Ok... I’m completely lost.... can someone help me out with this one.... and I’m sure I’m missing the bleeding obvious.... but I’ve been given
1760 AMSL for aerodromes elevation
QNH 1005
Temp 25 c
What is the pressure height....
The answer is 2000 to enter the landing weight chart )

As for the other question -

(a) draw a picture - if you do this ALWAYS you will help yourself greatly.

(b) from the picture, the 1013 surface is below the MSL QNH surface.

How about you run the question and then post a scan of your answer for comment to get maximum value from the forum ? If you don't get 2000 ft for Hp, try again with the picture and keep doing it until it all falls into place for you.

Ummmm wowzers... I’m the fool.... I realised just what I was doing wrong....
So my equation was PH = ele + (qnh - 1013) x 30..... ) hmmmm...
I gave it another shot with the correct one this time.....
PH = ele + (1013-qnh) x30... ahhhhh... think it’s time to go to bed and get some rest ......
John.... you have the patience of a saint my friend.... I bet you were just shaking your head....

I bet you were just shaking your head

Not at all, everyone takes a while to get their head around this flying stuff.

However, the comment about drawing a picture is really important. It is my conviction that a great many people try to use formulae without really understanding the derivation, limitations, etc. A picture usually makes everything a whole lot easier to follow (and not make simple formula mistakes).

I just got stuck on this same topic.

RPL/PPL Volume 1 - pg 188
"The 1013 pressure level must be (1030 - 1013) x 30 feet above sea level".

This algorithm has the QNH and the ISA back to front.

On pg 190 it lists it as "Pressure Height = Elevation + (1013 - QNH) x 30"

Had me stumped for a bit...

This algorithm has the QNH and the ISA back to front.

Not quite - the equation in the text is talking directly about the previous diagram, rather than presenting the generalised equation shown on p190. One needs to keep apples and oranges separate in the discussions, methinks.

Keep in mind, though, that the 30 ft/hPa is a very rubbery approximation. The variation of pressure with height in ISA is an exponential, not a linear, equation. Therefore, the rate varies. 30 ft/hPa is accurate only for one altitude and will be less, at lower, or greater, at higher, levels. For light aircraft, operating at lower levels, 30 ft/hPa is a reasonable approximation.

For the lower levels in ISA, the pressure rate varies from around 27 ft/hPa at sea level to around 37 ft/hPa at 10000 ft.

I just reworked out the equation in both directions and the difference is actually not in the question, its that the algorithm was reworked to produce the same result.

a)
1013 - 1030 = -17
-17 x 30 = -510
-510 + 1500 = 990ft

OR

b)
1030 - 1013 = 17
17 x 30 = 510
1500 - 510 = 990ft

The green guy (me) mixing the equations as i tried to memorise how it worked, produced the error.

I would be inclined to argue though, that due to the pressure height being 510ft BELOW the elevation, that -510ft would be the more accurate answer making example 'a' the correct example of the equation.

Hence the general equation on p190. If a general equation doesn't work for all examples then it would be flawed and of little practical use.

What I don't comprehend is why folks waste their effort with an equation when a quick sketch takes much the same time and reduces the likelihood of error in working the solution.

I agree that 'drawing a picture' is far better than using a formula. I fact, I recommend that at the BAK/PPL level, the picture is definitely the best option. I introduced the formula simply as an alternative that may be some help at CPL level when the principle is more likely to be well understood.

Because CPL students are more likely to be under time constraints, I encourage CPL students to use the formula because it allows the answer to be obtained without even picking up a pencil.

However, let me make it clear that it's not worth getting your 'nickers in a knot' over it. As John points out, both methods are very much approximations. Since CASA introduced the 'type-in' answers, many students are stressing out unnecessarily about unrealistic degrees of accuracy. If you're finding a pressure height it's probably to allow you to use such things as take-off or landing charts. Those chart are designed with very generous margins for safety.

How far do we take it? The elevation is the elevation of a particular point on the aerodrome and unless the field is perfectly level, that's almost certainly wrong in the first place.

We can only hope that the examiner applies a reasonable margin to the marking of the 'type-in' type question.

Equations are fine providing they are used correctly and appropriately and with due consideration of any inherent limitations. Unfortunately, my observation has been that students often get things wrong with using equations.

For those who have equation use under control, go for it.

However, in my view, drawing a sketch helps out the other folks considerably.