# GyroKit library and GyroRotor program

##### Newbie
Your u is perpendicular to NFA and w is aligned with the NFA.

Yes when derivatives are computed in the no-feathering coordinates frame (S). Others derivatives (X,Y,Z,L,M,N) are computed in the body fixed coordinates (F), which is the one we use for stability purpose. See the following figure :

Bramwell has his u aligned with the direction of flight

You are right. It is clearly stated page 141 and this explains why in these axis, they can write : ∂Ct/∂û =∂Ct/∂μ.

Unfortunately these axes can only be used for the pure longitudinal motion but not for 3D motion, because body m.o.i. Ixx, Izz and Ixz are not constant in wind axis (in fact Iyy is no more constant when sideslip angle varies). They are only constant in the body fixed coordinates.

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#### Jean Claude

##### Junior Member
As you know, I did not follow courses in mechanics. Can someone explain to me what the axis XY? Thank you.

The delay in return to the original A.o.A of the frame is different between the real flight and wind tunnel testing.
In a real flight, a lift loss produces a fall, able to change in itself the angle of attack.
In a wind tunnel, a lift loss does produces no fall. Return is then lower.
Which of the two results you want to mention?

##### Newbie
Of course, Jean. This is mentioned on my green curve.

Fine. So you should be able to compute the thrust line offset of a Magni tailless gyroplane where the short modes just become instable.

Can you say a few words about the way you have modelized rotor damping ?

##### Newbie
Can someone explain to me what the axis XY?

Do you mean in my Figure ?

In aeronautical convention, generally speaking, X axis is pointing forward, Z axis is pointing downward and Y axis is pointing sideway to starboard.

There are mainly three coordinates system : 1) inertial coordinate system, 2) body-fixed coordinate system and 3) wind coordinate system.

Origin of these coordinates system are the center of mass of the aircraft.

1) Inertial coordinate system :
The Z axis is parallel to the direction of gravity, and direction of X and Y axis doesn’t matter for mechanic equations.

2) Body-fixed coordinate system :
Axes of the body-fixed coordinate system (or fuselage coordinate system) rotate with the aircraft. These axis are generally chosen so that m.o.i. Ixy=Iyz are zero which is equivalent to say that the XFZF plane is a plane of symmetry of the aircraft.

3) Wind coordinate system
The X axis of the wind coordinate system is align with the freestream velocity, and points forward. The Z axis of the wind coordinate system is pointing downward and belongs to the XFZF plane (plane of symmetry of the aircraft).

In rotary wing aircraft you must also define coordinates system relative to the rotor. It is more convenient to study rotor behavior to have the Z axis pointing upward.

To do so, I decided in GyroRotor program to swap the X and Y axis. So in GyroRotor program the YS axis is pointing forward and belongs to the XFZF plane (therefore XS=YF), the ZS axis is pointing upward (the ZS axis is the NFA) and the XS axis is pointing sideway to starboard.

Finally, in GyroRotor program the aircraft velocity is supposed to be in the YSZS plane (the slidesleep angle is zero).

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#### Jean Claude

##### Junior Member
These axis are generally chosen so that m.o.i. Ixy=Iyz
I agree with your fig. but I still not understand xy (or yz). Can you showed this axis on your fig ? Thank you.

So you should be able to compute the thrust line offset of a Magni tailless gyroplane where the short modes just become instable.

