Author Topic: Arm Designs (Read 12878 times)

gamefreak · « **on:** January 22, 2008, 04:48:02 PM »

Im having trouble deciding between have a double bar design( two bars spaced x amount apart extend equally and attach to a seperate part still spaced x apart, the end moves in a circular pattern) with a DOF or two on the end,or just having multiple arm parts( a basic arm).

any ideas on the specific uses on the two?

ed1380 · « **Reply #1 on:** January 22, 2008, 05:15:57 PM »

if you have the 2 rods paralel to each other and they are the same length, then the ends will always be parallel. (base and arm for example)

gamefreak · « **Reply #2 on:** January 22, 2008, 06:47:53 PM »

Alright, and if I put weight on the back end could I reduce the required torque to rotate the arm(theoretically down to 0, realistically maybe down to half)?

ed1380 · « **Reply #3 on:** January 22, 2008, 06:50:45 PM »

sure. the weight just has to be on the oposite side of the claw(load) when related to the fulcrum

gamefreak · « **Reply #4 on:** January 22, 2008, 07:23:44 PM »

alright, so what exactly would it do to the torque required, I know its in Admins Static tutorial, relating to moments(admin you need to add units, some of us havent taken physics, and half of those dont want to look things up). I happen to live in the backwards US which doesnt like metrics, so i need some help on what to do with my numbers.....Lets say I have an arm that 5 inches(12.7 cm) away from the fulcrum has a mass of 1 pound, which is 453.59 grams and lets say my arm weighs 200 grams, so in total 653.59 grams which is 23.05 oz, I have an 69.43 oz/in(HS-635 HB) servo and a 49 oz in(HS-311) servo. So i know that these servos cant handle it, but can someone help me on the math since there is a lack of units....

It is technically possible for me to create gears for this, my school finally got a laser cutter, but the gears would be out of wood and I would have no way to mount them(yay for noobishness)

ed1380 · « **Reply #5 on:** January 22, 2008, 07:50:20 PM »

if you put the counter balance the same distance from the fulcrum it will balance and make the servos work alot easier

now about the eqations and stuff. look at admins arm tutorial. it tells how to even include the arm weight

gamefreak · « **Reply #6 on:** January 22, 2008, 08:01:07 PM »

yet again their are no units, I find it difficult when there is absolutely no units for the measurements, and if i use the wrong units then the calculations are wrong. I would also not want to use the calculators because I would rather do it myself to make sure it all works.

ed1380 · « **Reply #7 on:** January 22, 2008, 08:17:42 PM »

scroll halfway down
http://www.societyofrobots.com/robot_arm_tutorial.shtml

units dont matter. as long as you keep everything in inches, lb, etc. and stick to them it should be ok

gamefreak · « **Reply #8 on:** January 22, 2008, 08:59:35 PM »

alright while without doing to much math, if i use my 69.43 oz/in servo, I could use a 4 inch arm with 1 lb on the end(just barely)

but what im confused about is the counterweight, the motor has to have enough torque to lift it as well, so if I put 2 lbs 2 inches away on the far end, would my servo only require to lift a maximum of 1 lb/in (assuming that 1 lb is the total weight of the arm on the gripper side)

Also, wouldnt the point where the servo translates its motion to the arm matter as well? if I used a basic lever then I wouldnt need a counter weight, if i put the 69.43 oz/in servo down and had a string going from it to a point 2 inches away from the fulcrum on the side opposite the gripper, then I could lift a total weight of 128.85 oz, since the servos rotation is simply pulling on a string I no longer have to worry about the inch part of the equation and I could simplify the entire process, correct?(only if the gripper end was one inch long)

So if I used a continous rotation servo and put a spool on it, then depending on the size of the spool it would also change the results, if i used a spool with a half inch radius then I would get double the 128 but half the speed, right?So using the spool way, i would need a spool with a 1 inch radius, going up to to a point 2 inchs away from the fulcrum opposite the gripper, and the gripper attached 1 inch away from the fulcrum, but then the only weight that im not sure of is the weight of the two arm beams.

