Mast angle and moments

perbgyro

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Does anyone have any calculation examples for the affect of moment changes with respect to mast angle? I'd like to calculate the affect of moving weight around on the pod to predict the new mast angle based on a previous hang test.
Thanks for any help.
 

kolibri282

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Paul,

are you trying to find the new mast angle that is required to make the rotor thrust pass somewhere near cg such that the attitude of the aircraft is the same after you moved a substantial mass around?
 

perbgyro

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Given the current measurements, weights, cg and mast angle, I want to predict what a change in weight (movement of items) would do to the mast angle. I'm not concerned with CG y axis movement, just x axis movement.
For example, current mast angle is x and my rear keel with the rudder and stab are to be shortened by 12". That will shift the CG forward and change the mast angle, but by how much and will I need to change the rotor head cheek plates.
It's easier to do the calculation and get an idea as opposed to loading up the gyro and traveling to a place to do a hang test, which I would do anyway. It's just easier to fly to the location of the hang test than to trailer.
It's a statics problem and it has been a while since I did that stuff.
 

groundhog

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what you are looking for is how to do a wieght and balance for any aircraft.Look it up in any pilot handbook.basically pick any point (some like the very front so that there is no negatives) multiply the weight by distance from that point .Remember if you are moving the tail closer to centre you are reducing its leverage or effectivness, never a good idea.
 

perbgyro

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It's not just weight and balance for any aircraft because any fixed wing doesn't hang like a pendulum. Now translate that into a mast angle change.

The rudder movement is by specification. Currently it is set for having a nose pod of which I no longer have, so the new position is for a configuration with just an instrument pod.
 

Doug Riley

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Let B equal the weight of the ballast you are moving.

Let GFW equal the gross frame weight (loaded weight of the gyro, less the weight of the rotor)

Let H equal the distance between the aircraft's CG and the teeter bolt in the initial configuration.

Let d equal the horizontal distance that you move B.

The distance you move the CG when you move B is (B/GFW) x d. Call that smaller distance D.

The change in dangle angle will be arc tan D/H.
 
Last edited:

Doug Riley

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Sorry, can't type to save my life. Small c should have been been d - the distance you moved the ballast.
 

kolibri282

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my rear keel with the rudder and stab are to be shortened by 12"
Paul,

since you are moving the (H?)-stab as well one should perhaps not
only consider the static case (Doug has given you the solution for
this one) but also have a brief look at what this means in flight.

Let all variables with index 1 refere to the case of the longer keel,
then moment equilibrium in flight is:

1) Mf - Ft1*lt1 - FR*xR1 + mt*g*lt1 + mG*g*xG = 0

where
- Mf...... is the moment generated by the fuselage
............alone (without the H-stab contribution)
- Ft1......is the H-Stab force
- lt1.......is the distance from H-stab to cg
- FR.......is the rotor force
- xR1.....is the distance from cg at which the rotor
............force passes throug the x-axis
- mt......is the mass of the tail unit
- lt1......is distance to the cg of the tail unit
- mG.....is the mass of the gyro sans tail unit
- xG......is the distance of mG to cg
- g........is the acceleration of the earth (9.81 m/s^2)


Note that mG does not appear in the picture since it cancels
anyway and is very close to cg so would clutter the picture.
Also note that xR1 = dxR1 in the picture.

For the new keel length, moment equilibrium looks like this:

2) Mf - Ft2*lt2 - FR*xR2 + mt*g*lt2 + mG*g*xG = 0

Let me redo the static case for which Mf and Ft are zero.
(I'll need the solution later on)

if we subtract 1) from 2) we get:

3a) -FR*(xR2-xR1) - mt*g*(lt2-lt1) + [mG*g*xG - mG*g*xG] =0

where the last two terms in square brackets cancel.

If we solve for xR2:

3b) xR2= mt*g/FR*(lt2-lt1) + xR1

(lt2-lt1) is d in Dougs formula, so xR2 is the new cg
position while mt*g/FR*(lt2-lt1) is the distance you
have moved the cg (thats D in Dougs post). For your
case of lt2 < lt1 the new cg postion is a bit closer to
cg and you would have to subtract the angle Doug
calculated from the one you started out with.


If we subtract 1) form 2) in flight we get:

Mf - Mf + Ft1*lt1 - Ft2*lt2 { -FR*(xR2-xR1) - mt*g*(lt2-lt1) } + [mG*g*xG - mG*g*xG]= 0

the mG terms again cancel but the two terms in curly braces are 3a
which also cancels since we have calculated xR2 to give static
equlibrium which leaves us with:

Ft1*lt1 - Ft2*lt2 = 0

If the angle of the H-stab remains unchanged then Ft1=Ft2 and
the gyro will not be in equilibrium in flight or in other words
the flight attitude will be different, perhaps even grossly so.

To me that means you want to be careful when taking off for the
first flight since you have substantially changed your aircraft.

I would also second James' view that shortening the lever arm of
your H-stab is perhaps not a good idea. With most pusher gyros
the lever arm is very short anyway.
 

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groundhog

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at the risk of sounding pissy,most aircraft hang off the center of lift.wieght and balance only resolve the vertical components then doug reconsiled the fixed point to the new hang location using good ol trig.I was merely trying to get you to sneak up on realizing you already knew how to solve the problem if you took it one step at a time.
 
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