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Founder
26th Nov 2006, 15:15
Say that you've got a normal stall speed of 100 kt at 1000 kg, if you increase the mass to 2000 kg, what will the new stall speed be?

Does anyone know how to calculate this?

fireflybob
26th Nov 2006, 15:53
Long long time since I have thought of this but sees to recall it's a square root relationship - if you double the weight, the stall speed increase by a factor of the square root of 2 = (about) 1.4 so in this case the new stall speed is 140 kts.

No doubt those more educated can explain this!

A good book, although quite old, is "Mechanics of Flight" by A.C. Kermode which explains much of this stuff!

BOAC
26th Nov 2006, 16:01
At the stall:-

Lift=weight (dare I use that expression?)
Speed = Stall speed

Lift (Weight) is proportional to Speed (Stall speed) squared

Double the weight and fireflybob has it.:)

Founder - look up the equation of lift (http://www.grc.nasa.gov/WWW/K-12/airplane/lifteq.html)?

Lift =CL(Wing 'lift coefficient') x 1/2 x density x V squared x wing area

Of course, afore I gets leapt on, all at 1g and low speed:) (and of course, we all know the answer is really 141.421356 knots :p )

.......and Genghis - that was TIC as gravity has been pretty constant in most low-speed stalls I have done:)

Genghis the Engineer
26th Nov 2006, 16:04
At first approximation, ignoring CG effects, pitch authority limitations, ASI calibrations, and assuming that the stall is measured consistently from 1g flight, using the same stall definition and from a 1 knots/second deceleration:


New stall speed = Old Stall speed * SQRT (new weight / old weight).

G



N.B. BOAC - weight is fine, so long as you assume that gravity is constant, which is a good enough assumption for most purposes.

cwatters
26th Nov 2006, 16:18
I thought I was seeing double. We just did this thread recently..

http://www.pprune.org/forums/showthread.php?t=253067