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? |
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 |
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) ***a= Sin-1(13W/12T)***:mad: 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 |
Wrong again let me get a piece of paper:ugh: :ugh: :ugh: :ugh: :mad: :ouch: :*
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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 :ok: :ok: |
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 |
...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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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: |
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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