Hi all,
Im Having trouble in Density heights figuring out when to + or - the result from the end of the equation onto the pressure height. Im currently Mid CPL and have used a method since RPL for this however its not longer working. Particular in Bobs performance book question 1 on page 32
An Aerodrome has an Elevation of 1250ft. If the QNH is 1003hPa and the ambient air at the field is 7Deg what is the density height of the field?
To sum it up
PH=1550
Oat=7
IsaT= 12
IsaD= -5
(-5 x 120) = -600 ```` now here i would have thought it would be: -600 - 1550 = -2150
however that is not the correct answer can anyone please tell me why this is -600 + 1550 = 950
and in future how to know when to + or -

First point is that a sketch provides useful visual reasoning input.

Second - you are almost there. Take a moment to consider the temperature deviation when you are running this sort of question -

(a) if the temperature deviation from ISA is zero, then DH = PH

(b) if the temperature deviation from ISA is positive, then the air is hotter than ISA, that is to say, less dense, and that reduced density will occur higher in the air column. So DH > PH

(c) if the temperature deviation from ISA is negative, then the air is colder than ISA, that is to say, more dense, and that increased density will occur lower in the air column. So DH < PH

In the example you have provided, the deviation is -5, as you have noted. So, (c) applies - the deviation is negative (ie minus), so the air is colder than ISA and more dense than ISA. This more dense air occurs lower in the air column, so DH < PH and you end up with 1550-600 = 950 for your answer.

Often, thinking it through is more productive than applying rules of thumb which you have to recall correctly.

ah thank you, yes this was what i had previously been doing however I must have mixed myself up somewhere. It definitely helps viewing it all written out like that. Thanks for your help