oriol
Well-Known Member
Hi!
I am following Anderson's introduction to flight to calculate static stability in my glider. In practice all theory regarding static stability is only a matter of calculating moments. Thanks god, because I enjoy drawing force diagrams and computing forces, it is much more fun than dealing with abstract calculus. However there is something striking in Anderson's diagrams. He locates the center of gravity way behind C/3. Perhaps this is something arbitrary, and at the end of the calculations the numbers can shift the final location of the C of G?
From my limited knowledge RC aircraft pilots make sure the battery is located in C/3, so that the aircraft flies smooth. For small sport real airplanes or ultralights, I assume that engineers do the same, the center of gravity of the aircraft at C/3 of MAC.
Here below are Anderson's diagrams. The last one is of course mine. The obvious difference with mine and his, is that mine's is a high wing, the center of gravity is located at C/3, and at C vertical distance from the chord line. I follow the same approach of Anderson, the chord line is horizontal and the relative wing is tilted alfa with respect to the aircraft, to ease the calculations.
I have written the calculations below, because it allows to find a value for it (tail incidence)? The only purpose of writing the calculations, is to show how the values are derived?
Beyond the doubt about the rationale behind anderson's center of gravity. My second doubt is to find if the below calculated tail incidence is too high?
I understand though, that tail incidence is set for minimum drag at cruise. However tail incidence has to be compatible with the below case too.
The wing moment around the center of gravity is;
MCGW = MacW - Lcosalfa*(C/3-C/4) + Lsinalfa*C - Dcosalfa*C - Dsinalfa*(C/3-C/4)
MCGW = MacW - (C/3-C4)(Lcosalfa+Dsinalfa) + C(Lsinalfa- Dcosalfa)
MCGW = MacW - (C/12)(Lcosalfa+Dsinalfa) + C(Lsinalfa- Dcosalfa)
Anderson simplifies, given that alfa is very small cosalfa=1, sinalfa=0, all sin disapear. L is much bigger than D, so D goes out too;
MCGW = MacW - (C12)(Lcosalfa)
Then he divides both sides of the equality by qSC, to use dimensionless coefficients;
CMCGW = CMacW - (C/12)(CLw)
Given that CL = awb*alfawb, the above equation can be rewritten as,
CMCGW = CMacW - (C/12)*awb*alfawb
The tail moment around the center of gravity is;
MCGT = MacT - Lcosalfa*(5C-(C/3-C/4)) - Lsenalfa*C/2 - Dcosalfa*C/2 +Dsinalfa*(5C-(C/3-C/4))
MCGT = MacT + (5C-(C/3-C/4))(Dsinalfa - Lcosalfa) - C/2*(Lsenalfa+Dcosalfa)
MCGT = MacT + 4,91C*(Dsinalfa - Lcosalfa) - C/2*(Lsenalfa+Dcosalfa)
Again we follow the same philosophy, sin and D terms disapear plu Cmtail because it is considered to be too small;
MCGT = -4,91C*L
Again both sides are divided by qSC;
CMCGT = -4,91*St/Sw*Clt
Considering;
alfat = alfawingbody - incidence tail,
CLt = at*alfat = at (alfawb -it -E), E is downwash
E = Eº + dE/dalfa*alfawb, Eº is downwash at l=0
CMCGT = -4,91St/Sw * (at*alfawb(1-dE/dalfa) - at*(it - Eº))
The criteria necessary for longitudinal balance and static stability is that Cm0 (L=0) must be positive.
If I evaluate the above equation for L=0, I have;
CMcg = CMcgw + Cmcgt
CMcg = CMacW - (C/12)*awb*alfawb - 4,91St/Sw * [(at*alfawb(1-dE/dalfa) - at*(it - Eº)]
considering L=0;
CM0 = CMacW + 4,91/Sw * at*(it - Eº)
CMacW = 4,91*St/Sw * at*(it - Eº)
CMacW = -0,35*Pi/4 = -0,274
Stail = 1,4m2
at = 0,0786
it ?
E0 = 0
0,274 = 4,91*1,4/10*0,0786 * it
it = 0,35 / 0,05502 = 4,98º
Finally we have reached the end!
Is 4,98 degrees a value strikingly high? Is that high, because the simplifications made by Anderson, eliminating all the terms that were orders of magnitude smaller?
Billski came to my rescue when I stumbled a wall, I was unable to calculate downwash? Anderson's says that downwash is obtained through wind tunnel data; he does not provides data for it. Billski's adviced me to use Dommasch's formula; E = 36,5/(AR*Cl), which in my case results in 6,36º for Cl cruise.
In theory with 4,98º I am within a reasonable margin for downwash at cruise. Even so, I wanted to hear from you? Because Anderson locates the center of gravity in a very odd spot, plus he considers a low wing etc. It is the first time that I calculate those things, so I am worried about doing something wrong?
As usual, any thoughts will be greatly appreciated!
Thanks for helping me, having so much fun learning!
