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# Density altitude/pressure altitude help

Hi,

I am currently doing some revision on aircraft performance and want to confirm my understanding on calculating density altitude.

You first need to work out the pressure altitude. So for example the airfield is 399 feet amsl.

PA =airfield elevation +(30x(1013-ambient pressure). I was wanting to clarify is ambient pressure the QNH pressure setting.

PA = 399 +(30x(1013-1012))

=>399+30

PA=429 feet

So now I can work out the density altitude.

DA=PA + (120 x isa dev)

Is the term isa deviation the difference in the actual temperature against the ISA standard lapse rate of 2 degrees C?

Is the lapse rate calculated from the QNH height or the known elevation. So the value of 399 feet amsl would be used. So if for 1000 feet its 2 degrees for 499/500 1 degree drop from standard 15 degrees C could be used.

So would the isa deviation be the 14 degrees - the actual temperature / forecasted temperature. So say the outside air temperature at the airfield is 10. The difference would be 4. The DA would then be:

DA= 429 +(120x4)

DA=429 +(480)

DA = 909 feet.

Thanks for your help. I am just really wanting to check my understanding of the above.

I am currently doing some revision on aircraft performance and want to confirm my understanding on calculating density altitude.

You first need to work out the pressure altitude. So for example the airfield is 399 feet amsl.

PA =airfield elevation +(30x(1013-ambient pressure). I was wanting to clarify is ambient pressure the QNH pressure setting.

PA = 399 +(30x(1013-1012))

=>399+30

PA=429 feet

So now I can work out the density altitude.

DA=PA + (120 x isa dev)

Is the term isa deviation the difference in the actual temperature against the ISA standard lapse rate of 2 degrees C?

Is the lapse rate calculated from the QNH height or the known elevation. So the value of 399 feet amsl would be used. So if for 1000 feet its 2 degrees for 499/500 1 degree drop from standard 15 degrees C could be used.

So would the isa deviation be the 14 degrees - the actual temperature / forecasted temperature. So say the outside air temperature at the airfield is 10. The difference would be 4. The DA would then be:

DA= 429 +(120x4)

DA=429 +(480)

DA = 909 feet.

Thanks for your help. I am just really wanting to check my understanding of the above.

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