Reduction System for Gate-opener motor 1

[Joaquin Osses]

I've been studying this topic for a couple of days now. I started by asking some questions regarding output Torque and gear ratios relations. Now I came with the whole picture and some calculations that I been working on. My goal is to find out if I'm in the right direction, and also point-out relations between equations that I'm not sure of. To make this more clear and clean, I divide the system into three sections; A, T and B.

In section A its the electric motor and worm/spur torque transmission.

On another topic with a lot of help, I realise that the power provides by the motor is crucial to generate the output torque and angular velocity that your system requires. No matter the radius, ratios or N of teeth of the spur; if the power isn't enough, the gate won't move at the desired velocity.

So, the problem in question: My goal is to find out if this system will be able to move a gate of 600 kg with a velocity of 0.33 m/s, and how much time it will take to reach that velocity.

The technical data of the motor is:

P = 1/2 hp ; RPM = 1450 ; 220V/50Hz

SECTION A:

P = 0.5 hp = 373 (kg * m2 /s3) Pin = Tin * win

win = 1450 rpm = 151.84 (rad/seg)

Tin = 373 (kg * m2 /s3) / 151.84 (rad/seg) = 2.456 N*m

This, of course, is in an ideal situation, without considering losses. (how can I add an estimation of electric motor losses?)

The worm/spur system has a gear ratio of 23:1.

Considering the efficiency of the system of 80%, we have:

Tout = 23 * (0.8 * Pin) / win = 45.2 N*m

Now, Tout also can be calculated by Tout = 0.8 * Pin / wout

angular velocity applied on the spur is: wout = (0.8 * Pin) / Tout

wout = 0.8 * 373 (kg * m2 /s3) / 45.2 N*m = 6.6 rad/sec --> 63 rpm

To consider real circumstances, its the moment of inertia has something to do within this case? I mean, is this Tout enough to accelerate the spur to 63 rpm??

SECTION T:

In this section, I'm wondering what happens with force transmitted in the distance "d" to the pinion. I'm not sure if I need to consider losses in this section.

SECTION B:

Here I have a lot of question about what's happening. The Pinion has 17 teeth and is module 4.

From gear design, I understand that module, and axial pitch determines the Lead and lead angle of the gear. This has a direct relation with the surface contact (involute) in the system and frictional forces.

Is the torque Tout the same as calculated for the worm/spur system? What about the radius of the pinion? If I increase the radius, the Torque will increase as well? The radius of the pinion is = 0.025 (m)

Tout = F * r

F = 45.2 N*m / 0.025 m = 1808 N

Does this mean that the force applied to the rack is 1808 N? Is this force enough to accelerate until 0.33 m/s?

F = m * a = 600 kg * 9.8 m/s2 = 5880 N ; here I need to consider a friction factor according to the wheels and ground. (0.1)

F = 5880 N * 0.1 = 588 N; this means that I need to generate 588 N force to move the gate, but what about the acceleration? if the gate has 4 m, how much time will take to reach the 0.33 m/s

I need a last push to understand this system entirely.

 

[3DDave]

Force times time = mass times change in velocity.
Rearranging, the time is the mass times the change in velocity divided by the force.

Alternatively, divide the kinetic energy in the gate at the expected speed by the expected output power from the gearbox. Both will be approximate because the motor torque won't be uniform starting from 0 RPM to the desired output speed.

I expect the efficiency of that worm is closer to 60%. I think many are higher, but worms are closer to sliding on an incline than the lever-pushing-lever action of most spur gears. Example calculation

https://www.powertransmission.com/articles/0615/Calculation_of_the_Efficiency_of_Worm_Gear_Drives/

 

[Joaquin Osses]

3DDave, thanks for your post.
The idea is to determinate the force (work) needed to accelerate the gate at the desired speed.

you told me the following:

F = m * a --> where a = ∂v/∂t

F * t = m * change in speed; now, this change in speed is from 0 - 0.33 m/s?

Now, from my calculation F = 588 N, which is the force to move the gate.

t = (600 Kg * 0.33 m/s) / 588 (Kg *m /s2) = 0.3 (s); this teel me that the gate will reach the desired speed at 0.4 s.

Second,
KE = 0.5*600 Kg * 0.33∧2 m2/s2 = 32.7 J

As you suggested, if I divide the KE of the gate by the Toutput:

32.7 (kg⋅m2⋅s−2) / 45.2 (kg*m*s-2) *m )= 0.72 , Is this the efficiency of the system? what does it mean exactly?

 

[3DDave]

You didn't calculate the force provided by the motor. You used a friction force from the wheels holding this up. That force is subtracted from the force provided by the motor.

 

[Joaquin Osses]

Oh got it.
Fnet = Fpin - Ffr

The frictional force is Ffr = 600 Kg * 9.8 m/sec2 * 0.1
Ffr = 588 N

My desired speed is 0.33 m/sec and assuming a time of 4 sec to achieve it;
a = ∂v/∂t = 0.33 m/sec / 4 sec = 0.0825 m/sec2

Now, the net force, which is the force what drives the gate to that acceleration;

Fnet = m * a = 600 Kg * 0.0825 m/sec2 = 49.5 N

Last, the pinion force is:
Fpin = Fnet + Ffr = 48.5 N + 588 N = 637.5 N

But I still have a doubt, though. What about the 1808 N that I calculated?
In the calculation above I assumed 4 sec to achieve a speed of 0.33 m/sec. That gives me 637.5 pinion force.
But if the 1808 N is the tangential force of the pinion which is transmitted to the rack, and with this force, I define the Fnet, acceleration and the time to achieve the 0.33 m/sec. Is this correct?