I will explain it better.
My Stepper mottor will move an endless screw that will move the cylinder A and the cylinder A will move to the cylinder B.
Remember that cylinders are tires and I want to use them as closed hydraulic cylinders. That is my question. Is this possible to do?
of course it is possible. liquids don't compress as easily as gases so this is simple replacement of the media. if you cannot get all air bubbles out, those bubbles will be compressed and cause in softer action (less positionally accurate). long term compatibility is another story. there may be some corrosion due to water contact or failure of seals if they get dissolved or hardened by oil. but... if that happens, that is what maintenance is for...
btw. what do you mean by "tire"? is that a typo for air?
I want to use a pneumatic cylinder as a hydraulic.
I'm going to buy this: https://www.amazon.com/-/es/gp/product/B07DYPM7X9/
And this: https://www.amazon.com/-/es/gp/product/B08FR11BLK/
And this oil: https://www.amazon.com/-/es/gp/product/B001FAE4T2
I ask this question before buying the cylinder, because I have never had one. I plan to buy 2 pneumatic cylinders with oil inside and use them as manual actuators, that is, I pushed it on one side and on the other side the strength goes out.
My question is: it is possible to do that?
Note: I know the difference between tire and hydraulic, the thing is that it is easier to buy pneumatic cylinders than the hydraulics, besides that they do not sell Small Hydraulic Cylinders (in Amazon, for example). That's why I want to comprise the tire.
Why not use direct coupling or a gear reducer?
I do not want to put the actuator in the joint, but to stay in the base. That is, the torque must be transmitted by X via (eg hydraulics).
Which one would you recommend? I am looking for the simplest way to do it.
I am building a robot arm on my own. My main problem is the mechanical transmission. I plan to use pulleys and metal strings, but when I bought it I realized two things: they weigh more than the metallic structure and are not accurate.
I think the most convenient thing is to use hydraulic pistons powered by a steppeper motor with a screw (linear movement), but, I'm still not quite sure.
Do you recommend that one or another better?
I have a somewhat basic question about mechanisms, I watch videos and read, but they do not give me the answer I am looking for, they only focus on clearing the formula to know the RPM of the last pulley.
I know that if two pulleys are fitted and the drive pulley is smaller in diameter than the take-up pulley then the take-up pulley will have higher torque and lower RPM compared to the drive pulley. If it is twice the size, then in theory it will have twice the strength.
My question is regarding the attached image. Imagine that pulley 4 has the same diameter as pulley 2, and pulley 2 has a transmission ratio of 2: 1 (twice the size, twice the torque, less speed) relative to that of the driving pulley (the 1), so ... will pulley 4 have 4: 1 torque, or am I wrong?
Because if you don't have a higher torque, why are you using multiple pulleys instead of just one with a bigger belt?
Regards.
Hi.
I found powerful motors, and not so expensive (Stepper Motor Nema 34, Torque 12 Nm). But, my current problem is in relation to the transmission.
That is, the engine is powerful, but it is also heavy, I do not want to put it in the air next to the arm because the torque it would apply would be greater because it would carry its own weight.
1. I must use reducers, to multiply the torque.
2. I can't use gears because they must be close to the joint, and using longer gears would make the arm more expensive.
3. If I use toothed pulley reducers it would allow me to do what I want, but it would be very unsightly since to increase torque they must have a large diameter. That is, it is not an option for aesthetics.
Can you think of a solution that allows me to have the motors in the base and the movement is transmitted to the body? Multiple small diameter pulleys and threads were cut out for me, but I don't know if it's very precise.
At least two small motors, for the two axes (one that rotates and the other that advances). I imagine it as a ball, so I recommend you see the design of BB-8, a company sells one like that. You need the battery: https://www.amazon.com/Pisugar-Portab…s/dp/B07RC649ZC
I would put a gyroscope and accelerometer, a little horn, a vibrator. You can add a thousand things to it. I guess what I say is basic, I think you should ask a more precise question, what is the design you have?
Now it does give me like you, but my question is: why is it multiplied by 180 rad and not by 180 degrees (unless the formula is so)?
If the arm is down, resting
That is my other question, because 180 gives 0. It is assumed that to move the motor it must perform X torque.
Obviously it won't have exactly 180 degrees, maybe 177 or 178, but if 180 gives me 0, it makes me doubt that formula.
Unless in theory it is 0, and in practice the torque is very small due to the losses in the mechanical system.
This video is related (in case someone in the future has the same problem): Degrees and anatomical and geometric angles do not get confused anymore
Can you clarify which calculations you used? I don't care about the image.
Sorry to bother you so much, but how does speed affect the formula?
If the engine revolution is 2000 RPM and a 10:1 speed reducer is used the speed would be 200 RPM. Is that what you mean?
In relation to the other "positional accuracy", you mean my arm knows exactly where each joint is?
1. sure but keep in mind other factors like velocity and positional accuracy
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What do you mean by this?
reaction forces due gravity will need to be taken by some mechanical device,
Thanks for the clarification.
Question 1 (Just to recap):
1. Calculate the torque (T = r * F * sinθ).
2. I am looking for a speed reducer that will increase the torque of the motor to be able to lift the weight of the arm.
3. I calculate the efficiency of the reducer, that is, if it is 80%, I multiply it by 0.8
I know it won't be 100% accurate, but it will help me understand which motor to buy for which joint. True?
Question 2:
In the case that it is a movement of turning the arm, that is, not raising or lowering it or moving it from right to left, just turning it. The arm would be straight (I assume it is a 180 degree angle, if I am not wrong), that is, the torque needed to turn it would be a lot less, so I can look for a smaller motor. True?
