Welcome to EDAboard.com

Welcome to our site! EDAboard.com is an international Electronic Discussion Forum focused on EDA software, circuits, schematics, books, theory, papers, asic, pld, 8051, DSP, Network, RF, Analog Design, PCB, Service Manuals... and a whole lot more! To participate you need to register. Registration is free. Click here to register now.

Register Log in

can 360 g-cm servo lift 90 pounds?

dl09

Full Member level 4
Joined
Feb 12, 2020
Messages
219
Helped
1
Reputation
2
Reaction score
1
Trophy points
18
Activity points
1,626
can a servo rated at 360 gram-centimeters of torque lift 90 pounds, if the servo has rod that is 35 cm long and the 90 pound weight is attached to the other end?
 

KlausST

Super Moderator
Staff member
Joined
Apr 17, 2014
Messages
17,443
Helped
3,938
Reputation
7,874
Reaction score
3,812
Trophy points
113
Activity points
115,656
Hi,

Show a simple but complete drawing.

Klaus
 

FvM

Super Moderator
Staff member
Joined
Jan 22, 2008
Messages
46,966
Helped
13,929
Reputation
28,106
Reaction score
12,571
Trophy points
1,393
Location
Bochum, Germany
Activity points
273,873
g-cm is no torque unit. I guess you mean 360 p-cm, as you can easily calculate, the required torque to lift the weight is factor 4000 higher, at least on earth.
 

dl09

Full Member level 4
Joined
Feb 12, 2020
Messages
219
Helped
1
Reputation
2
Reaction score
1
Trophy points
18
Activity points
1,626
i found a servo on the internet rated at 180 kg.cm, could that lift 90 pounds? kg.cm is the unit they used.

- - - Updated - - -

if the servo had a rod 36 cm long and the 90 pound object was on the end of the rod?

- - - Updated - - -

here is a picture
Drawing (1).jpeg
 

KlausST

Super Moderator
Staff member
Joined
Apr 17, 2014
Messages
17,443
Helped
3,938
Reputation
7,874
Reaction score
3,812
Trophy points
113
Activity points
115,656
Hi,

It can be simple done by your own.

Read about "torque".
Then read about how to calculate pounds to kg with the help of gravity.
What direction do you want to move your 90 ponds object? Purely vertical?
The math is simple "multiply".

Klaus
 

dl09

Full Member level 4
Joined
Feb 12, 2020
Messages
219
Helped
1
Reputation
2
Reaction score
1
Trophy points
18
Activity points
1,626
Hi,

It can be simple done by your own.

Read about "torque".
Then read about how to calculate pounds to kg with the help of gravity.
What direction do you want to move your 90 ponds object? Purely vertical?
The math is simple "multiply".

Klaus
i know gravity is a force. therefore pound and kilogram are used to measure force. i kilogram equals 2.2045 pounds. torque is equal to force times lever arm. lever arm equals distance from fulcrum to point of application of force times sine angle between axis of rotation and lever arm. i have already read about torque. so to move 90 pound object vertically multiply 90 pounds by 36 centimeter? well that would be 1469 kg.cm. is this calculation right?
 

KlausST

Super Moderator
Staff member
Joined
Apr 17, 2014
Messages
17,443
Helped
3,938
Reputation
7,874
Reaction score
3,812
Trophy points
113
Activity points
115,656
Hi,

i know gravity is a force.
No. Gravity is acceleration. With the unit "m/s^2"

therefore pound and kilogram are used to measure force.
No. Kilograms is a mass with the unit "kg"
And there is
* pound-force (which is a force) with unit "N"
and there is
* pound-mass (which is a mass and can directly calculated as kg)

well that would be 1469 kg.cm. is this calculation right?
Seems to be correct.

Klaus
 

FvM

Super Moderator
Staff member
Joined
Jan 22, 2008
Messages
46,966
Helped
13,929
Reputation
28,106
Reaction score
12,571
Trophy points
1,393
Location
Bochum, Germany
Activity points
273,873
so to move 90 pound object vertically multiply 90 pounds by 36 centimeter? well that would be 1469 kg.cm. is this calculation right?
Numerically correct. I notice that servo vendors still use outdated g-cm or oz-in torque units. Seems to be an unbreakable habit.
 

wwfeldman

Advanced Member level 2
Joined
Jan 25, 2019
Messages
693
Helped
169
Reputation
338
Reaction score
167
Trophy points
43
Activity points
5,020
can a servo rated at 360 gram-centimeters of torque lift 90 pounds, if the servo has rod that is 35 cm long and the 90 pound weight is attached to the other end?
no
360 gm-cm will not lift 1469 kg-cm (1469000 gm-cm)
 

dl09

Full Member level 4
Joined
Feb 12, 2020
Messages
219
Helped
1
Reputation
2
Reaction score
1
Trophy points
18
Activity points
1,626
torque produces angular acceleration, if i use a 180 kg.cm servo,connected to a rod and with the 90 pound mass on the other end, will the 90 pound mass move, but at a smaller angular acceleration than if i use a 1469 kg.cm servo.
 

wwfeldman

Advanced Member level 2
Joined
Jan 25, 2019
Messages
693
Helped
169
Reputation
338
Reaction score
167
Trophy points
43
Activity points
5,020
180 kg-cm is 180,000 gm-cm
you need more than 1,469,000 gm-cm

this servo simply does not provide enough torque to move your mass

did you (or a friend) ever put a wrench on a nut and not be able to turn it?
or not be able to open a jar of something?
that's because you could not provide enough torque.

there will be no acceleration
your 90 pound mass will not move
 

KlausST

Super Moderator
Staff member
Joined
Apr 17, 2014
Messages
17,443
Helped
3,938
Reputation
7,874
Reaction score
3,812
Trophy points
113
Activity points
115,656
Hi,

Yes it will move, downwards.
With a little less acceleration than earth gravity.

Klaus
 

c_mitra

Advanced Member level 5
Joined
Nov 13, 2012
Messages
3,390
Helped
852
Reputation
1,704
Reaction score
803
Trophy points
1,393
Activity points
25,958
Yes, you need gears.

Use gears such that 360 gm.cm gets converted into 35x90 lb.cm (slightly more).

You can also use pulleys.

There are other ways too.
 

wwfeldman

Advanced Member level 2
Joined
Jan 25, 2019
Messages
693
Helped
169
Reputation
338
Reaction score
167
Trophy points
43
Activity points
5,020
the OP needs 1,469,000 gm-cm of torque to balance the torque presented by the load (90 lbs or about 41 kg at 35 cm)

clearly more is needed to cause some angular acceleration to get started, then 1,469,000 gm-cm
is needed to maintain whatever angular velocity was achieved.

with the 180 kg cm servo motor described, there is 180,000 gm cm of torque
if you put in a gear ration of 10 to 1, you will boost the torques to a little less than 1,800,000 gm cm
(less because of the need to also rotate the gears and overcome axle friction etc, and for the twist in the drive shaft)

lets say that motor and gear set is enough. because of the 10 to 1 gear ratio, the angular acceleration will be 1/10th what
it might have been, had there been no need for gears.

since the motor and gears are only marginally able to exert enough torque, the lift rate will be painfully slow.

pulleys and ropes introduce the same sort of extra losses - rotating the pulleys, axle friction,
rope stretch (different materials will have different stretch, but there's losses in any case)

so, OP, what other things are going on in this project that have a bearing on the question?
 

Toggle Sidebar

Part and Inventory Search

Welcome to EDABoard.com

Sponsor

Top