Cheers,
Oriol
I am following Anderson's introduction to flight to calculate static stability in my glider. In practice all theory regarding static stability is only a matter of calculating moments. Thanks god, because I enjoy drawing force diagrams and computing forces, it is much more fun than dealing with abstract calculus. However there is something striking in Anderson's diagrams. He locates the center of gravity way behind C/3. Perhaps this is something arbitrary, and at the end of the calculations the numbers can shift the final location of the C of G?
From my limited knowledge RC aircraft pilots make sure the battery is located in C/3, so that the aircraft flies smooth. For small sport real airplanes or ultralights, I assume that engineers do the same, the center of gravity of the aircraft at C/3 of MAC.
Here below are Anderson's diagrams. The last one is of course mine. The obvious difference with mine and his, is that mine's is a high wing, the center of gravity is located at C/3, and at C vertical distance from the chord line. I follow the same approach of Anderson, the chord line is horizontal and the relative wing is tilted alfa with respect to the aircraft, to ease the calculations.
I have written the calculations below, because it allows to find a value for it (tail incidence)? The only purpose of writing the calculations, is to show how the values are derived?
Beyond the doubt about the rationale behind anderson's center of gravity. My second doubt is to find if the below calculated tail incidence is too high?
I understand though, that tail incidence is set for minimum drag at cruise. However tail incidence has to be compatible with the below case too.
The wing moment around the center of gravity is;
MCGW = MacW - Lcosalfa*(C/3-C/4) + Lsinalfa*C - Dcosalfa*C - Dsinalfa*(C/3-C/4)
MCGW = MacW - (C/3-C4)(Lcosalfa+Dsinalfa) + C(Lsinalfa- Dcosalfa)
MCGW = MacW - (C/12)(Lcosalfa+Dsinalfa) + C(Lsinalfa- Dcosalfa)
Anderson simplifies, given that alfa is very small cosalfa=1, sinalfa=0, all sin disapear. L is much bigger than D, so D goes out too;
MCGW = MacW - (C12)(Lcosalfa)
Then he divides both sides of the equality by qSC, to use dimensionless coefficients;
CMCGW = CMacW - (C/12)(CLw)
Given that CL = awb*alfawb, the above equation can be rewritten as,
CMCGW = CMacW - (C/12)*awb*alfawb
The tail moment around the center of gravity is;
MCGT = MacT - Lcosalfa*(5C-(C/3-C/4)) - Lsenalfa*C/2 - Dcosalfa*C/2 +Dsinalfa*(5C-(C/3-C/4))
MCGT = MacT + (5C-(C/3-C/4))(Dsinalfa - Lcosalfa) - C/2*(Lsenalfa+Dcosalfa)
MCGT = MacT + 4,91C*(Dsinalfa - Lcosalfa) - C/2*(Lsenalfa+Dcosalfa)
Again we follow the same philosophy, sin and D terms disapear plu Cmtail because it is considered to be too small;
MCGT = -4,91C*L
Again both sides are divided by qSC;
CMCGT = -4,91*St/Sw*Clt
Considering;
alfat = alfawingbody - incidence tail,
CLt = at*alfat = at (alfawb -it -E), E is downwash
E = Eº + dE/dalfa*alfawb, Eº is downwash at l=0
CMCGT = -4,91St/Sw * (at*alfawb(1-dE/dalfa) - at*(it - Eº))
The criteria necessary for longitudinal balance and static stability is that Cm0 (L=0) must be positive.
If I evaluate the above equation for L=0, I have;
CMcg = CMcgw + Cmcgt
CMcg = CMacW - (C/12)*awb*alfawb - 4,91St/Sw * [(at*alfawb(1-dE/dalfa) - at*(it - Eº)]
considering L=0;
CM0 = CMacW + 4,91/Sw * at*(it - Eº)
CMacW = 4,91*
CMacW = -0,35*Pi/4 = -0,274
Stail = 1,4m2
at = 0,0786
it ?
E0 = 0
0,274 = 4,91*1,4/10*0,0786 * it
it = 0,35 / 0,05502 = 4,98º
Finally we have reached the end!
Is 4,98 degrees a value strikingly high? Is that high, because the simplifications made by Anderson, eliminating all the terms that were orders of magnitude smaller?
Billski came to my rescue when I stumbled a wall, I was unable to calculate downwash? Anderson's says that downwash is obtained through wind tunnel data; he does not provides data for it. Billski's adviced me to use Dommasch's formula; E = 36,5/(AR*Cl), which in my case results in 6,36º for Cl cruise.
In theory with 4,98º I am within a reasonable margin for downwash at cruise. Even so, I wanted to hear from you? Because Anderson locates the center of gravity in a very odd spot, plus he considers a low wing etc. It is the first time that I calculate those things, so I am worried about doing something wrong?
As usual, any thoughts will be greatly appreciated!
Thanks for helping me, having so much fun learning!
Cheers,
Oriol
