# Tag Info

5

The updated image solves the problem. You did not consider the end-effector coordinate frame earlier. Also, the crosses (going into) in the diagrams should be replaced by dots(coming out), because the crosses don't hold the right hand rule in case you are using a right hand coordinate system.

4

A frame in that configuration or any other witch holds the props in a square without being unbalanced would work, the reason that most people do it with cross bars is because the FC batteries and ESCs must go somewhere. For your proposed solution you would have to balance all of that equipment around the ring running power and control lines all around ...

4

It looks like 80/20 to me: http://www.8020.net/ I've used it many times...pretty useful stuff.

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Add a coordinate system that matches the previous coordinate system exactly. The last rotary joint will be the parameter for the next-to-last coordinate system, and the link length will be the parameter for the prismatic joint. Note: a simpler approach is possible, but this gets the job done easily.

3

To answer your questions directly, the X-shape is not the only design that will work. There are many variations of frame design around a number of rotors that can range from 3 to (in some cases) 8. Since it hasn't been mentioned in the other answers, I want to point out that perhaps the biggest reason for the X shape is to keep the weight in the center, ...

3

Wow, this is a spectacularly detailed question, so thanks for that, welcome to Robotics, and I'm sorry you're having trouble with this. The trouble that you're having is that you're inverting the wrong matrix! Allow me to explain. You have two matrices, $R_r$ and $R_{\ell}$. To be more specific, as you have, these are the matrices that convert world ...

3

As pointed out in my earlier comment, this is actually simpler than you may think. Remember, $qs$ and $qr$ are fundamentally different, where the former represents orientation (in reference to the outside world) and the latter represents rotation (irrespective of any reference coordinate system). You're right in saying that you don't need to do any ...

3

The mistake is in the second matrix $T_B$ in the last column this column means that you want to transfer the Frame by -2 in it's X direction and by -2 in it's Y direction which in your case is like this: The X vector of the second Frame (A-Frame) is the Y Vector of the world Frame (O-Frame) The Y vector of the second Frame (A-Frame) is the opposite of X ...

3

The original T-slot frame system, was developed in the 80's by item GmbH and was quickly developed into the MB Building Kit System, hence I've always known it as item bar. Since then though, there have been many other manufacturers producing similar systems. Examples include Bosch Rexroth (as mentioned by Guy Sirton), 80/20 (Andrew Capodieci's suggestion), ...

3

A homogeneous transformation matrix $H$ is often used as a matrix to perform transformations from one frame to another frame, expressed in the former frame. The translation vector thus includes [x,y(,z)] coordinates of the latter frame expressed in the former. Perhaps that this already answers your question, but below is a more elaborate explanation. The ...

3

First you need to integrate the output from the gyro to get the actual X, Y and Z angles. angleX = gyroAngleX + gyroInputX angleY = gyroAngleY + gyroInputY However this value will drift over time so you will need to use a complementary filter or kalman filter. Personally, I would recommend a complementary filter because it is much simpler to implement. ...

3

Look into a complementary filter. It isn't the correct way to go out this but it will give you usable data for attitudes around level. It's also worth mentioning that you will not be able to track yaw. There is no way to account for bias/noise with the two sensors you've listed. complementary filter: http://www.pieter-jan.com/node/11

2

You've written your equations as if the weight vector you drew was positive, but then used a negative weight vector in your calculation. If you flipped $W$ around in your drawing you'd get \begin{align} W_x &= -W \sin(\theta) \\ W_z &= W \cos(\theta) \end{align} ** edit ** To expand, a more consistent approach is use to proper vector math. In ...

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Here are the step that are in my course for using D-H convention: Step 1 Define the z axis of every joint Step 2 Define the origin frame R0 (O0, x0, y0, z0) Step 3 For i in [1, n] (n=number of joint) Step 3.1 Define Oi that belongs to zi, and to the common normal to zi-1 and zi Step 3.2 Define xi such as xi is normal to the plan formed by zi-1 and ...

1

The IMU senses deviations from gravity within the inertial frame. Essentially, IMU measures the specific force $f_b$ given in the base frame as: $$f_b = R^{bn}(a_{ii}^n-g^n),$$ where $g^n$ is the gravity in the navigation frame (NED), $a_{ii}^n$ is the inertial acceleration expressed in NED frame and, finally, $R^{bn}$ is the rotation matrix from NED to ...

1

Depending on which matrix you convert to an axis angle representation. If you do it successively, you end up with 3 axis with 3 angles in 3 different frames ($R_0$, $R_1$ and $R_2$). It is trivial to notice, that in this the axes will coincide with axes of the successive frames. If you choose to take the result of the multiplication and write it in axis ...

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If I understand your question correctly you are asking about what rules there are for assigning coordinate frame axes compatible with DH parameters? There are definitely rules when assigning coordinate frames. In order to guarantee that there exists a homogenous transformation from frame 1->0 must follow the properties Axis $x_1$ is perpendicular to axis $... 1 It sounds to me like you have a system with an unknown parameter that you want to estimate. That is the purview of parameter estimation literature. If this parameter to be estimated is not changing over time (as yours is not), then you can use old-school methods like LS or similar, as you suggest. This particular brand of parameter estimation is known as ... 1 suppose you calculate the positions of your joints$p_i, i=1\ldots6using your homogeneous matrices. The best solution in my experience is to plot the initial configuration e.g. like this: // calc p for time t=0 f=figure(1); plot3([0, p_1(1)], [0, p_1(2)], [0, p_1(3)]); hold on; plot3([p_1(1), p_2(1)], [p_1(2), p_2(2)], [p_1(3), p_2(3)]) .... // for all ... 1 I mentor several FTC (First Tech Challenge) Robotics Teams. These teams build medium size robots (about 50cm cube starting size). For this size, there are four Building Systems that are the most common Tetrix Matrix Actobotics 8020 Extrusions (There are lots of other suppliers too) These solutions are all very flexible allow a large about of options on ... 1 Turns out I was transforming before interpolating the data, and my interpolation function assumed that the two streams of data were taken at the same time stamps. A simple mistake. So I was comparing data from different points in time. Consequently, it makes sense that resulting fitted transformation was incorrect. @Chuck -- The resulting (correct) ... 1 I found this video helpful when learning about the DH method. DH is all about describing the differences between coordinate systems using rotation and translation about/along the X and Z components of the coordinate system. EDIT It seems that your frames {0} and {1} are not in the correct location if your wanting to follow the DH convention. When your on ... 1 Each parameter is for a simple transform such that when all 4 are combined you go from one frame to the next. It can be helpful to understand each of the 4 simple transforms, and you should do that if you haven't already. 1 Your first step works because it is implied that your frames are "aligned", meaning that: $$\theta_{X_0} = 0 \\ \theta_{Y_0} = 0 \\ \theta_{Z_0} = 0 \\$$ In general (as with any integration!), you have some starting angles(initial condition),\theta_{(X,Y,Z)_0}\$, and you proceed to update from there, such that: $$\theta_{X_N} = \theta_{X_{N-1}} + \... 1 I'm not sure what you mean by: R_A^C must for a 2x2 matrix be defined as [xa \cdot xb , xa \cdot xb ; ya \cdot yb , ya \cdot yb] because I don't know where you are getting the names or formulas from. If you have:$$ R_A^B = \left[ \begin{array}{ccc} a & b \\ c & d \end{array} \right] \\ R_B^C = \left[ \begin{array}{ccc} e & f \\ g & ...

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