##### Newbie
I agree with your fig. but I still not understand xy (or yz)

I better understand your question now. xy or yz are not axes. Mathematically, moments of inertia are express as a matrix :

Where :

and

Body axes are chosen so that Ixy and Iyz are zero. See this link

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#### kolibri282

##### Super Member
Small progress, I hope

Small progress, I hope

To validate results there seem to be two options. One is to see how well they match measured values or a second approach is to compare results calculated with different sets of formulae. naca 6763 gives curves for a helicopter with a solidity of 0.1 so I scaled the blades of the sample helicopter of that report, which had a solidity of about 0.05, to 0.1 to avoid the hassle of having to correct the results for sigma=0.5. I have attached my file for the 6763 values and here is one of the rare cases where there is an error in a naca report. The dt42dmu value is outright wrong, which I found out by calculating finite differences using the complete coefficients of naca-716. The dt42dmu value in this file is pure guesswork. All the result values have been calculated using the finite differences.
Next I computed cQ/sigma values using the formulae of 2655, Bramwell and 6763, the results are in the first picture. They are close enough for me to conclude that validation was successful. Next I computed values for the KD-1. Looking at those one has to keep in mind that the 6763 results are only valid for the case of sigma=0.1. Funny enough these seem to match Jean's results best. Following the discussion of the dcTs/dmu values I wonder whether the rather higher 2655 values might not in the end turn out to be correct.
I have attached the data file for the 6763 sample helicopter with sigma=0.1 (just in case someone wants to verify the results) and the trim results: alfaNf= -14.11° thetaFus= -6.6°. The advance ratio is mu=0.3. I'll be working from a tablet for a few days now so no new results next week. Would it perhaps help if I implemented my R-6763 helicopter in Jeans program, any thoughts?

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• cQ_comp_2655_Bram_6763_100215Snap2212.JPG
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• fd_comp_JF_JH_1-5_3_cQ_100215Snap2212.JPG
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• StabParaComp_JF_JH_all_derivatives_100215Snap2205.xls
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• dcQsdmu6763.txt
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• RotDatR-6763_s1.txt
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#### Jean Claude

##### Junior Member
Can you say a few words about the way you have modelized rotor damping ?

Everything is in the "Retard angulaire disque".
This value is obtained by the reasoning below:

##### Newbie
Juergen, be aware that all my derivatives WRT to μ must be calculated with :

∂/∂μ = ∂Ct/∂û - tan α_NF ∂/∂ŵ.​

I will recompute my values tomorrow and compare with yours.

##### Newbie
Jean Claude, if you look to Bramwell page 155, equation 5.82 gives the longitudinal flapping coefficient derivative with respect to pitch rate. This equation, neglecting μ^2 and with your notation leads to : θ = 16 δ / γ ω .

With a Lock number γ=4.54 for Magni rotor and ω=36, this equation gives the same value as you : θ = 0.0978 δ.

I wonder how you calculated the mean blade angle of attack. If one consider that the total surface of the blade is concentrated at r=0.75 R and neglecting the forward speed of the gyroplane, the mean blade angle of attack α must fulfill : P/2 = 1/2 ρ S a α (r ω)^2 with S the area of the blade. Unfortunately this expression is wrong. The good one to be consistant with the good value of θ is 1/4 ρ S a α (r ω)^2 (half of the previous value). Any idea why ?

The value of θ gives approximately 85% of rotor damping. The other 15% are coming from the rear force H (see Bramwell equation 5.87). In case of blade twist we can add 30% more (see my post #55).

#### Jean Claude

##### Junior Member
Jean, The mean a.o.A of blade to 0.75 R simply comes from my rotor spreadsheet. Given my approximations I am amazed that my results are as close to yours.

If one consider that the total surface of the blade is concentrated at r=0.75 R and neglecting the forward speed of the gyroplane, the mean blade angle of attack α must fulfill : P/2 = 1/2 ρ S a α (r ω)^2 with S the area of the blade. Unfortunately this expression is wrong. The good one to be consistant with the good value of θ is 1/4 ρ S a α (r ω)^2 (half of the previous value). Any idea why ?
The blades does not work in calm air but in the induced flow of the rotor. So, 5,7/rd slope in your formula is incongruous . This clearly appears to you by observing the blade pitch of a hovering helicopter

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#### raghu

##### Senior Member
You are right. It is clearly stated page 141 and this explains why in these axis, they can write : ∂Ct/∂û =∂Ct/∂μ.

