Motor sizing help

Skella83

Joined Aug 7, 2021
4
I have a human gyroscope ride like the pic attached and would like to power it with a motor.

At the moment it's a manual machine and you turn the wheel to rotate the ride. I haven't measured the force required to turn the wheel yet but I'm guestimating it's approx 20kg. The radius of the wheel is approx 20cm and I would like it to rotate at 60rpm.

How do I go about sizing the motor and ensuring the motor has enough torque? Once I know how to calculate it I can take proper measurements of the forces needed to rotate the wheel.

Any help appreciated.

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Joined Jul 18, 2013
24,986
To get in the ball park, use the 20cm radius of the wheel and the 20kg force to initially move.

Skella83

Joined Aug 7, 2021
4
To get in the ball park, use the 20cm radius of the wheel and the 20kg force to initially move.
That's where I need help. From what I think I understand its something like f x r so would be the 20kg in newtons? So let's round up to 200n. Then take the 10cm radius in m so 200 x 0.1 = 20nm.

I also want to direct drive. The spindle is 25mm but I don't know how I can measure the force needed to turn the spindle.

Joined Jul 18, 2013
24,986
So you have the load requirements, or am I missing something?
You said the radius of the wheel was 20cm or did you mean dia'?

Skella83

Joined Aug 7, 2021
4
So you have the load requirements, or am I missing something?
You said the radius of the wheel was 20cm or did you mean dia'?
Yes my mistake It's the radius so now it's 200 x 0.2 = 40nm

So I know the force required at 20cm radius but how do I work out the force directly at the spindle?

The 20kg force is just a guesstimate at the moment I will take a proper measurement with a pull scale. But the idea is to remove this manual wheel and directly mount a motor. I can use the pull scale at the 20cm radius I.e pull on the existing wheel to get my measurement but how do I then calculate the force needed directly at the spindle?

Last edited:

Joined Jul 18, 2013
24,986
That is the effort required to move the spindle. You are using the wheel and its radius to figure it out.
For precise measure, you could use a spring scale etc.

LowQCab

Joined Nov 6, 2012
1,916
The amount of Power required is going to be the same as if this was not Gyroscope related.
Figure out how much the maximum weight of the Machine, and its passenger, is going to be.
then figure-out how much of that weight, is how far from the center of rotation.

In other words, a 400lb Lead Bowling-Ball mounted exactly in the center,
is much easier to accelerate in a different direction, or accelerate from a stop,
than an empty hollow sphere 6-feet in diameter, that also weighs 400lbs.
This is because the empty sphere has its weight distributed a long distance away from the center.

You need to estimate how far from the center of rotation,
the average amount of weight is.

This requires pages of complex Math equations that I have no clue about.
But I do have an excellent practical "feel" for.

You seem to be talking in Metric, so the term You need is Newton-Meters.
In Imperial Units, it would be Foot-Pounds.
Both are measurements of Torque.

You need to apply Torque, at a very, very LOW number of revolutions per minute (RPM).
1 revolution per second is an extremely fast speed for a Human to endure,
not to mention the incredible forces placed on the structure of the machine.

A ~25mm Shaft is nowhere near strong enough to withstand the
massive amount of Torque that would be required to cause a
1-revolution per second reversal in direction of a 400lb Gyroscope.
It will simply "snap-off", and be sheared into 2 pieces.
This is, of course, assuming that you can find
a "Gear-Motor" that can generate that level of Torque,
that weighs less than ~200 pounds.

You need a very low RPM Gear-Motor that has a "Fluid-Coupling"
which will limit the Torque that can be applied in a very smooth manner.
This Motor would be mounted to the Frame of the Ride, below the Shaft in question,
and its ~60-RPM output should be coupled with
a Chain-Drive providing around a ~60-to-1 Gear-Reduction-Ratio.

The Motor should have a spring-loaded Speed-Control-Lever,
and a Foot-operated Safety-Switch,
which both require constant manual actuation by an attentive operator.

I have seen one of these "Fluid-Coupled" Gear-Motors working,
but have no clue as to what company may build such a device.

These types of Drives are used on many Carnival-Rides to prevent equipment breakage.

I think that what You want to do is a VERY BAD IDEA.
There are very good reasons why this Machine did not come with a Motor already installed.
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Skella83

Joined Aug 7, 2021
4
I think that what You want to do is a VERY BAD IDEA.
There are very good reasons why this Machine did not come with a Motor already installed.
A lot of the newer versions of this ride come with motors. It's been done and is a GREAT idea. There were actually problems with the manual ones that operators could spin them too fast or at inconsistent speeds.

