Acceleration resistance in crane design

AI Thread Summary
The discussion centers on an equation for calculating acceleration resistance in crane design, specifically for trolleys and gantries. The equation incorporates load moment of inertia, motor moment of inertia, and angular velocity, with a focus on unit conversions to kilowatts. Participants clarify the derivation of the equation from power definitions and the relationship between torque and angular acceleration. The constant 3.65 x 10^5 is explained as necessary for converting units to kW. Overall, the conversation emphasizes understanding the underlying physics and unit consistency in crane design calculations.
Ridzuan
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Hi everybody
Does anyone familiar with this equation?

Acceleration resistance,
= {(load MOI/mech. efficiency) + (motor MOI) x rpm^2} / {3.65 x 10^5 x acceleration time}
= {(kgm^2/ƞ) + (kgm^2) x rpm^2} / {3.65 x 10^5 x second}

= final value unit is in kilowatt (kW)

It used to calculate the acceleration power for trolley and gantry in crane design. I am looking for the origin/principal of this equation... thanks guys
 
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First, I would guess your equation should read:

{(load MOI/mech. efficiency) + (motor MOI)} x rpm^2 / {3.65 x 10^5 x acceleration time}

It comes from the definition of power ##P = T\omega##, where ##T## is the torque and ##\omega## is the angular velocity (i.e. rpm).

The torque, if converted entirely to acceleration, is ##T = I\alpha##, where ##I## is the mass moment of inertia and ##\alpha## is the angular acceleration.

The angular acceleration is ##\alpha= \frac{\omega}{t}##, where ##t## is the time.

So, ##P = T\omega = I\alpha \omega = I\frac{\omega}{t}\omega = \frac{I\omega^2}{t}##.

Thus ##I## = (load MOI/mech. efficiency) + (motor MOI), ##\omega## = rpm/2 and ##t## = acceleration time.

The constant 3.65 x 10^5 = ##2^2 \times 1000 \frac{W}{kW} \div \left(\frac{\pi}{30}\frac{rpm}{\frac{rad}{s}}\right)^2##. So it is for unit conversions. The ##2^2## is from ##\omega## = rpm/2, because the acceleration time is measured from 0 to rpm, so we used the average rpm during the process, i.e. rpm/2.
 
Dear sir, thank you for your explanation. it is very helpful. I appreciate it so much.
just wonder, what came out from the (kgm^2 x rpm^2) / t... to be specific, what unit it produce that enable it to be convert to kW?
 
The basic SI units of your formula are:
\frac{kg.m^2 \left(\frac{rad}{s}\right)^2}{s}= \frac{kg.m^2}{s^3}
The basic SI units that define the Watts are:
W = \frac{J}{s} = \frac{N.m}{s} = \frac{\left(kg\frac{m}{s^2}\right)m}{s} = \frac{kg.m^2}{s^3}
So you can see that they have equivalent dimensions, i.e. ##\frac{mass \times length^2}{time^3}##
 
jack action said:
First, I would guess your equation should read:

{(load MOI/mech. efficiency) + (motor MOI)} x rpm^2 / {3.65 x 10^5 x acceleration time}

It comes from the definition of power ##P = T\omega##, where ##T## is the torque and ##\omega## is the angular velocity (i.e. rpm).

The torque, if converted entirely to acceleration, is ##T = I\alpha##, where ##I## is the mass moment of inertia and ##\alpha## is the angular acceleration.

The angular acceleration is ##\alpha= \frac{\omega}{t}##, where ##t## is the time.

So, ##P = T\omega = I\alpha \omega = I\frac{\omega}{t}\omega = \frac{I\omega^2}{t}##.

Thus ##I## = (load MOI/mech. efficiency) + (motor MOI), ##\omega## = rpm/2 and ##t## = acceleration time.

The constant 3.65 x 10^5 = ##2^2 \times 1000 \frac{W}{kW} \div \left(\frac{\pi}{30}\frac{rpm}{\frac{rad}{s}}\right)^2##. So it is for unit conversions. The ##2^2## is from ##\omega## = rpm/2, because the acceleration time is measured from 0 to rpm, so we used the average rpm during the process, i.e. rpm/2.

Dear Jack Action
You mentioning the 3.65 x 10^5 is constant for unit conversions to kW. i am unable to figure out what unit to be converted to kW. can you help me? thx
 
The units in your formula are all basic SI units (kg, m, s), except for rpm. Once you converted rpm to rad/s (see post #2), you have all SI basic units. In the end, the final SI unit is kg.m2/s3. In post #4, I showed that 1 kg.m2/s3 = 1 W. Then in post #2, W is converted to kW.
 
jack action said:
The units in your formula are all basic SI units (kg, m, s), except for rpm. Once you converted rpm to rad/s (see post #2), you have all SI basic units. In the end, the final SI unit is kg.m2/s3. In post #4, I showed that 1 kg.m2/s3 = 1 W. Then in post #2, W is converted to kW.
Thanks Jack
 
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