Thanks JF. And that is what confused me about your earlier assertion that ∂Ct/∂û was not equal to ∂Ct/∂μ in Bramwell. I knew it was but could not explain it until I noticed the different axis system.

I guess the formal way to adjust for this would be to rotate the X,Y,Z, M etc Bramwell derivatives by alphaNE and slid slip angle to arrive at your values. If you are doing just the longitudinal subset then a single rotation by alphaNE (which your tan formula dies in effect) would be sufficient.

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#### raghu

##### Senior Member
Juergen, be aware that all my derivatives WRT to μ must be calculated with :

∂/∂μ = ∂Ct/∂û - tan α_NF ∂/∂ŵ.​

I will recompute my values tomorrow and compare with yours.

JF, should this (though with more terms) not apply to the other derivatives as well relating to u and w? Thanks!

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##### Newbie
Juergen, here are my last values. We are not ok on hc and a1 (red values). Concerning hc, I think your value is calculated in the TPP. Am I right ? Mine is computed is the NFP.
For a1, I also put values from Bramwell which are consistent with mine. Be aware that there is a mistake in Bramwell's equation 5.64 p 152. The sign in front of the second term must be positive.

I also add in the file some derivatives WRT to q and omega.

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• Derivatives 7 10.png
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• StabParaComp_JF_JH_all_derivatives_100215Snap2205.xls
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##### Newbie
JF, should this (though with more terms) not apply to the other derivatives as well relating to u and w? Thanks!

Yes. There is a mistake in my previous message. The good equation is :

∂/∂μ = ∂/∂û - tan α_NF ∂/∂ŵ​

which must be apply on every terms, i.e. :

∂Ct/∂μ = ∂Ct/∂û - tan α_NF ∂Ct/∂ŵ
∂Ch/∂μ = ∂Ch/∂û - tan α_NF ∂Ch/∂ŵ
∂Cq/∂μ = ∂Cq/∂û - tan α_NF ∂Cq/∂ŵ
∂a1/∂μ = ∂a1/∂û - tan α_NF ∂a1/∂ŵ

...

#### kolibri282

##### Super Member
It is great to see, Jean, that we are now more than half way through the process of matching linear derivatives and I feel confident that we will find out the reasons for any remaining differences. I will be able to start on it on Sunday.

So long,

Juergen

#### raghu

##### Senior Member
Concerning hc, I think your value is calculated in the TPP. Am I right ? Mine is computed is the NFP.

also Bramwell hc derivatives are based around the discAxis/TPP.

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#### raghu

##### Senior Member
∂/∂μ = ∂/∂û - tan α_NF ∂/∂ŵ​

...

In addition JF unless I am missing something, don't you need to modify all the
∂/∂ŵ derivatives as well?

Brawell's w is rotated by alphaNF from yours. Brawells w is 0 in steady flight as u is aligned with the direction of flight. Your w is parallel to NF axis. For ∂ct/∂ŵ the effect will be negligible but for the other derivates it would be more significant.

#### Jean Claude

##### Junior Member
also Bramwell hc derivatives are based around the discAxis/TPP.
I agree with Raghu. Bramwell writes: Referring to Fig. 5.5, let αD be the disc incidence in steady flight.

##### Newbie
Raghu, you are right. If we denote ∂/∂ŵB derivative with constant aircraft speed U, then we get :

∂/∂ŵB = tan α_NF ∂/∂û - ∂/∂ŵ​

I have updated my values. There is a slight change for high incidence.

Most derivative in Bramwell are actually based around TPP. I found this more complicated because TPP rotates with μ and α_NF while NFP doesn't. For instance when computing Mq you have to calculate ∂tc/∂q (which is zero if we neglect ∂λ/∂q), ∂hc/∂q and take into account the fact that TPP has changed and calculate ∂a1/∂q. In the NFP, you just have to calculate ∂hc/∂q.

#### Attachments

• Derivative 9 10.png
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• StabParaComp_JF_JH_all_derivatives_100215Snap2205.xls
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