Are servos powerful enough?
So basically I did some calculations and im predicting my robot will weigh 2.5 pounds at the most. Based on this, I did some simple physics to come up with the servo torque I needed. I am using two HS-485HB servos modified for continuous rotation; each one has a specified stall torque of 83.3 oz-in. My robot wheels have a diameter of 3 -7/8". I was wondering if anyone had the knowledge to check if the servos had enough torque to move at 2.5 pounds? Some equations to consider are torque = force (N) X distance(m), ·weight = mass(kg) x acceleration due to gravity(9.81N/kg), and acceleration of my robot would be assumed zero since it is·moving at constant velocity. the·static friction coefficient is·around 0.5-0.8; I remember using this somehow, but have forgotten...
so 2.5 pounds is 11.12 newtons
torque = radius of wheel x weight(N)
torque = 0.049 meters x 11.12 Newtons
torque = 0.54488Newton meters
0.54488 N x m = ·77.162 ounce inches
2 x HS-485 servos for differential drive = 166.6 ounce inches
therefore, my robot has little over twice the torque needed
can somebody please check this, I haven't gotten far enough into physics to know if im right
Also, does anybody know how to reduce dead reckoning error as far as weight is concerned, like using two caster wheels instead of one and positioning the bulk of the weight of the robot between the servos.
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"You do not really understand something unless you can explain it to your grandmother."
-Lucky[size=-1][/size]
so 2.5 pounds is 11.12 newtons
torque = radius of wheel x weight(N)
torque = 0.049 meters x 11.12 Newtons
torque = 0.54488Newton meters
0.54488 N x m = ·77.162 ounce inches
2 x HS-485 servos for differential drive = 166.6 ounce inches
therefore, my robot has little over twice the torque needed
can somebody please check this, I haven't gotten far enough into physics to know if im right
Also, does anybody know how to reduce dead reckoning error as far as weight is concerned, like using two caster wheels instead of one and positioning the bulk of the weight of the robot between the servos.
▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔▔
"You do not really understand something unless you can explain it to your grandmother."
-Lucky[size=-1][/size]
Comments
acceleration is equal to force divided by 2.5 pounds
Rolling drag may be an option but its going to be very small= .0005
The bigger concern is acceleration and deceleration.·Your trying to accelerate or brake 2.5 pounds
Consider at one inch from center your·servo can lift 83.3 ounces = 5.20625 Pounds.
Now move that out to almost two inches (1.975) and put on 1.25 pounds.
Thats and equivilant of 39.5 ounces or 2.47lbs of torque
Just a different approach.
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