end
end

Pease indent your code in the normal way, to make it more readable. I have done so.

The script does not generate the red lines. Therefore I don;t know how to do a rotation that the red lines supposedly represent.

Please describe the mechanics of your finger more precisely so that we can understand how you generated the points you have genrated, and what exactly is going to rotate, relative to what. I suspect figures would help.

You said your finger has 2 joints. Does that mean there are 3 segments, and the proximal segment is mounted on a platform? Are the joints hinge joints, axial joints, ball joints, or something else? If they are hinge or axial joints, then each joint is characterized by a single angle. If they are hinge or axial joints, then please specify (as a vector) the axis of each joint. What is the length of each segment, measured from joint center to joint center, and the length of the distal segment (which has no joint center at its tip)?

One possibility that I can imagine is that you have a 3-segment finger with 2 hinge joints, and you want to add an axial joint at the base of the proximal segment, where that segment attaches to a mounting platform. This would allow the proximal segment (and the rest of the finger) to rotate about the long axis of the proximal segment. Tell us if that is your plan.

I recommend that you make your code more readable by adding clear useful comments and by defining variables that correspond to the physical quantities of interest, such as: L1, L2, L3=length of proximal, middle, and distal segments; theta12, theta23 = angles between segments 1 and 2, and between segments 2 and 3; n1, n2 = unit vectors for the axis orientations of joints 1 and 2; r1, r2 = position vectors for the current joint center locations. Such definitions will allow the writer, readers, and users of the code to have a much clearer understanding the model. Then, if you add rotation at the base, you will need to add theta01 (scalar), n0 (vector), and r0 (vector) for the angle, axis orientaiton, and joint center location, respectively.

If the joints between segments are ball joints, then you should consider quaternions to represent joint angles.

The script calls plot3() 16,471 times (=91*181). A single blue point is plotted on each call, at location x,y,z=pp, p_x, pb, where pp, p_b, px are scalars. This works, but it is not very efficient. A more efficient approach would be to make vectors of length 181, and call plot3() 91 times. The mesh() command could be useful and could prodce a visually pleasing result.

I am somewhat familiar with fitting finger reaching data. I wrote code to fit a convex hull and the best-fit cone around the points which a subject could reach with the tip of the thumb, while the rest of the hand was fixed. Reflective markers were placed at the joints and thumb tip. A 3D motion capture system was used to collect the set of reachable points. Thumb joint workspace, in steradians, was also computed.

[Ahmed]

Dear William Rose

Thank you very much for your prompt response.

To be honest, I don't use Matlab daily; I use it for this kind of problem. Below is a model of my finger:

As you can see, I have a base and two revolute joints. I wanted to draw all the points the fingertip could reach from P to B, considering that the first joint should be rotated with every PB movement.

Simply this is the problem.

Best regards

[William Rose]

@Ahmed, THis is very helpful. I will list what I think you want to do and you tell me if I am correct.

1. The system has two revolute (uniaxial) joints which we label joint 1 and joint 2. The angles at the joints are theta1 and theta2.
2. We want the coordinates x,y,z of the finger tip, as theta1 and theta2 vary through their range of motion.
3. The finger tip location (x,y,z) should be expressed in the coordinate system of (a) the fixed base (C0=base coordinate system=x0,y0,z0)) or (b) the first segment that rotates relative to the base (C1=(x1,y1,z1)).

Whis is correct: 3(a) or 3(b)?

If 3(a) is correct, then please provide the offset from the base to the center of joint 1. This distance is not given on the diagram.

The other question which the diagram does not answer is: What is the orientation of the axis of joint 2, in coordinate system C1? The diagram shows theta2 as a rotation about axis z2. But z2 is an axis in the coordinate system of the second (final) rotating segment. We need to know the orientaiton of joint 2 in C1 coordinate system.

My guess, from inspecting the diagram, is that the axis of joint 2 is perpendicular to z0 and z1. (z0 and z1 point in the same direction.) Is that correct?

If that is correct, then we may choose, without loss of generality, to define x1 so that it is parallel to the axis of joint 2.

