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Calculation of gradient?

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Calculation of gradient?

Old 22nd Nov 2006, 15:54
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Calculation of gradient?

Given:
Weight 50 000 kg
L/D = 12
Thrust = 4 x 21 000 N
Assumed g = 10 m/s2

Does anyone know how to calculate the climb gradient if in a steady, wings level climb?
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Old 22nd Nov 2006, 17:08
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Weight 50 000 kg
-0: L/D = 12
Thrust = 4 x 21 000 N
Assumed g = 10 m/s2

Does anyone know how to calculate the climb gradient if in a steady, wings level climb?

I asume Mass= 50000kg
Thus W= 50000*10= 500000 N.
Being a= CLB Angle
-1: L= W+Cos(a)
-2: Sin(a)= (T-D)/W

From 0: L=D*12
thus with 1: D*12=W+Cos(a)
D=(W+Cos(a))/12
In 2:
Sin(a)=(T-(W+Cos(a))/12)/W
a= Sin-1 (T-(W+Cos(a))/12)/W

When you find the CLB Angle "a" CLB GRAD= TAN(a)*100

ie 45 CLB angle= 100% CLB GRAD (for every ft you advance horizontaly you climb one ft verticaly)
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Old 22nd Nov 2006, 18:32
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Sorry Iīll write it again itīs difficult to write formulas here


I asume Mass= 50000kg
Thus W= 50000*10= 500000 N.
Being a= CLB Angle
-1: L= W*Cos(a)=
-2: Sin(a)= (T-D)/W

From 0: L=D*12

thus with 1: D*12=W*Cos(a)

D=(W*Cos(a))/12 = W/12Sin(a) **1/Sinx= Cosx

In 2:

Sin(a)=(T-(W*Cos(a))/12)/W

Sin(a)=(T-(W/12Sin(a)))/W

Sin(a)=((12TSin(a)-W)/12Sin(a))/W

Sin(a)=((12TSin(a)-W)/12WSin(a))

1=(12TSin(a)-W)/12W

12W= 12TSin(a)-W

13W= 12TSin(a)

13W/12T=Sin(a)

***a= Sin-1(13W/12T)***

When you find the CLB Angle "a" CLB GRAD= TAN(a)*100

ie 45 CLB angle= 100% CLB GRAD (for every ft you advance horizontaly you climb one ft verticaly)
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Old 22nd Nov 2006, 18:45
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Wrong again let me get a piece of paper
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Old 22nd Nov 2006, 18:59
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Devil Calculatio of Gradient

Save all the hassle,
Just take the page out of Aerad Flight Information Supplement the Grey Book.
Page (xv) (xv1) Gradient to rate of climb/descent
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Old 22nd Nov 2006, 19:25
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Quick and easy...

The formula is sin(a) = (T-D)/W where a = angle of climb, T = total thrust, D = drag and W = aircraft weight.

For these questions at ATPL level you assume L = W so you have:-

W = 50000kg x g (10 m/s/s) = 500000 N;
L = W = 500000 N;
L/D = 12 so D = 500000/12 = 41667 N (5 s.f.);
T = 4 x 21000 = 84000 N

Plug all of the above into the formula and you get sin(a) = 0.085 (2 s.f.). Multiply this by 100 and you get the climb gradient as a percentage:-

Climb gradient = 8.5 %.

Hope this helps,

V1R
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Old 22nd Nov 2006, 19:44
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...and watch out for exam questions that may require, without clearly stating, an engine out climb gradient. i.e. "Determine the second segment gradient for the purpose of obstacle clearance".
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Old 22nd Nov 2006, 22:23
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Thumbs up

L/D = 12 so D = 500000/12 = 41667 N (5 s.f.); #$%"&
I didnīt saw that one! Tot D=W/L/D more practical indeed
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Old 12th Jul 2011, 21:30
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EASY way :)

Given: Aeroplane mass: 50 000kg. Lift/Drag ratio: 12. Thrust per engine: 60 000N. Assumed
g: 10m/sē. For a straight, steady, wings level climb of a twin engine aeroplane, the allengines
climb gradient is:

just :

(T - D)/W = sin alpha !!!

T (thrust) = number of engines (2) * they thrust (60 000 N) = 120 000 N
W (weight) = aeroplane mass (50 000kg) ! * 10g (m/sē) = 500 000 N
D (drag) = W/ (Lift/Drag) -> 500 000 / 12 = 41666

so !

-> (120 000 - 41666) / 500 000 = 0,156668

result * 100 (to get %) = 15,6668 -> 15,7% !
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