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Founder
22nd Nov 2006, 16:54
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?

ICEMAN757200
22nd Nov 2006, 18:08
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:confused:
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):O

ICEMAN757200
22nd Nov 2006, 19:32
Sorry Iīll write it again itīs difficult to write formulas here:eek: :} :ugh:

I asume Mass= 50000kg:confused:
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)

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):O

ICEMAN757200
22nd Nov 2006, 19:45
Wrong again let me get a piece of paper:ugh: :ugh: :ugh: :ugh: :mad: :ouch: :*

PENNINE BOY
22nd Nov 2006, 19:59
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 :ok: :ok:

Vee One...Rotate
22nd Nov 2006, 20:25
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:-

Hope this helps,

V1R

Alex Whittingham
22nd Nov 2006, 20:44
...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".

ICEMAN757200
22nd Nov 2006, 23:23
L/D = 12 so D = 500000/12 = 41667 N (5 s.f.); :mad: #\$%"&
I didnīt saw that one!:ugh: Tot D=W/L/D more practical indeed:D :ok: :zzz:

titinius
12th Jul 2011, 22:30
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

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% !