You can find videos all.over YouTube of powered versions.

After analysising several videos the actual rpm is more like 30rpm.

I have measured the shaft and it is 30mm not 25mm. its not going to snap or break off. The motor is there to replace what a human is doing manually no more no less. Plus there will be acceleration and de-acceleration of the motor 100% of torque would never be applied especially when changing direction.

I think a fluid coupler is over kill for such a small ride.

The rides like this out there that are already powered come in a host of different configurations. Some use a direct drive hydraulic motor. Some use dc motors some use ac motors. Some are belt driven Some are chain driven.

For ease of design added attachment I would prefer direct drive. I have found some hydraulic motors that have from 100 Newton meters of torque uo to over 580. With rpms from 10 - 100. Lower the rpm higher the torque.

What I require is someone that can help figure out what the direct load torque requirement would be. I can figure it out myself using a distance from the center of the shaft. Ie using a pulley as I can physically measure the force required in the physical ride using a pull scale. I can't use a pull scale directly on the shaft so need to know what maths is used to work this out.

I also don't need lectures on the safty or operation of such a device. I know everything health and safety wise what the ride requires and all rides have to go through an adips inspection. New rides or modified rides also need a full design review and pre use inspection. Rides will not be issued an adips DOC if it is not safe.

I'm specifically asking how I work out the required torque of the motor.

ericgibbs

Joined Jan 29, 2010
15,342
I think that what You want to do is a VERY BAD IDEA.
There are very good reasons why this Machine did not come with a Motor already installed.
Hi 83,
I agree with this quote clip ,I would not fit a motor.
E

LowQCab

Joined Nov 6, 2012
1,916
If You just have to do this,
the simple way to calculate the required Torque is to insure that it remains below what an average Man could
exert on the provided Hand-Wheel.

It would appear from the picture that the Wheel has roughly a 10-inch radius, lets call it 12-inches for convenience.
Probably the most force that would be exerted at the Rim of the Wheel is around ~10-pounds,
this would equate to roughly ~10-pounds/feet of Torque.
Now, find a Hydraulic-Motor that is rated for a maximum of ~10ft.lbs. of Torque at a nominal, standardized,
low-range of Hydraulic-Pressure, the maximum of which is regulated by the remote mounted Pump and Fluid-Resevior.
A spring-loaded Forward/Stop/Reverse Valve can be mounted to the Hydraulic-Motor.

An Electric-Motor will not last very long in this type of service unless it is specifically designed for the application.

And it's STILL a bad idea.
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Joined Jul 18, 2013
24,986
Considering the low RPM required, the most efficient way is a GB allowing a much smaller motor to be used, the torque rating of the motor will be reduced by the ratio of the GB.

shortbus

Joined Sep 30, 2009
9,322
A ~25mm Shaft is nowhere near strong enough to withstand the
massive amount of Torque that would be required to cause a
1-revolution per second reversal in direction of a 400lb Gyroscope.
It will simply "snap-off", and be sheared into 2 pieces.
So then why doesn't the axle on my zero turn mower snap in half? It's only ~20mm diameter.

shortbus

Joined Sep 30, 2009
9,322
A spring-loaded Forward/Stop/Reverse Valve can be mounted to the Hydraulic-Motor.
Where is this forward/reverse coming from? After looking at many videos on Youtube they all seem to be going in only one direction. Stop would be handled by a brake.

LowQCab

Joined Nov 6, 2012
1,916
"" So then why doesn't the axle on my zero turn mower snap in half?
It's only ~20mm diameter. ""

Because the Tires can Slip or Spin,
and because the Transmission is probably Hydrostatic-Drive,
and because somewhere in the Drivetrain there's a fairly weak Clutch of some sort,
or a Belt-Drive which can slip.

And because, the Diameter of your Tires is probably less than ~20-inches, not ~48",
this is the most important point,
along with the fact that your Tire/Wheel combination probably weighs
less than ~15-lbs.,
instead of somewhere around ~100-lbs., plus the weight of a child.
And because your Lawnmower Transmission is not fighting against a giant Gyroscope.
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LowQCab

Joined Nov 6, 2012
1,916
"" Where is this forward/reverse coming from? ""

The question is, why would You not want Reverse ?
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shortbus

Joined Sep 30, 2009
9,322
The question is, why would You not want Reverse ?
Where is a gyro used that does reverse?