Calculate minimum RPM to self-balance a CMG on two legs

  • Thread starter Thread starter demej00
  • Start date Start date
demej00
Messages
1
Reaction score
0
TL;DR Summary
I would like to have a formula for determining the minimum RPM a mechanical gyroscope needs to balance itself on two legs.
Here is a photo of a rough drawing of my apparatus that I have built many times and works.
Diagram Side.webp
Diagram Front.webp

I would like to have a formula to give me the RPM necessary for the gyroscope to balance itself on the two legs (screws). I asked Claude to give me a formula and it gave me the following:

Let me calculate the required RPM foreffective stabilization. I'll use the principles of gyroscopicprecession and the moment of inertia.
First, let's calculate the keyparameters:

1. Moment of inertia of the gyroscopedisk (I):
- Radius (r) = 32.5 cm = 0.325 m
- Mass (m) = 3.5 kg
- For a disk: I = ½mr²
- I = ½ × 3.5 × (0.325)² = 0.185kg⋅m²

2. Required torque to stabilize (τ):
- Bicycle mass (M) = 13.5 kg
- Assume center of mass height (h) ≈0.5 m
- Assume small lean angle (θ) ≈ 5°= 0.087 radians
- τ = Mgh × sin(θ)
- τ = 13.5 × 9.81 × 0.5 ×sin(0.087)
- τ = 0.575 N⋅m

3. Required angular momentum (H) forstable precession:
- For good stability, precession rate(Ω) should be slow, say 0.1 rad/s
- H = τ/Ω
- H = 0.575/0.1 = 5.75 kg⋅m²/s

4. Required angular velocity (ω):
- H = Iω
- ω = H/I
- ω = 5.75/0.185 = 31.08 rad/s

5. Converting to RPM:
- RPM = (ω × 60)/(2π)
- RPM = (31.08 × 60)/(2π)
- RPM ≈ 297 RPM

However, this is the minimum RPM forbasic stabilization. For robust stability accounting for externaldisturbances, you typically want:
- A safety factor of 3-5×
- Therefore, recommended operating RPM:900-1500 RPM


This relatively low RPM requirement isdue to the large diameter and mass of your gyroscope. The largemoment of inertia means you don't need extremely high speeds toachieve the necessary angular momentum.


My question is do you think this is a reasonable answer to my question as to how to calculate the minimum RPM necessary to self balance the CMG on two legs?
My experimental results show that I can balance it at about 600 rpm but only for 10 seconds or so but that could just be due to imbalances in the wheel (I had to balance the bicycle wheel but still out of balance at high rpm) and other friction in the circular bearing, improper CG alignment over the bearing, etc.

I can balance the wheel indefinitely by swinging a counterweight perpendicular to the rotor using two sensors and an Arduino at very low rpm, around 200RPM. But I want to balance it indefinitely without using any active mechanism.

Thanks for any consideration you may give me.
 
Engineering news on Phys.org
Welcome to PF.

The speeds seem reasonable.
Why, and for how long, must it balance on two points?

If the points are mounted on the Earth's surface, then the Earth will rotate under it, and it will fall over. Alternatively, it could be operated as a gyrocompass.
 
How did you find PF?: Via Google search Hi, I have a vessel I 3D printed to investigate single bubble rise. The vessel has a 4 mm gap separated by acrylic panels. This is essentially my viewing chamber where I can record the bubble motion. The vessel is open to atmosphere. The bubble generation mechanism is composed of a syringe pump and glass capillary tube (Internal Diameter of 0.45 mm). I connect a 1/4” air line hose from the syringe to the capillary The bubble is formed at the tip...
Thread 'Calculate minimum RPM to self-balance a CMG on two legs'
Here is a photo of a rough drawing of my apparatus that I have built many times and works. I would like to have a formula to give me the RPM necessary for the gyroscope to balance itself on the two legs (screws). I asked Claude to give me a formula and it gave me the following: Let me calculate the required RPM foreffective stabilization. I'll use the principles of gyroscopicprecession and the moment of inertia. First, let's calculate the keyparameters: 1. Moment of inertia of...
Thread 'Physics of Stretch: What pressure does a band apply on a cylinder?'
Scenario 1 (figure 1) A continuous loop of elastic material is stretched around two metal bars. The top bar is attached to a load cell that reads force. The lower bar can be moved downwards to stretch the elastic material. The lower bar is moved downwards until the two bars are 1190mm apart, stretching the elastic material. The bars are 5mm thick, so the total internal loop length is 1200mm (1190mm + 5mm + 5mm). At this level of stretch, the load cell reads 45N tensile force. Key numbers...
Back
Top