If you were to attach to a servo something that weighs 15-20 pounds (6.80 - 9.07) and 18 inches long (0.457) , what kind of torque would you need to rotate this?
It depends. There is a good force and torque primer here.
You first need to find the force required to move that object, then find the torque. With the torque now you can search for your motor.
salad7:
If you were to attach to a servo something that weighs 15-20 pounds (6.80 - 9.07) and 18 inches long (0.457) , what kind of torque would you need to rotate this?
If the something is mounted on good bearings, maybe not a lot of torque. If the something is partly buried in the ground, more torque would probably be needed.
The thing we are missing is how off-axis the mass is and how the axis is inclined to the
vertical - these details are all-important.
8 kg 0.5m off a horizontal axis exerts a torque of force x distance = 80 x 0.5 = 40 Nm (very large),
whereas with a vertical axis they exert no torque (there will be friction of the bearing though).
If the orientation can arbitrary go with the larger figure. If the is a robot arm you now understand why they
are counterbalanced to reduce the working torques needed.
Also you probably have to move the thing at some finite rate - thus there are
acceleration torques too - for this you need to calculate the MoI.
salad7:
If you were to attach to a servo something that weighs 15-20 pounds (6.80 - 9.07) and 18 inches long (0.457) , what kind of torque would you need to rotate this?
would you like to share what kind of modern servomotors you are using for that purpose ?
If say the weight was all at the opposite end of attachment to the servo and you had a length of 12" and a weight of 20 lb you would have 20 Ft.Lb. of force at the mounting point.
If the 20 lb's was spread evenly throughout the length of 12" you would have less than 20 Ft.lb's of force.
You are using 18" length, if the 20 lb was all at one end it would have about less than 1/3 ? more force so it would be 20+ Ft. lb's but don't quote me on that. If math is working for you go low tech, clamp a torque wrench in a vice attach a 18" long bar to the end of the wrench and 20 lb to the other end of the bar and take a reading. You could use a scale as well and a couple of 2x4 just subtract the weight of the 2x4 before adding your 20 lb's
|<18">|
Pivot point of 2x4 > +-----------------
| |
2x4, broom handle, whatever > | /
| /20lb\
scale > ===
davisdesigns:
If say the weight was all at the opposite end of attachment to the servo and you had a length of 12" and a weight of 20 lb you would have 20 Ft.Lb. of force at the mounting point.If the 20 lb's was spread evenly throughout the length of 12" you would have less than 20 Ft.lb's of force.
You are using 18" length, if the 20 lb was all at one end it would have about less than 1/3 ? more force so it would be 20+ Ft. lb's but don't quote me on that. If math is working for you go low tech, clamp a torque wrench in a vice attach a 18" long bar to the end of the wrench and 20 lb to the other end of the bar and take a reading. You could use a scale as well and a couple of 2x4 just subtract the weight of the 2x4 before adding your 20 lb's
|<18">|
Pivot point of 2x4 > +-----------------
| |
2x4, broom handle, whatever > | /
| /20lb\
scale > ===
Never mind, the 'schematic' does not show correctly, it did on preview...
If you make a low friction turntable, then a large servo could probably rotate the object.