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Raymer tells (Source: Raymer: Aircraft Design – A Conceptual Approach Chapter "5 THRUST-TO-WEIGHT RATIO AND WING LOADING" Paragraph “5.3 WING LOADING” Subparagraph “Climb and Glide”) that, when the aircraft is climbing, if the climb angle γ is small,

then $G=\frac{T-D}{W}$

(G is Climb Gradient as defined by Raymer words: "Rate of climb is vertical velocity,.. Climb gradient G is the ratio between vertical and horizontal distance traveled. This is approximately equal to the vertical climb rate divided by the aircraft velocity, or the sine of the climb angle γ." Source: Raymer: Aircraft Design – A Conceptual Approach Chapter "17 PERFORMANCE AND FLIGHT MECHANICS" Paragraph “17.3 STEADY CLIMBING AND DESCENDING FLIGHT” Subparagraph “Climb Equations of Motion”)

Then $\frac{D}{W}=\frac{T}{W}-G$ ( that is the first relation for D/W).

At the same time $D=q\cdot S\cdot c_{D0} + q\cdot S\cdot\frac{c_L^2}{π\cdot AR \cdot e}$

And substituting $c_L=\frac{W}{q\cdot S}$,

$$\frac{D}{W}=\frac{q\cdot S\cdot c_{D0}+q\cdot S\cdot\frac{c_L^2}{π\cdot AR\cdot e}}{W}=\frac{q\cdot c_{D0}}{\frac{W}{S}}+\frac{W}{S}\cdot\frac{1}{q\cdot π\cdot AR\cdot e}$$

(that is the second relation for $\frac{D}{W}$)

Equating the first and the second relation and solving for wing loading, we have:

$$\frac{W}{S}=\frac{\left(\frac{T}{W}-G\right)± \sqrt {\left(\frac{T}{W}-G\right)^2-\frac{4\cdot c_{D0}}{π\cdot AR\cdot e}}}{\frac{2}{q\cdot π\cdot AR\cdot e}}$$

MY FIRST QUESTION: In this equation are both, the bigger and the lower solution values for $\frac{W}{S}$ realistic for aircraft?

After, Raymer notes that the term within the square root symbol in the equation above cannot fall below zero, so the following must be true regardless of the wing loading:

$$\frac{T}{W}≥G+2\cdot\sqrt{\frac{c_{D0}}{π\cdot AR\cdot e}}$$

MY SECOND QUESTION: Raymer doesn’t consider that there are two solutions sourcing from the condition that the term within the square root symbol cannot fall below zero, the other one is: $$\frac{T}{W}≤G-2\cdot\sqrt{\frac{c_{D0}}{π\cdot AR\cdot e}}$$. Why shouldn’t be considered? Could the second solution represent something realistic?

I HAVE A SUPPLEMENTARY QUESTION: Relating to my questions above, I have to report two comments Raymer writes immediately after the last equation and inequality.: "This equation says that no matter how "clean" your design is, the T/W must be greater than the desired climb gradient! ....... Another implication of this equation is that a very "clean" aircraft that cruise at a high speed despite a very low T/W will probably climb poorly. A 200-mph airplane that flies on 20 hp can't be expected to climb as well as an airplane that requires 200 hp to reach 200 mph (unless the latter weighs ten times as much)" Could anyone explain me better the last sentences, in particular, in which case T/W is very low with high speed? If we have high speed doesn't T/W should be high too?

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    $\begingroup$ I converted your equations into MathJax to make the question more readable. Could you please check that I converted everything truthfully? $\endgroup$ – Peter Kämpf May 5 '16 at 7:09
  • $\begingroup$ In inequality inside squared root "greek pi"times"AR"times"e" should be at denominator, "CD0" should be at numerator. Anyway, I'm really grateful for your work of conversion to MathJax, because I tried to use it, but I need getting use to that. $\endgroup$ – d.pensopositivo May 5 '16 at 8:24
  • $\begingroup$ Finally, I think I wrote correctely the formulae for my intention of question. Since now, if someone understood better my questions , can try to answer. $\endgroup$ – d.pensopositivo May 5 '16 at 10:15
  • $\begingroup$ You should include your last comments into your question (and then remove the comments by clicking the (X) that appears when you hover the comment with the mouse). You may edit your question by using the editlink between the question tags and your signature block. $\endgroup$ – mins May 5 '16 at 15:04
  • $\begingroup$ I saw "second grade equation" in the title and thought I'd be able to follow along. Sadly not... :( $\endgroup$ – FreeMan May 5 '16 at 16:05
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Raymer's G is a little unusual. In effect, it is the tangent of the flight path angle $\gamma$, and since his equations are for small values of $\gamma$, we can assume that it is approximately equal to $sin\gamma$ or $\gamma$ itself when expressed in radians. $$G \approx \gamma$$ Next, the expression $\pi\cdot AR\cdot e$: This is part of the square root of the lift coefficient for minimum drag. The complete equation is: $$c_{L\:(minimum\:drag)} = \sqrt{c_{D0}\cdot\pi\cdot AR\cdot e} = c_{L\:(md)}$$ Next, I need to express that thrust-to-weight ratio differently. As long as we keep $\gamma$ small: $$\gamma = \frac{T}{W} - \frac{c_D}{c_L} => \frac{T}{W}-\gamma = \frac{c_D}{c_L}$$ Now I can rewrite the equation of your first question: $$\frac{W}{S} = q\cdot\frac{\frac{c_D}{c_L}\cdot \frac{c^2_{L\:(md)}}{c_{D0}}}{2} ± q\cdot\frac{\sqrt {\left(\frac{c_D}{c_L}\right)^2\cdot\left(\frac{c^2_{L\:(md)}}{c_{D0}}\right)^2 - 4\cdot c^2_{L\:(md)}}}{2}$$ and when I restrict the equation to the minimum drag point where $c_{D0} = c_{Di}$, I can write $$\frac{W}{S} = q\cdot c_{L\:(md)} ± q\cdot\sqrt {c^2_{L\:(md)} - c^2_{L\:(md)}}$$ Fancy that! The expression under the root becomes zero when I look at the polar point of minimum drag! The only variable term under the square root is the ratio $\frac{c_D}{c_L}$, and it reaches its minimum at the $c_{L\:(md)}$ point. When I move away from this polar point, I should get two valid solutions.

