Moment of Inertia of helicopter blades.

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Discussion Overview

The discussion revolves around calculating the moment of inertia for a model of helicopter blades. Participants explore various methods and approximations for determining the moment of inertia based on a specific model involving a square base and blades, considering both theoretical and practical aspects of the calculations.

Discussion Character

  • Technical explanation
  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • One participant describes a model with a square wooden base and four blades, seeking guidance on calculating the moment of inertia given known density, mass, and thickness.
  • Another participant suggests approximating the rotor blades as rods and provides a formula for angular inertia, while noting that real helicopter blades are hinge-mounted.
  • A participant introduces additional geometric complexities by mentioning triangular shapes at the center of the blades and questions whether to add their moment of inertia to that of the rectangular blades.
  • There is a discussion about the moment of inertia of a rectangular plate, with one participant providing the formula and expressing discomfort with integration for complex shapes.
  • One participant attempts to derive the moment of inertia for the triangular sections, sharing their integration process and seeking verification of their approach.
  • Another participant reassures that the middle square is accounted for in the calculations provided earlier and references an old discussion for further context.
  • A later reply questions if the moment of inertia can be simplified by multiplying by 2, indicating uncertainty about the calculations.

Areas of Agreement / Disagreement

Participants express various methods and approximations for calculating the moment of inertia, but no consensus is reached on the best approach or the correctness of the calculations presented.

Contextual Notes

Participants mention limitations in their calculations, including assumptions about the shapes involved and the integration process. There is also a reference to the complexity of deriving certain formulas from textbooks.

eptheta
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I have built a model of rotating blades of a helicopter and want to mathematically calculate it's moment of inertia. I am unsure on how to do that.
Its a simple model:
a square wooden base with uniform thickness and 4 equally long blades made of styrofoam from the midpoints of the 4 sides of the square.
The rotational axis is through the centre of the square

Lets just say i know the density, mass and thickness of the base and the blades, how do i get moment of inertia for a specific length of blades ?

Thank you
 
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What if i have something like this:
http://img706.imageshack.us/img706/418/moin.png
I have those extra triangles at the center to deal with. can i just get the MOI of the triangles and add it to the MOI of the rectangular blades ?
yeah, i know I'm approximating a LOT, but it'll have to do.
Can anyone help ? Can anyone show me the derivation of it as well ?
 
Last edited by a moderator:
eptheta said:
I have built a model of rotating blades of a helicopter and want to mathematically calculate it's moment of inertia. I am unsure on how to do that.
Its a simple model:
a square wooden base with uniform thickness and 4 equally long blades made of styrofoam from the midpoints of the 4 sides of the square.
The rotational axis is through the centre of the square

Lets just say i know the density, mass and thickness of the base and the blades, how do i get moment of inertia for a specific length of blades ?

Thank you

The moment of inertia of a rectangular plate might get you even closer depending how you want to set it up. Its usually given in textbooks. I personally don't like to do the integration on some of these figures.

The moment of inertia of a rectangular plate is 1/12m(a^2 + b^2) where a is the width and b the length of the plate. Note that this is the moment of inertia taken from the center of mass of the plate.

I see your picture. So a and b are b and l...

So really one plate would have length 2l +b and width b... and there are two of them.
 
Last edited:
The moment of inertia of a rectangular plate might get you even closer depending how you want to set it up. Its usually given in textbooks. I personally don't like to do the integration on some of these figures.

The moment of inertia of a rectangular plate is 1/12m(a^2 + b^2) where a is the width and b the length of the plate. Note that this is the moment of inertia taken from the center of mass of the plate.

I see your picture. So a and b are b and l...

So really one plate would have length 2l +b and width b... and there are two of them.
From what you've said, if i take two rectangular plates, then after adding the two MOIs i' have an extra square in between. And i don't think i can just subtract that...

Also, do you know the derivation of 1/12 M(a2+b2) or a link to it ? I can't find it in my textbook.

I'm not sure if this is correct, but i tried my best to do the integration based on the baterials densities. I havn't added limits yet, but can you verify this for me ?:For the triangle in the middle for each blade:
{σ : density of triangle}
m=1/2 r2hσ
r2=2m/hσ
therefore I=∫2m/hσ
=m2/hσ dm=dx*b*h*rho where dx is a small distance on the wing.
{ρ: density of blade}
therefore ∫r2 dm
substituting for dm:
=∫x2*b*h*ρ dx
=1/3 b*h*ρ*x3

The dimensions are right [ML2] but I'm really not too sure...
Thanks
 
eptheta said:
From what you've said, if i take two rectangular plates, then after adding the two MOIs i' have an extra square in between. And i don't think i can just subtract that...

Also, do you know the derivation of 1/12 M(a2+b2) or a link to it ? I can't find it in my textbook.

I'm not sure if this is correct, but i tried my best to do the integration based on the baterials densities. I havn't added limits yet, but can you verify this for me ?:


For the triangle in the middle for each blade:
{σ : density of triangle}
m=1/2 r2hσ
r2=2m/hσ
therefore I=∫2m/hσ
=m2/hσ


dm=dx*b*h*rho where dx is a small distance on the wing.
{ρ: density of blade}
therefore ∫r2 dm
substituting for dm:
=∫x2*b*h*ρ dx
=1/3 b*h*ρ*x3

The dimensions are right [ML2] but I'm really not too sure...
Thanks

You don't have to worry about the middle square its taken care of if you use the sides I gave you.

And I am going to admit I don't like doing the math on these. So here is an old discussion on this board.
https://www.physicsforums.com/showthread.php?t=57119
 
Really ? So i just multiply it by 2 ? i.e I= 1/6 M(a2+b2)
 

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