Clearly doin' something stooopid..
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Clearly doin' something stooopid..
Morning fellow Ppruners
Apologies for the hassle but was wondering if someone could help me with this question... just can't seem to get to the right answer... its a BGS question bank classic so I am sure some of you will have seen it before...
A 50 ton twin engine aeroplane performs a straight, steady, wings level climb. If the lift/drag ratio is 12 and the thrust is 60 000N per engine, the climb gradient (@) is: (assume g = 10m/s2)?
50 tonnes = 50,000kg which x 10 gives 500,000N of weight
2 x engines so 120,000N of thrust
So I assume you now use sin@ = (T-D)/W or Sin@ =(T/w)-(Cd/Cl)?
but can't seem to get to one of the answers offered.... exam on Monday gulp
Apologies for the hassle but was wondering if someone could help me with this question... just can't seem to get to the right answer... its a BGS question bank classic so I am sure some of you will have seen it before...
A 50 ton twin engine aeroplane performs a straight, steady, wings level climb. If the lift/drag ratio is 12 and the thrust is 60 000N per engine, the climb gradient (@) is: (assume g = 10m/s2)?
50 tonnes = 50,000kg which x 10 gives 500,000N of weight
2 x engines so 120,000N of thrust
So I assume you now use sin@ = (T-D)/W or Sin@ =(T/w)-(Cd/Cl)?
but can't seem to get to one of the answers offered.... exam on Monday gulp
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Nope... I don't think it was but I get 9-10 degrees as well ... must be doing something wrong... the explanation under the "info" tab was a bit strange in that is said something along the lines of the drag = 500,000/12= 41,667... then "you have the formula so do the rest!".... as the aircraft is in a climb the weight does not = lift so surely this is wrong?
I'm sure that sin@ = (T/W) - (Cd/Cl) is correct; in which case the answer would seem to be inv sin (120/(50 x 9.81)) - (1/12)
i.e. inv sin (0.2446 - 0.0833) = inv sin 0.1613 = 9.28 deg.....
What were their answers?
i.e. inv sin (0.2446 - 0.0833) = inv sin 0.1613 = 9.28 deg.....
What were their answers?
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Beagle that is exactly the same calc I did... think one of the answers was 15.7 but none were the 9-10 degrees we both arrive at.... think another one of the answers was 3... 
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I notice you're using sin and inv.sin in these calculations. I don't know much about the calculation you're doing, but a simple thing to check is that you've got your calculator in the correct mode - make sure it's set to Degrees, not Radians or Grads - this could be giving you an error.
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Its asking for a climb gradient... not climb angle.... climb gradient is sin @ which = 0.15666 so 15.7% which is one of the answers... thanks for your help.... lesson learnt RTFQ... Good luck to everyone else sitting exams on Monday
Last edited by Grass strip basher; 1st Apr 2006 at 11:48.
Ah - of course. The Eurocratic JAA which loves to quantify aircraft climb values in the same way they measure road gradients...
Why is 'g' expressed as 10 rather than 9.81 m/s/s?
Why is 'g' expressed as 10 rather than 9.81 m/s/s?
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Ah - of course. The Eurocratic JAA which loves to quantify aircraft climb values in the same way they measure road gradients...
For example: RoD = Gradient * GS as opposed to Sin(Deg) * GS !!
Bandit Man - because 'g' varies with latitude and location - hence to use the value to the degree of accuracy you quote is pointless. The value you quote is only 0.034% less than the average value of 9.81 m/s/s, whereas using 10 m/s/s incurs an error of around 30 times greater than that.
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In the question you have to assume that Lift does indeed equal weight. Thus you get (120000-41667)/500000 = 0.15666
Now using small angle approximation sin a = tan a = opp/adj
Thus you have a ratio of 15.6/100 or 15.7%.
Tan-1 and Sin-1 don't come into it.
I expect to see one in the exam on Monday, but with different numbers
Now using small angle approximation sin a = tan a = opp/adj
Thus you have a ratio of 15.6/100 or 15.7%.
Tan-1 and Sin-1 don't come into it.
I expect to see one in the exam on Monday, but with different numbers
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Just thought I'd remark that by stating the aeroplane is in a "straight, steady, wings level climb", the question presents all the information one needs to make the assumption that lift does in fact equal weight. The straight and level stuff indicates that the aircraft is being subjected to no acceleration forces. If lift were greater than weight, it would accelerate upwards, i.e. would climb at an ever-increasing rate.
Good luck tomorrow, Grass strip basher!
Good luck tomorrow, Grass strip basher!
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Going slightly off thread here really but using g=9.81 is also pointless since 2 of the numbers you are given are only accurate to one significant figure and the other only accurate to 2 sig. figs. 10 is perfectly reasonable and banditman's point is a good one.
