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.