Acceleration resistance in crane design

[Ridzuan]

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

 

[jack action]

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.

 

[Ridzuan]

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?

 

[jack action]

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. $\frac{mass \times length^2}{time^3}$

 

[Ridzuan]

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

 

[jack action]

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.

 

[Ridzuan]

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