Admin · « **Reply #9 on:** January 22, 2008, 09:45:30 PM »

hmmmm I have units everywhere . . . for example:
http://www.societyofrobots.com/robot_arm_calculator.shtml
lets you choose your own units . . . and do all your math for you . . .

and google can even do unit conversions for you, such as:
4 pounds * 3 inches to newton meters

now for the counterweight . . . the point of the counterweight is to reduce the amount of torque an arm needs to lift . . . the disadvantage is that it increases required dynamic torque (reduces arm acceleration)

so if you have a 4 inch arm lifting a 1 lb weight on one end, and a 4 lb counterweight at 1 inch away on the other end, then that means your actuator will not require any static torque (but will require dynamic torque, for acceleration)

now for the parallel bars . . . it doesn't give your arm a lifting advantage, but it is however a way to make your arm more rigid but also lighter at the same time, in exchange for being more complicated and taking up more space. There is a mathematical way to calculate which design is better, but that'd require another tutorial . . .

ed1380 · « **Reply #10 on:** January 22, 2008, 10:00:37 PM »

Quote from: Admin on January 22, 2008, 09:45:30 PM

hmmmm I have units everywhere . . . for example:
http://www.societyofrobots.com/robot_arm_calculator.shtml
lets you choose your own units . . . and do all your math for you . . .

and google can even do unit conversions for you, such as:
4 pounds * 3 inches to newton meters

now for the counterweight . . . the point of the counterweight is to reduce the amount of torque an arm needs to lift . . . the disadvantage is that it increases required dynamic torque (reduces arm acceleration)

so if you have a 4 inch arm lifting a 1 lb weight on one end, and a 4 lb counterweight at 1 inch away on the other end, then that means your actuator will not require any static torque (but will require dynamic torque, for acceleration)

now for the parallel bars . . . it doesn't give your arm a lifting advantage, but it is however a way to make your arm more rigid but also lighter at the same time, in exchange for being more complicated and taking up more space. There is a mathematical way to calculate which design is better, but that'd require another tutorial . . .

and it makes the ends paralel

gamefreak · « **Reply #11 on:** January 22, 2008, 10:05:34 PM »

since the point( at least with this project) of the counter weight is to reduce the required torque for motion wouldn't it be conter intuitive to balance it, wouldnt you want something like 1 lb 4 inches away, and 2 lb 1 inch away, that way if there is the expected 1 lb weight you only have to lift 2 lbs, and if the weight is not present, you only have to lift two pounds.

Also, to go along with the stream of ideas of my previous post, if you attached the motor( via a lever and translational movement) on the side of the gripper, two inches in, wouldn't that be the same as applying translational movement to a point two inches away from the fulcrum on the opposite side of the gripper.

Admin · « **Reply #12 on:** January 22, 2008, 10:16:01 PM »

Quote

wouldnt you want something like 1 lb 4 inches away, and 2 lb 1 inch away, that way if there is the expected 1 lb weight you only have to lift 2 lbs, and if the weight is not present, you only have to lift two pounds.

really up to you . . . if your arm was idle (not lifting anything) 90% of the time, in terms of energy efficiency, you wouldn't want the arm lifting its counter weight. but having a heavier counterweight also lets your robot lift heavier objects . . .

Quote

if you attached the motor( via a lever and translational movement) on the side of the gripper, two inches in, wouldn't that be the same as applying translational movement to a point two inches away from the fulcrum on the opposite side of the gripper.

got a pic?

gamefreak · « **Reply #13 on:** January 24, 2008, 08:09:24 PM »

Alright, the top pic is of a servo(yellow) with a spool on it, with a string(red) going from the 2 inch side of the fulcrum to the spool, the bottom one is the servo attached to a solid rod going two inches towards the gripper, would these be exactly the same in terms of torque needed?

And what does everyone think of having a block for the arm to pick up be the letters SOR

Admin · « **Reply #14 on:** January 24, 2008, 08:23:29 PM »

Quote

would these be exactly the same in terms of torque needed?