[William Rose]

@Ahmed, Is it true that the axis of rotation of joint 2 is perpendicular to the two black bars that make up segment 1? That is what it looks like in the diagram you provided.

[Ahmed]

Dear William Rose

Really appreciate your help.

In the following, I have provided you with my answers to your question (in bold italics).

1. The system has two revolute (uniaxial) joints which we label joint 1 and joint 2. The angles at the joints are theta1 and theta2. Yes true.
2. We want the coordinates x,y,z of the finger tip, as theta1 and theta2 vary through their range of motion. These coordinates (x,y and z) are usually known because we want to move P to the target point, and the target point coordinates are always known.
3. The finger tip location (x,y,z) should be expressed in the coordinate system of (a) the fixed base (C0=base coordinate system=x0,y0,z0)) or (b) the first segment that rotates relative to the base (C1=(x1,y1,z1)).

Whis is correct: 3(a) or 3(b)? 3(b) is correct. It is possible to express the fingertip in terms of C0, but I have referred to C1 in my calculation.

If 3(a) is correct, then please provide the offset from the base to the center of joint 1. This distance is not given on the diagram.

The other question which the diagram does not answer is: What is the orientation of the axis of joint 2, in coordinate system C1? The diagram shows theta2 as a rotation about axis z2. But z2 is an axis in the coordinate system of the second (final) rotating segment. We need to know the orientaiton of joint 2 in C1 coordinate system. Z2 is perpendicular to the two black bars and the Z1 axis, so they have a 90-degree twist angle.

My guess, from inspecting the diagram, is that the axis of joint 2 is perpendicular to z0 and z1. (z0 and z1 point in the same direction.) Is that correct? Yes this is correct

If that is correct, then we may choose, without loss of generality, to define x1 so that it is parallel to the axis of joint 2. Sorry to which axis are you referring?

Many thanks for your help

Best regards

[William Rose]

@Ahmed,

The attached document gives an analysis of a general system with a base, two joints, and two moveable segments. I derive the general equations for the x,y,z coordinates of the reachable points, in the base, or global, coordinate system. Then I use the gerneal equations to get specific equations for your system.

I also made a figure that is similar to your but which has some important differences, which you will see when you compare my figure to yours. One of the differences is that in my analysis and figure, the coordinate axes of the base and both segments are all parallel (x0 || x1 || x2, etc) when the joint angles are zero.

When analyzing systems such as this, it is essential to clearly define a coordinate system in the base and in each moving segment. The fingertip, for example, is at a fixed location in the segment 2 coordinate system2. Segment 2 and its coordinate system move with respect to segment 1 when the relevant joint angle changes. Likewise for segment 1 with respect to the base. Therefore we use the vector and matrix equations for translation and rotation of axes, to transform the fingertip position from segment 2 coordinates to segment 1 coordinates to base coordinates. I have not written any Matlab code, but I have provided equations that make it possible to write code that is relatively short, simple, and clear, which will generate a 3D mesh in base, or global, coordinates, showing the reachable fingertip positions.

Figure 1. A system with a base, two joints, and two moving segments. There are three coordinate systems: C0 (base), C1 (fixed in Segment 1, purple) and C2 (fixed in Segment 2, yellow). The x, y, z axes of each coordinate system are shown with red, green, and blue arrows.

[William Rose]

@Ahmed,

Attached script shows one way to compute and plot the finger workspace.

1. Define the segment lengths.
2. Loop over the joint angles: joint 1 angle=0 to 2Pi; joint 2 angle sufficient to reach the centerline and to reach the level of the top of the fixed support post. Each time through the inner loop, compute the fingertip position p0, in the base (i.e. world) coordinate system: p0=R1*(R2*p2+r1)+r0. Save the x,y,z coordinates of each p0, for plotting in step 3.
3. Plot the surface of reachable points. Also plot the base, for visual reference.

I used z0=height of fixed support post=20 cm. This height was not given in your diagram.

[Ahmed]

Dear William Rose

My apologies for taking so long to respond to your prompt reply.

I will review your answers and get back to you.

