Using the clear weather preset (i.e., not real time weather) to do some flight

model work, I have another question regarding the atmospheric modeling. The

screenshot below shows the situation shortly after taking off at SKBO (Bogota)

with standard sea level pressure, but at a temperature of 42 deg C (using the

AGL altitude calculation selection). The temperature at the airplane’s

position in the screenshot is 41.4 deg C or 314.55 K. This would mean θ

(temperature ratio) = 314.55/288.15 = 1.0916. The ambient pressure at the

airplane’s position is 734.665 hpa. With sea level pressure being 1013.25 hpa,

this means δ (pressure ratio) is 734.665/1013.25 = 0.72504. Since δ=σθ, σ

(density ratio)=δ/θ, σ = 0.72504/1.0916 = 0.6642. **Since σ is defined as
ρ/ρ0, ρ = 0.6642 * 0.00237689 = 0.001579 slugs/ft3. However, MSFS says the
ambient density, ** Where did this value come

**ρ,**is 0.001758 slugs/ft3.

from? A further check on the density calculation can be made using the ideal

gas law directly, p = ρRT, or ρ = p/RT. In Imperial units, R = 1716.5619 foot-

pounds/slug.oR, p = 734.652 hpa = 1,534.354 lb/ft2, and T = 41.44 C = 566.26 R

for the condition above. So ρ = 1534.354/(1716.5619*566.26) = 0.001579

slugs/ft3, the same value calculated above. Since dynamic pressure is normally

calculated from 1/2ρVtrue2, this can have a large impact on the calculated

lift and drag forces in a testing environment where preset/customized weather

settings are used instead of real weather.