Let's Make Robots!

Determine servo strength

Could we get a tutorial on servos?   I know I seen somewhere on here that explain how  to determine how much a servo can lift and also how much it can lift by the length to the next servo.  I'm not sure if its CTC was the one but could you please explain it all.  I would love to know this as well as other on this site and viewers also.  This would be a major issues for alot I'm sure.  thanks

Comment viewing options

Select your preferred way to display the comments and click "Save settings" to activate your changes.

Here you'll find some information

http://www.societyofrobots.com/actuators_servos.shtml

If I remember correctly the also have some calculations

How about 44 oz/in... That's 44 oz at 1" arm.

44/1=44

Now, put it on a 2" arm...

44/2=22

etc.etc. --You sure it is not devision?

Τ = F.d   (Torque = Force x Distance)
44oz.in = 44oz x 1"
Also, F = Τ/d. As the distance decreases, perpendicular force increases - this is why it's much harder to stop a motor that has a small shaft (d = radius in this case) with your fingers compared to a motor that has a bigger wheel attached, even though the torque generated by the motor is the same.

If torque worked in reverse (oz/in, etc), then mechanical advantage would also work in reverse. For example if you wanted to undo a wheel bolt to change a car tyre you'd find it easier to use a shorter handle...
Or to take another example it would be easier to lift someone on a seesaw if they were further away from the pivot instead of closer.

Chris, you had me puzzled there for a moment. Always nice to challenge the things you thought to be true, all your life. Oh well, this one can stay true.

I suppose there is a valid point in the oz/inch notation though, as Chris points out. When describing servo "Strength", you would want to know how much force can be expected at the end of a given arm connected to the servo. And you'd want to be reminded that the strength decreases with longer arms. A 44 oz/inch servo would have enough strength to yield 44 oz at the end of a one inch arm, or half that at double the arm: 22 oz / 2 inch. But the slash (indicating a division) still is confusing. It'd better be something like:
Strength = 44 oz @ 1 inch = 22 oz @ 2 inch.

And there you have the magic word: strength:
expressed as force is measured in oz, lb, kg or preferably N.
expressed as torque is measured in oz*inch, lb*yard, kg*cm or preferably N*m.

Strength is not a quantity that physicists recognise. force is. And torque is derived from force and length, by means of multiplication.

oz/in or kg/cm  is just that, a given weight at a given distance. I just looked up a standard servo and it rates at 44oz/in @ 4.5v. This simply means that if you stuck on a little arm that was 1" from the tip of the arm to the center of the servo, the end of that arm would lift 44oz. If that arm is 2" long, it will lift 22oz. If the arm is 4" long it will lift 11oz at the tip. The bottom line is this: Arms are a LOT heavier than you would think. Most small arms use at least 2 heavy-duty servos working together for the base.

(I will grow over this. I promise. Not today.)

The forward slash in "oz/inch" would appear to indicate "oz per inch". But that is not the unit for torque! That is a quotient, where torque is always expressed as a product: "oz multiplied by inch". The right way to indicate a multiplication might be different in your custom, but multiplication is never a slash!
Better options would be:

  • oz*inch
  • oz x inch
  • oz·inch (this dot · is what mathematicians use for multiplications, they speak the most universal language in the universe, take a hint!)

(Even more) strictly speaking, the above discusses dimension, rather than unit. As Chris mentioned, torque can also be expressed in different units. Force by armlength: Newton·meter, or N·m or just Nm. Force is hardly ever expressed in Newton in the daily conversation. We'd rather borrow from the force that we experience daily: weight. In Holland we experience it in kilogram, in Kansas they feel the pounds. Same goes for units of length of course.

Now I am not picking on just you LMR-boys. I like you guys. Everybody in the world seems to be conspiring against the dimension of torque. Most notably the people selling us servos. They should know better! They should stop confusing the newbies.

Servo vendors of the world, grow up! Torque is a pro·duct! Show that you care!

Thank you all, I feel much better now.

Thanks alot man.  Tonight I can finally figure my bot out.  If you or anyone else have any more to share please do.

http://letsmakerobots.com/node/2390

Go to the above link. There is an awesome post about a lynx arm. At the bottom of the post, there is a link to a flickr gallery of the whole assembly process. You can see each part in detail. In addition, you should google that lynx arm and find out what servos it is using. This way, you can get an idea of how strong of a servo will move how much of an arm.

This is for the peoples that don't want to search googles endless to figure it all out.  I'm know someone had a few ideas how to figure it out.  Also you must know that the equation on wiki might be extreme to alot, so we must make it as simple as we can for new comer please.

Google? That was a link to Wikipedia. No searching involved, just math.

Math is math. Put a pound on a foot-long lever, you get one foot-pound of torque. Two pounds on a half-foot lever gets you 1 foot-pound of torque. Put one foot-pound of torque into a two-foot lever gets you half-a-pound of force at the end of said lever.

edit0 = CtC beat me to it ;)

edit1 = If you would do LMR a kindness: Please edit your post and add the tags "torque, tutorial, howto, math" and any others you deem appropriate. So it can be more easily findable by others in the future. Thanks!