Thanks again

[Ahmed]

Dear William Rose

Thank you very much for your detailed response to my question.

According to your answer, you solved the model using algebraic equations and coordinate transformations. My solution, however, was based on a geometric approach.

There is a maximum finger rotation of 60 degrees, but no further rotation is possible. In order to obtain the mesh shown below, I modified the m file.

This is exactly was I looking for because if you observe on both sides, the rotation appears.

The code is the same as shown below, but I modified the angles to offer the same workspace as the finger arm.

I should study your way more to explain it during my presentation. Can I have your permission to use your code and the provided solution?

Many thanks again

Best regards

%mapFingerReach.m W. Rose 2022-10-23

%Map the points reachable by a 2-joint finger gripper.

%See changeOfCoordinates.docx.

%See Matlab Answers question by Ahmed, 2022-10-22.

clear;

L0=20; Lx1=48.22; Lz1=24.67; L2=166.69; %segment lengths (cm)

r0=[0,0,L0]'; %location of C1 origin in C0 coords

r1=[Lx1,0,Lz1]'; %location of C2 origin in C1 coords

p2=[0,0,L2]'; %location of fingertip in C2 coords

a2min=-asin(Lx1/L2)/2; %minimum angle at joint 2

a2max=acos(-Lz1/L2); %maximum angle at joint 2

N1=60; %number of joint 1 angles, -1

N2=30; %number of joint 2 angles, -1

X=zeros(N1+1,N2+1); Y=X; Z=X; %allocate arrays for position values

%% Calculate x,y,z positions for ranges of joint angles

for i=0:N1

a1=(pi/3)*i/N1; %joint 1 angle

R1=[cos(a1), -sin(a1), 0; sin(a1), cos(a1), 0; 0, 0, 1]; %joint 1 rotation matrix

for j=0:N2

a2=a2min+(a2max-a2min)*j/N2; %joint 2 angle

R2=[cos(a2), 0, sin(a2); 0, 1, 0; -sin(a2), 0, cos(a2)]; %joint 2 rotation matrix

p0=R1*(R2*p2+r1)+r0; %fingertip position in C0 coords

X(i+1,j+1)=p0(1);

Y(i+1,j+1)=p0(2);

Z(i+1,j+1)=p0(3);

end

end

%% Plot surfaces showing base and reachable points

[Xb,Yb]=meshgrid([-220:20:220,-220:20:220]); %x, y grids for base

Zb=zeros(size(Xb)); %z grid for base

figure

mesh(X,Y,Z,'EdgeColor','r'); hold on

mesh(Xb,Yb,Zb,'EdgeColor','k','FaceColor',[.5,.5,.5])

xlabel('X (cm)'); ylabel('Y (cm)'); zlabel('Z (cm)');

axis equal %scale plot so 3D proportions are realistic

[William Rose]

@Ahmed, YOur figure looks good. The top edge of the surface in your figure looks like the arc of a circle, rather than a point. I expect it to be a point. Here is the image I get, using angle2=0 to 60 degrees. I also changed the base grid to [-50:50:250], and I changed N1 from 60 to 20, because 3-degree angle increments are sufficiently fine. Note that the top edge of the surface is a point, in the diagram below.

Yes you may use my explanation and code. Thank you for asking. Please cite the explanation as

Rose, W. "Change of coordinates", lecture notes, University of Delaware, October 2022.

and the code as

Rose, W. (2022) mapFingerReach.m, [Matlab code]. https://www.mathworks.com/matlabcentral/answers/uploaded_files/1169183/mapFingerReach.m.

[Ahmed]

Dear Sir,

In reality, it should look like the curve of a circle. I have mechanical constraints in the gripper that cause this. The gripper totally has six fingers.

Regards

[William Rose]

@Ahmed,

I chose to limit angle 2 to values that reach the centerline, and which go out far enough to reach the level of the top of the fixed base of support.

You could go father than the centerline with angle 2. Then you will reach across the centerline. Here is a plot with angle 2 minimum (a2min) = -pi/4 = -45 degrees:

[Ahmed]

Thank you very much, Sir

It has been noted, and thank you for your support.

Regards