Determining Torque for this problem

[Sudesh]

Hi,

Could someone please help me with this.

I need to move an object which weighs 300 g horizontally, which is linked to a linear actuator consisting of rack and pinion arrangement which is driven by a servo motor. The speed at which the object should move is 3 m/s and the distance is 0.5 m.
How can I determine what Torque is required to move the object?

 

[russ_watters]

Hi,

Could someone please help me with this.

I need to move an object which weighs 300 g horizontally, which is linked to a linear actuator consisting of rack and pinion arrangement which is driven by a servo motor. The speed at which the object should move is 3 m/s and the distance is 0.5 m.
How can I determine what Torque is required to move the object?

Can you provide a diagram of the device?

 

[JBA]

Including how is the 300 gram load supported vertically.

 

[Sudesh]

Can you provide a diagram of the device?

This is the link for a similar setup -

 

[Sudesh]

Including how is the 300 gram load supported vertically.

I have attached the diagram above

 

[JBA]

A rack and pinion is designed for moving the object. If you are hanging the 300 gram load on the end of the rack without any other support then you are creating a vertical load on the end of the rack that cause increasing drag and friction of the rack bearings that they are not designed to support.

 

[Sudesh]

A rack and pinion is designed for moving the object. If you are hanging the 300 gram load on the end of the rack without any other support then you are creating a vertical load on the end of the rack that cause increasing drag and friction of the rack bearings that they are not designed to support.

Yes, I understand that. Hence, the load would be supported by another rod which is not part of this mechanism.
We can neglect any frictional and efficiency losses too for the calculations if it makes easier for solving.

 

[JBA]

If you are ignoring friction then the required torque is determined by the time in which you want to move the load your required distance.

 

[Sudesh]

If you are ignoring friction then the required torque is determined by the time in which you want to move the load your required distance.

Okay. Time is not a major constraint for me. The main criteria is the motor should have less than 50 Hz frequency (if this information helps)
The speed for moving the load = 3 m/min and distance = 0.5 m

 

[Sudesh]

If you are ignoring friction then the required torque is determined by the time in which you want to move the load your required distance.

Calculating from this the time = 0.5/3 min which equals to ~ 10 sec

 

[JBA]

s = 1/2 x a x t² can be reconfigured to determine a and f = mass x a will give the actuator force required to move your load .5 m in 10 sec. and then the required torque is that force x the pitch radius of the pinion gear driving the rack. At the same time the maximum speed of the rack is dependent upon the rpm of the motor driving the pinion gear x the pitch radius x 2 x π; so it is necessary to first determine the required pinion gear diameter to provide the maximum speed required before calculating the required torque of the motor. The required rpm x torque will then give you the required motor power.
It is important to understand that by using this method the speed of the rack will not be constant throughout its travel.

 

[Sudesh]

s = 1/2 x a x t² can be reconfigured to determine a and f = mass x a will give the actuator force required to move your load .5 m in 10 sec. and then the required torque is that force x the pitch radius of the pinion gear driving the rack. At the same time the maximum speed of the rack is dependent upon the rpm of the motor driving the pinion gear x the pitch radius x 2 x π; so it is necessary to first determine the required pinion gear diameter to provide the maximum speed required before calculating the required torque of the motor. The required rpm x torque will then give you the required motor power.
It is important to understand that by using this method the speed of the rack will not be constant throughout its travel.

Okay. I have these dimensions for the rack & pinion -
SPUR PINION
Diametrical Pitch (= T/d) 24
Number of Teeth 12
Pressure Angle 20 deg
Max. Pitch Diameter 0.5 inch
Face Width 0.25 inch
Bore 0.25 inch
Hub Projection 0.31 inch
Hub Outer Diameter 0.39 inch
SPUR RACK
Pressure Angle 20 deg
Face Width 0.25 inch
Thickness 0.25 inch
Pitch Line to Back 0.208 inch
Length 22 inch

So is the pitch radius equal to = 0.5/2 which equals to 0.25 inch?

 

[JBA]

That is correct.

 

[Sudesh]

s = 1/2 x a x t² can be reconfigured to determine a and f = mass x a will give the actuator force required to move your load .5 m in 10 sec. and then the required torque is that force x the pitch radius of the pinion gear driving the rack. At the same time the maximum speed of the rack is dependent upon the rpm of the motor driving the pinion gear x the pitch radius x 2 x π; so it is necessary to first determine the required pinion gear diameter to provide the maximum speed required before calculating the required torque of the motor. The required rpm x torque will then give you the required motor power.
It is important to understand that by using this method the speed of the rack will not be constant throughout its travel.

I have come up with this calculations - is this correct ?

s=1/2*a*t^2
time t = 10 sec ⇒ 0.16 min
Therefore a = 2*3 (m/min) / (0.16)^2
⇒ a = 234.3 m/min^2
⇒ a = 0.065 m/s^2

F=ma
⇒ 0.3 kg * 0.065 m/s^2
⇒0.0195 N

T = F* pitch rad
where Pitch rad = 0.25 in ⇒ 0.00635 m
T = 0.0195 * 0.00635
⇒ 0.0001238 Nm

Max s = rpm * radius * 2π
max s = 3 m/min
⇒ s = 9.8 ft/min
radius = 0.25 in ⇒ 0.02 ft
Therefore rpm = s / (radius *2π)
⇒ 9.8 / (0.02 * 2 π)
⇒ 78.02 rpm

Is this correct? How do I calculate Power from this?

 

[JBA]

Your only problem is in the first equation "s" should be the distance not the speed which is commonly denoted as "v" (velocity), so your above equation "a = 2*3 (m/min) / (0.16)^2" should be "a = 2* 0.5 (m) / (10 sec)^2 = .010 m/sec^{^2}" and your remaining equations need to adjusted accordingly.

For calculating power. Power = Torque x rpm.

 

[Sudesh]

Your only problem is in the first equation "s" should be the distance not the speed which is commonly denoted as "v" (velocity), so your above equation "a = 2*3 (m/min) / (0.16)^2" should be "a = 2* 0.5 (m) / (10 sec)^2 = .010 m/sec^{^2}" and your remaining equations need to adjusted accordingly.

For calculating power. Power = Torque x rpm.

Okay. Is this correct?
Therefore :
F=ma
⇒ 0.3 kg * 0.01 m/s^2
⇒0.003 N

T = F* pitch rad
where Pitch rad = 0.25 in ⇒ 0.00635 m
T = 0.003 * 0.00635
⇒ 0.00001905 Nm
⇒ 0.0001686067 lb.in

Max s = rpm * radius * 2π
max s = 3 m/min
⇒ s = 9.8 ft/min
radius = 0.25 in ⇒ 0.02 ft
Therefore rpm = s / (radius *2π)
⇒ 9.8 / (0.02 * 2 π)
⇒ 78.02 rpm

Power = Torque * rpm
P = 78.02 * 0.0001686067 lb.in
HP = 0.0131

 

[JBA]

One problem I am having in evaluating your calculations is that you keep jumping back and forth between SI and US/Imperial units from one calculation to the next and my standards of measure are US/Imperial. Also I have been making the mistake of believing you were already familiar with the basic formulas required, my fault, not yours. With that in mind:

1. For your F = m*a calculation: As I understand it in SI to convert your mass (kg) x acceleration (m/s^2) result to force (Newtons) your result must be multiplied by 9.8
In US units the formula is (weight / g) x acceleration i.e. F =(.6614 lbm / 386.16) * .3937 (in/sec^2) = 0.000673906 lbf2. The correct equation for Hp in US units is HP = torque (ft-lb) * rpm / 5252