Yeap, on condition that the distance from the center of rotation is the same. I'd go for the rigid link, but placed in back.

gamefreak · « **Reply #15 on:** January 25, 2008, 06:43:53 AM »

why have a rigid link at all, and why in the back?

ed1380 · « **Reply #16 on:** January 25, 2008, 07:04:13 AM »

rigid cause then you can control both up and down motion

in the back- umm IDK. i think so that it wouldnt get in the way of anything

Admin · « **Reply #17 on:** January 25, 2008, 08:24:12 AM »

Quote

in the back- umm IDK. i think so that it wouldnt get in the way of anything

yup

gamefreak · « **Reply #18 on:** January 28, 2008, 08:49:31 PM »

Alright i've been doodling out the current design and have come up with some numbers, with a 63.5 oz/in servo connected two inches away from the fulcrum i would have 127 oz/in, if im expecting one pound of weight 4 inches away from the fulcrum on the other side then that is 64 ounces, so i get 63 ounces of lifting power, but if i put a 1 pound weight over the rigid linkage conneccting the servo to the arm, then that is 32 ounces on one side, 64-32 is 32 oz, 127-32, 95 oz(almost 6 pounds) of excess lifting power(this is without the weight of the arm linkages themselves), but this arm is very small, about 6 inches tall, and maybe 9 inches long, should i consider scaling it up since i have a large amount of lifting power left over?

Admin · « **Reply #19 on:** January 29, 2008, 08:19:36 AM »

Nah thats not right, at least based on your drawing.

Your servo is connected at an angle, so you need to multiply your lifting force by cos(angle).

Can you draw out a free body diagram (FBD)? Basically a drawing of the arm with all the forces, torques, angles, and lengths labeled.

gamefreak · « **Reply #20 on:** January 29, 2008, 07:03:11 PM »

alright, here is an idea drawn out

Im also considering moving the servo so that the arm is straight up.

Admin · « **Reply #21 on:** January 30, 2008, 09:00:16 AM »

two small comments . . . how do you plan to connect that servo? If your servo horn is .5", that means your robot arm will only be able to move up/down 2x.5" = 1"

And of course, a longer servo arm decreases lifting force, so you'd have to calculate that too.

Also, you may want to increase the parallel bar distance from 1" to maybe 1.5" or 2" so it doesn't collide at the extremes.

gamefreak · « **Reply #22 on:** January 30, 2008, 08:46:21 PM »

I intend for a 1 inch servo arm, that way my math stays the same, and i get 2", so double that, the end effector moves up/down 4 inches, I am still considering the distance of the parallel bars to be 2 inches.

I still have a large amount of foam left over from my last project, would it make a good prototyping material? what about building?

ed1380 · « **Reply #23 on:** January 30, 2008, 10:07:13 PM »

if you're actually going to put some stress on the test arm, i dont think so.

gamefreak · « **Reply #24 on:** January 31, 2008, 10:59:34 AM »

I dont really expect to lift 5 or 6 pounds, it would probably lift up alphabit blocks and spell out things for the judges.

ed1380 · « **Reply #25 on:** January 31, 2008, 03:06:23 PM »

all the styrafoam i can think of would be too weak to trust.

use some hdpe (sp?) or aluminum

Admin · « **Reply #26 on:** January 31, 2008, 03:20:38 PM »

if you want to prototype it, try wood, cardboard, or legos

gamefreak · « **Reply #27 on:** February 01, 2008, 11:20:43 AM »

My final building material will probably be wood, but the stiff styrofoam(like the kind you used for your boat), wouldnt even be good for trying the differnt distances for the beams?

Admin · « **Reply #28 on:** February 02, 2008, 09:16:04 AM »

If you just want to try out different distances, you should CAD your design

But yea, the pink foam I used on my robot boat, its really stiff so it'll work . . . it can't hold much weight tho . . .

gamefreak · « **Reply #29 on:** February 03, 2008, 05:02:55 PM »

Alright, Heres an idea that recently hit me, to rotate the arm lengths it involves the torque of a motor since it is a weight over a distance, but what if I had an arm length of 0 inches? Such as the motor that will rotate the arm at its base around the z axis(for me, z is up), how do I calculate the torque neccasary for that since its impossible to divide anything by 0 since it technically results in infinity.

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Author Topic: Arm Designs (Read 12878 times)