I guess these two wing loadings allow to fly with the specified flight path angle at the given polar point and thrust-to-weight ratio. Since the weight is prescribed by the thrust-to-weight ratio, the solution is actually for two different wing areas. The smaller wing requires a higher speed and the larger wing a smaller speed than what you get at the lowest drag point. Between both, the speed and wing area should result in configurations which achieve a higher flight path angle $\gamma$, and outside of this interval no level flight should be achievable with the resulting wing areas and speeds.

I cannot follow how you arrive at the second question. The term under the square root should be $$\left(\frac{T}{W}\right)^2 - 2\cdot\frac{T}{W}\cdot G + G^2 ≥ \frac{4\cdot c_{D0}}{\pi\cdot AR\cdot e}$$ But when I insert again as explained above, I can write for the term under the root: $$\left(\frac{c_D}{c_L}\right)^2 - \frac{4\cdot c^2_{D0}}{c^2_{L\:(md)}}$$ so the condition for a positive root is $$\frac{c_{D0}+c_{Di}}{c_L} ≥ \frac{2\cdot c_{D0}}{c_{L\:(md)}}$$ or $$ \frac{c_{D0}}{c_L} + \frac{c_L}{\pi\cdot AR\cdot e} ≥ 2\cdot\sqrt{ \frac{c_{D0}}{\pi\cdot AR\cdot e}}$$ which is certainly true; both sides of the equation are equal at the lowest drag polar point and away from it the left side is bigger and the right side stays constant.

If you just isolate the term under the root and look at it by itself, I doubt you have something that corresponds to a realistic configuration. You need to extract both numerator and denominator to keep the physical analogies intact. What Raymer does is to look at the term under the root in isolation and then postulates that the first summand must be larger than the second. He puts both on the two sides of an inequality and extracts the roots on both sides separately. Then he moves G over to the other side. There is no second version of the inequality possible.

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  • $\begingroup$ Peter Kämpf first of all, thank you for your answer, it was very helpful. When you say: "The ratio CD/CD reaches its maximum at the CL(md) point, so the first summand of the root will always be smaller than the second, except for the polar point at minimum drag when both are equal." CONTINUE $\endgroup$ – d.pensopositivo May 10 '16 at 12:06
  • $\begingroup$ CONTINUE What do you mean? First I suppose you mean "CD/CL" instead of "CD/CD". If it so, it seems to me that, at min drag CD/CL is min, not max, (CD is numerator) then, out of that point of minimum drag we have 2 solutions. $\endgroup$ – d.pensopositivo May 10 '16 at 12:08
  • $\begingroup$ @d.pensopositivo: You are right, thanks for catching this. Corrected. If the root becomes zero, it does not matter if there is a + or a - ahead of it. Therefore I think there is only one solution. Or two with the same value, if you want to see it this way. $\endgroup$ – Peter Kämpf May 10 '16 at 13:46
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    $\begingroup$ Anyway as I showed above, Raymer considers the existence of three cases: 1) Two different solutions : when T/W≥G+2⋅(cD0/(π⋅AR⋅e))^(1/2) 2) Two coincident solutions : when T/W=G+2⋅(cD0/(π⋅AR⋅e))^(1/2) 3) No solutions : when T/W≤G+2⋅(cD0/(π⋅AR⋅e))^(1/2) . My objection in "my second question" remind that such condition is coming from a resolution of a second grade disequation i.e. (TW−G)^2−4⋅CD0/(π⋅AR⋅e)≥0, then T/W≤G-2⋅(cD0/(π⋅AR⋅e))^(1/2) is another part of condition coming from case 1) (unless there are other constrains not explicitly mentioned there). $\endgroup$ – d.pensopositivo May 15 '16 at 18:01
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    $\begingroup$ I return on my second question. In order to have real solutions $(\frac{T}{W}-G)^{2}-\frac{4*Cdo}{\pi *AR*e}\geq 0$. Then for simplicity of reasoning I define $(\frac{T}{W}-G)=x$ and $\frac{4*Cdo}{\pi *AR*e}=b$. The condition becomes $x^{2}-b\geq 0$, but this (second grade) inequation is verified in both following zones: $x\geq \sqrt{b}$ and also $x\leq - \sqrt{b}$ i.e. $\frac{T}{W}-G\geq 2\ast \ \sqrt{\frac{CD0}{\pi \ast AR\ast e}}$ and also $\frac{T}{W}-G\leq-2\ast \ \sqrt{\frac{CD0}{\pi \ast AR\ast e}}$. Raymer excludes the second one, I don’t know why. $\endgroup$ – d.pensopositivo May 17 '16 at 9:24

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