G'day jhurditch,
I'm no engineer but I think I can answer this one.
Let's say:
- the Q400 in question weighs the same at sea level as at FL150
- lift is equal to weight in level, unaccelerated flight
Therefore we can say that the lift must be the same at both levels.
The lift equation is:
L = CL˝ρV2S
Where:
CL = the coefficient of lift of the wing. This is determined by the designer.
ρ = (rho) the density of the air. We can find this in ICAO Doc 7488. 1.225 kg/m3 at sea level.
V = the true airspeed. According to my Jepp CR-3 (Same as an E6-B) 102 KIAS is equal to 128 KTAS.
S = the surface area of the wing. This is set by the designer (64.1 m3 for the Q400)
Since CL and S are fixed we can ignore them and just say that Lift is proportional to ˝ρV2 which happens to have it's own name: q, or "dynamic pressure". q is the total air pressure at a point minus the static pressure or, in other words: indicated airspeed!
So this is all a roundabout way of saying that Lift is proportional to IAS, therefore the wing at the stalling angle (or any other angle) will produce the same lift at the same IAS. Therefore the stall IAS must be the same at all altitudes!
Am I also correct to say that IAS will reduce with altitude?
The 128 knot figure you calculated would be the TAS which, in nil-wind conditions, would equal the groundspeed. So the Q400 was indicating 102 knots but it was actually moving over the ground at 128 knots. So the answer to your question is: if the pilot maintains the same TAS as s/he climbs then yes, the IAS will decrease as the air density decreases. However, the beauty of it is that if you maintain the
same IAS up to higher altitudes you will get a
higher TAS! The decreasing air density also affects the engine's ability to maintain the same IAS, but that's another issue...
Like I said: I'm no engineer. This is just me figurin'
