RunRyder - Force calculations

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CAD - Engineering - Technical > Force calculations

FLYINFOOL
Veteran
Location: Cudahy, WI

Can anyone tell me the formula to calculate the centrifugal force generated at the blade grip bearings?

I will probably have to run it more that once to account for the weight of the various components and their distance from center.

I am asking because I want to know if, or how much overloaded the bearings were in the tail of a gasser heli that I bought used.
The original builder removed the thrust bearings and replaced them with another ball bearing insisting that the side load was no problem for the bearings.

Yes I did put it back the way it was supposed to be.

Jeff Borowski
Gohbee Field Rep
0133TD

07-23-2004 Over year old.

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bobkins
Senior Heliman
Location: Bocholt, Belgium

There are two formulas depending on what you know

The first one is F=m*r*w²
with F= Centrifugal Force (Newton)
m= mass (Kg)
r=the distance from the center (meters)
w= 2*3.14159265*n (rad/s) (n=rounds/s)

Or the second one : F=(m*v²)/r
with F= Centrifugal Force (Newton)
m= mass (Kg)
r=the distance from the center (meters)
v=w*r w= 2*3.14159265*n

I guess the first one is the easiest to use because w stays the same for
every distance from the center.

I hope this helps

Greetz Rob

07-24-2004 Over year old.

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rover
Senior Heliman
Location: Brandon,FL

Do not forget the value 'r' is to the center of mass from the center of rotation. If you use the linear velocity version of centripital acceleration you must use the velocity at the center of mass.

Dont forget to tell that guy that sold it to you that he is full of it. Thats why they make thrust bearings. the other bearings are radial bearings and can handle some side loads but nothing of that magnitude.

Dustin Kerr

08-03-2004 Over year old.

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helicopter34
Veteran
Location: New Jersey, exit 82

Yeah

Yeah like rover and bobkins said, F=m*r*w² applies for a point particle (infinitely small yet massive particle), to deal with bodies you have to integrate.

Like rover said, you can imagine all the weight of the rotating components existing (collapsing into) at the center of gravity and then apply
F=m*r*w²

Quote
'r' is to the center of mass from the center of rotation

If you actually care to be precise, first balance your tail blades and mark the COG of the blades, weigh 1 blade, install the blade and then measure from the COG to the axis of rotation. Then
(mass 1 blade)*(distance of COG to axis)*(2*Pi*rpm/60)²=axial force on bearing

I don't know what other components you are considering, the only things the blade grip bearings hold are the blades and like a bolt (which I guess you can neglict).

Make it idiot-proof, and someone will make a better idiot

08-06-2004 Over year old.

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bobkins
Senior Heliman
Location: Bocholt, Belgium

Hi

Just put the bearings in your tail is the best thing you can do if there are spaces for them. If you do the calculations you will come to the same conclusion. And there are no helicopters wich don't have bearings in they're tail that I can think of. So I would guess your helicopter isn't different from the rest.

Greetz Rob

08-06-2004 Over year old.

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FLYINFOOL
Veteran
Location: Cudahy, WI

True all of the helies have bearings in the tail, but they do not all have thrust bearings in the tail.
This is the first heli that I have hade that did have a thrust bearing and a radial bearing in the tail, instead of 2 radial bearings.
The purpose of this was not to decide if I should put in the thrust bearings, I will.
This was more to prove to the guy that built this that just because it works on a R30 dos not mean it will work on a gasser.

Thanks for your input

Now I just need an accurate scale to weigh this up and do the math.

Jeff Borowski
Gohbee Field Rep
0133TD

08-11-2004 Over year old.

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CAD - Engineering - Technical > Force calculations

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