Question about velocity and motors

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The discussion revolves around selecting a motor for moving a 159 kg load across a 0.45 m horizontal distance. Key calculations include determining maximum velocity, factoring in torque, RPM, and gear ratios. The user is working with a 15:1 gearbox and a pulley radius of 1 inch, leading to a calculated maximum force of 179.53 N and an acceleration of 1.13 m/s². The forum participants emphasize the importance of considering friction, inertia, and efficiency in the calculations. The user seeks further guidance to finalize the calculations before an impending deadline.
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I'm working on choosing an appropriate motor. I have the calculations I need for the required torque, power, max velocity, force, and acceleration. Now I need to know the maximum velocity that a specific motor can move 159 kg 0.45 m across a horizontal plane.

In all of my calculations for which motor was needed, I neglected to take friction and inertia into consideration because I did not need to be so exact. The friction is quite minimal anyway. It is a seat moving along a metal rail.

The motor specs are as follows:

Torque - 0.304 Nm
Speed - 4000 RPM
Power - 0.13 kW
Gearing - 60:1 (15*4)
Inertia - .031 kg-cm^2

I don't yet know what the radius of the gear attached to the gearbox will be. I imagine something like 5 cm. I was told that the equation would involve something like (2pi*r(RPM))/15, but I don't know how that will work in with the power, mass, and distance when calculating the max velocity.

I was thinking if using 1/2*130*0.45 would be possible to use for the kinetic energy and then solve for v with 29.25=1/2*159*v^2. But again, I'm not sure how to work in the Speed and gearing of the motor.

I appreciate the help. If I'm missing any info, let me know.
 
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Does anyone at least have some tips or ideas?

Thanks
 
Update:

Gearbox is 15:1. Pulleys are 1:1.
Radius of pulley is 1"

The pulley's RPM speed is (circumference*RPM)/gearing, right? In other words, (2pi*4000)/15=1676 RPM. The torque would then be 0.304*15=4.56 Nm.

So then how do I get the maximum velocity of the 159 kg along 0.45 m that this setup can provide?
 
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Hi FES, if this is your homework, it should go to the homework section.

Add a sketch or drawing if possible.

Think different, if you had to lift the same mass 0.45m in the vertical plane, frictionless and disregarding inertia, how would you solve the problem ? What would be the acceleration and kinetic energy ? And the maximum vertical speed ?

Now turn back to the horizontal plane and consider friction and inertia, what top speed would be attainable considering also the transmission efficiency?
 
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Hi John,

Thank you for your response. No, this is not homework, this is real-world for my job. I simply am unfamiliar with working with motors and have not done physics in several years, so I am having a mental block. Unfortunately, I cannot disclose this device, but think of those Concept 2 rowers you would find in a gym.

I am running out of time. The motor guy will be in on Monday and I must have a spreadsheet complete with this same calculation for every motor in a certain class by then for my CEO.

Ergo, if you wouldn't mind feeding me a bit more than a crumb, I'd greatly appreciate it. I am thinking what I need to do is use Force=torque/radius and find the acceleration from there with F/m.

4.56 Nm/.0254 m=179.53 N
179.53 N/159 kg=1.13 m/s

1.13*0.45 m=0.51 s

Do the RPMs still need to factor in somewhere?

Thank you.
 
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It looks like this is a gear\rack mechanism, a very simplified approach will be:

Considering the mass path is perfectly horizontal, negligible rotational inertia, no friction, starting at max power and max rpm...

Fmax = Tin x i x n / r

F -> Max Traction Force [N]
Tin -> Motor Input Torque [N.m]
i -> Gear ratio
n -> Efficiency
r -> Gear Radius [m]

amax = Fmax / m

amax -> Max Acceleration [m/s²]
m -> Mass

converting rpm -> rad/s => 1 RPM = 2*PI / 60s

w2 = w1 / i

w2 -> Gear speed [rad/s]
w1 -> Motor speed [rad/s]

vmax -> w2 x r

vmax -> Maximum Travelling Speed [m/s]

tmin = (2 x d / amax)^0.5

tmin -> Minimum Acceleration Time
d -> Travelling Distance [m]

P = Fmax x vmax / n

P -> Power [W]
 
I typed up a lengthy reply that failed to post and is now lost, so all I will say is thank you for all of your assistance, John. I'm grateful!
 

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