Example:
Greetings and thanks again for everything.
That is, the torque of the formula T = F * D is the maximum to hold it, but to move it you need more force, right? Thanks for that tip, it helps improve my logical reasoning.
yes... that would ensure 30Nm, Up to 19.8Nm would be used to balance out effects of gravity. remaining torque (at least 10.2Nm) would be used to lift the arm.
I'm researching and reading more on how to calculate it in 3D to make it a bit more accurate. But I have a question. I plan to use several stepper motors. For example, one that raises and lowers the entire arm, another that moves it from right to left, and another that rotates it. What I am analyzing is that each of my motors will perform a torque in 2D. That is, the formula T = F * D * sinθ
Will it be enough, or did I misunderstand?
I don't see or understand where the motor will torque on the Z axis.
Look at the same page as before with another calculator for torque as a vector product:
If I raise my arm in the middle, that is, without stretching it, it would be 0.5m.
But, it is a hypothetical case. Let's say the mass of the arm is 2kg (19.6 N) and the distance 0.5m.
T = 19.6N* 0.5m= 9.8Nm
1- I mean, I need a motor with a torque of more than 9.8 Nm to be able to lift that arm, right?
2- If my motor has a torque of 3Nm, and I put a reducer with a ratio of 10: 1, it means that it will have enough force to lift it, right?
For example I have no idea where 3Nm came from.
The stepper motor says it has a torque of 3 Nm, that's where it came from.
The 2 kg (or 19.6 N) came from the hypothetical arm weight with the motors.
0.5 m is the distance the arm must lift.
I am multiplying the force exerted by the weight of the arm by the distance it must lift. The result is the theoretical torque that the motor needs to raise said arm, right?
By the way, I don't understand this:
if motor is 4000RPM, you need gearbox with ratio 4000/9.55 which is 419:1.
if motor is 1600RPM, you need gearbox with ratio 1600/9.55 which is 167:1.
According to this page: https://woodgears.ca/gear/ratio.html
Set the different ratio. I quote:
Suppose you have a motor that turns 1200 RPM (revolutions per minute), and you need to turn something at 500 RPM. The ratio you need is 500:1200, or 5:12. However, simple gears with only 5 teeth tend to run a bit rough, so your best bet is to make (or obtain) gears with 10 and 24 teeth.
I say this because I think it is important to know at the time of purchase ... whenever I see a gearbox on pages like Amazon, they place a ratio like 10: 1 or i = 10.
I just assume that you have to multiply the force by 10 and divide the speed by 10, or the opposite.
I am honestly surprised that you wrote all that just to help me. I sincerely appreciate it. You did not judge or tell me to drop the idea (as other members of other forums have done). You clarified several points that I had doubts about.
Right now I am ready to buy the motors (Stepper Motors). Can you take a look at them so you can give me your opinion? I want to avoid losing money and buying what can work.
The arm will have 4 actuators:
2 actuators on the shoulders (3 Nm): https://www.amazon.com/-/es/23-Motor-…/dp/B00QG32Y86/
1 actuator in elbow (1.9 Nm): https://www.amazon.com/-/es/stepperon…/dp/B0781GNX8W/
1 actuator on the wrist (0.46 Nm): https://www.amazon.com/-/es/bipolar-2…/dp/B06XSYP24P/
In case of lack of strength I plan to use a speed reducer to increase the torque (as you said, it will be slower). Either a toothed pulley or gears.
The arm will measure a total of 100 CM (light material, like plastic, for now that will be test, then it will be carbon fiber). That is, 0.5 m from the shoulder to the elbow, and 0.5 m from the elbow to the wrist.
Let's say you only need to lift 2 kg (I know you need to calculate the weight of the motors and material, but let's say it's 2kg).
2 kg = 19.6 N
Torque = Force x Distance
T = 19.6 N * 0.5 m = 9.8 Nm
Note: that last was what I thought so far, but according to your formula it is different, with the engine torque (3 Nm / 0.5m = 6N). Please correct me if I have any mistakes.
My questions are:
1- Does that mean that I need to increase the torque of my motor by 3 Nm? How much more should it increase?
2- What do you think of the stepper motors?
Btw, I will be careful not to leave the motors frozen in X position when not in use. Thanks for that information, it was super valuable.
I will value other data like that. And... I value everything you wrote. Thank you.
Hi. I am new to robotics and I plan to create a robot arm on a human scale.
I have a problem with the calculation of the motor that I need. That is, I cannot find the correct formula to calculate:
1- Engine torque
2- Electricity consumption
3- Electric power
4- Reduction with gears
In other words, I want to know how many W my motor must have to move X kg, how much electricity it consumes, how much force can be increased with a reducer X.
I have found many formulas, and I honestly don't know which one to trust and if they are correct:
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W = M (Nm) x w (rad / s)
W = mechanical power
M = torque / torque
w = angular velocity
W = M * w
M = W / w
(Gearless at 3000 RPM)
w = RPM * 2 * pi / 60
w = 3000 * 2 * pi / 60 = 315 rad / s
M = 500 W / 315 rad / s = 1.5Nm
I don't know if that last one works, because it doesn't have the diameter. Because the force is not the same if the gear measures 0.3 m as if it measures 0.1 m, as you know.
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P = F * d / t
P = 196.13 N * d / t
P = 196.13 N * 0.1 m / s
P = 19.613 W
I don't know if it's correct, because I saw another formula for the same and it gave me something different:
Ec = ½ * m * v²
Ec = 0.5 * 20 kg * 0.1 m / s²
Ec = 1 J
W = Ec / T
W = 1 J / 10 s = 0.1 Watt
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Can you help me?
Regards.
