Author Topic: DH Parameter question!  (Read 736 times)

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Offline JjanniTopic starter

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DH Parameter question!
« on: October 08, 2013, 02:35:58 PM »
Hello,

I am having a problem that I hope to find some help for, this might be the right place!

I want to find a DH parameter table for a 3 degree of freedom robot, and made an illustration picture as linked below.

The problem is that when I found my solution, I stumbled upon someone who also have made his DH parameter table for this robot and they show different results.
He used it in a master thesis, so I can't convince myself that his answer is wrong too easily.

As shown in the picture, only the 3rd Link-parameters differ, and as you can see he does not account for the horisontal length of Link 3. Is this allowed? The axis definitions is made by me, I choose to put the x3-axis horisontal because this was the only way I could include the a3-length in the parameters. But is it possible to just ignore this?

I am going to use this for complex calculations later, so I hope to make the DH-parameters as easy to work with as possible.


Note: This is the ABB IRB140 robot, which is a 6DOF robot, but it is simplified for my problem to a 3DOF robot. The other guy as done the same thing, but he calculated the full 6DOF DH-table where the a3-length is included, but only used the DH-table as shown for the 3DOF version of the robot.

Thanks!
« Last Edit: October 09, 2013, 06:17:43 AM by Jjanni »

Offline bdeuell

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Re: DH Parameter question!
« Reply #1 on: October 08, 2013, 03:39:40 PM »
Please upload a higher resolution image, I am not able to read any of the labels/variables on the image you supplied.

Offline JjanniTopic starter

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Re: DH Parameter question!
« Reply #2 on: October 08, 2013, 04:15:21 PM »
Thank you for your interest.
« Last Edit: October 09, 2013, 06:15:55 AM by Jjanni »

Offline bdeuell

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Re: DH Parameter question!
« Reply #3 on: October 08, 2013, 05:33:52 PM »
I got the same results as you when calculating the DH parameters from your drawing.

I can't say if the other solution is wrong without seeing what they were intending to describe but I can say that the other solution is describing an arm where the origin of end effector is located on joint 2 (a3 = 0) ... not somthing I would expect to see. If it is assumed that both of you are trying to describe the same system shown in your drawing the other solution would need to have a non zero a3 value. Given that correction the two DH parameters would describe equivalent physical robots the difference being the other solution modeled link 3 being straight up.

Offline JjanniTopic starter

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Re: DH Parameter question!
« Reply #4 on: October 08, 2013, 05:46:56 PM »
Thank you so much, that is comforting considering all the work I have done based from these parameters.

We are both describing the same robot in the same starting configuration. What the other guy did was create a full 6DOF DH-table (all other joints are on the link described as a3 on my drawing), and then he said that we could simplify the robot to a 3DOF. So he just used the first 3DOF-parameters in his full DH-table to describe the new 3DOF robot.

My biggest question is if it is possible to align the last frame in any other way than having x3 being horisontal as shown in the picture?
I am not very good with this stuff yet, but if I understand correctly, the only way to implement the a3-length into the parameter table is to use the length in the direction of x3 OR the direction of z2. And z2 is forced to be in the direction showed in the picture, so we have to use x3 to get the a3-length.

If you can confirm or deny this, I would be very greatful.

Offline bdeuell

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Re: DH Parameter question!
« Reply #5 on: October 08, 2013, 06:55:50 PM »
I looked up the robot you are both working with. It appears the other solution is perfectly accurate to describe the first three joints of the robot but it is important to note that the fourth joint of this robot is a twisting joint ... thus the reason C3 is defined as it is. As I said before I can't say if this is what they were intending to describe without seeing their work. Your model does not match the one described in the DH parameters of the other solution. I'm sure if you look at the other full 6DOF solution you will find that they are represent your length a3 as d4 which is where this link would be incorporated into the DH parameters. The question is what point on the robot do you want to represent as the end effector in your simplified model?

My biggest question is if it is possible to align the last frame in any other way than having x3 being horisontal as shown in the picture?
I am not very good with this stuff yet, but if I understand correctly, the only way to implement the a3-length into the parameter table is to use the length in the direction of x3 OR the direction of z2. And z2 is forced to be in the direction showed in the picture, so we have to use x3 to get the a3-length.
Given the position you drew your robot in and the location you selected to represent the end effector in your simplified model the direction of x3 is fixed. As I alluded to previously it is possible to draw the same physical robot in a different configuration ie. the joints being in different starting positions to get different solutions for the DH parameters (having different angular offsets).

Offline JjanniTopic starter

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Re: DH Parameter question!
« Reply #6 on: October 08, 2013, 07:30:02 PM »
Yes they indeed implement the length that I called a3 in the full 6DOF model. This (full) DH-table that he made, I agree is correct. The only thing that seemed weird to me was the use of the table shown for the simplified 3DOF-model, without the needed modifications. He simply cut out the 3 first parameters of the 6DOF model and presented them as the DH table for the new 3DOF robot (in the same position as my drawing). His end effector point is the same as mine throughout his thesis.

My end effector will be on the end of the a3-link. My goal is to implement a time optimal control structure for a circle trajectory drawn by the end effector. I come from a control and optimization background, without too much knowledge of robotics. Since the first 3DOF can cover the entire workspace of the robot, I can simplify the model, at least that was my thought process.

So I guess I will continue using my DH table then.

Thank you again so much for taking the time to help, it is not easy finding answers to questions like these!
« Last Edit: October 08, 2013, 07:33:35 PM by Jjanni »

Offline bdeuell

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Re: DH Parameter question!
« Reply #7 on: October 08, 2013, 08:11:25 PM »
Using your DH parameters and simplified model should work fine to follow a circular trajectory with a point on the end of the arm, just keep the 5th joint on the arm straight and adjust your a3 dimension to reflect the total length of the arm.

I'm glad I was able to help.





 


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