Easy way to calculate stall speed?
Thread Starter
Join Date: Jan 2006
Location: EU
Age: 43
Posts: 364
Likes: 0
Received 0 Likes
on
0 Posts
Easy way to calculate stall speed?
Does anyone have an easy way to calculate stall speed when mass increases...
I'd also like to know how to calculate the increased stall speed relative to any angle of bank...?
And finaly how does the stall speed change relative to load factor?
anyone?
I'd also like to know how to calculate the increased stall speed relative to any angle of bank...?
And finaly how does the stall speed change relative to load factor?
anyone?
Join Date: Oct 2004
Location: Body
Posts: 130
Likes: 0
Received 0 Likes
on
0 Posts
If memory serves the rules of thumb are:
1. Stall speed increase for higher weight = old speed + 1/2 % increase in weight e.g. wt inc. 20% stall speed inc. 10% therefore if old speed is 100 kts new speed is 110 kts.
2. Stall speed inc. in turn = old speed x square root of load factor e.g. in 60 degree AOB turn load factor is 2, root of 2 = 1.414, if old speed is 100 kts new speed is 141.4 kts.
3. Stall speed increases if load factor increases, decreases if load factor decreases.
1. Stall speed increase for higher weight = old speed + 1/2 % increase in weight e.g. wt inc. 20% stall speed inc. 10% therefore if old speed is 100 kts new speed is 110 kts.
2. Stall speed inc. in turn = old speed x square root of load factor e.g. in 60 degree AOB turn load factor is 2, root of 2 = 1.414, if old speed is 100 kts new speed is 141.4 kts.
3. Stall speed increases if load factor increases, decreases if load factor decreases.
Join Date: Sep 2002
Location: La Belle Province
Posts: 2,179
Likes: 0
Received 0 Likes
on
0 Posts
Those rules of thumb are of course correct, but only as rules of thumb. There are other factors sometimes at work (like Mach number effects on the maximum lift coefficient available, or wing twist under load) which can break the simple rules. As long as you're not making huge corrections, they should be fine.
Join Date: Dec 2005
Location: SE England
Age: 50
Posts: 120
Likes: 0
Received 0 Likes
on
0 Posts
just to add on aswell, in terms of AoB, load factor (as memory serves) increases as the secant of the bank angle, ie. 1/cos(AoB), hence for 60deg its 1/cos60=1/0.5=2g, for 70deg its 1/cos70=1/.342=2.92g etc etc.
Join Date: Dec 2001
Location: England
Posts: 1,389
Likes: 0
Received 0 Likes
on
0 Posts
Lift is roughly proportional to the square of the speed. So to get double the lift (if the weight doubled) you would need to go root 2 (or 1.414) times faster. The equation is:
http://en.wikipedia.org/wiki/Lift_%28force%29
where L = Lift
V = velocity
and for the rest see the wikipedia page.
PS: I can't see any reason why this doesn't hold say 1kt above stalling speed.
http://en.wikipedia.org/wiki/Lift_%28force%29
where L = Lift
V = velocity
and for the rest see the wikipedia page.
PS: I can't see any reason why this doesn't hold say 1kt above stalling speed.
Thread Starter
Join Date: Jan 2006
Location: EU
Age: 43
Posts: 364
Likes: 0
Received 0 Likes
on
0 Posts
I've got a problem with my calculations... I've tried using the formulas provided above in this thread but I cant get it to work...
here's an example:
Vs should increase by 15% due to the bank but it should also decrease by 18% due to the weight loss from the original weight...???
What am I doing wrong?
here's an example:
Vs should increase by 15% due to the bank but it should also decrease by 18% due to the weight loss from the original weight...???
What am I doing wrong?
Join Date: Dec 2005
Location: SE England
Age: 50
Posts: 120
Likes: 0
Received 0 Likes
on
0 Posts
Almost impossible to calculate exactly, you would need to know the increase in wing area with flaps down in my opinion. Assuming by some flook that the wing area stays the same, I make it 156.6kts...
knocking out the constants in the lift equation and dividing one by the other, I get (Vs)^2/(Vsf)^2=0.32. So Vsf (wings level stall speed with flaps as in situation here and weight 6400kg)=135.7kts
Increase load factor for turn increases stall speed by 15%, therefore stall speed is 156.6kts.
Allowing for the fact that we are not given wing area so would expect a slight error, I'd go for 146 as the nearest. I think your 18% is a bit off - note that CLmax doubles.
knocking out the constants in the lift equation and dividing one by the other, I get (Vs)^2/(Vsf)^2=0.32. So Vsf (wings level stall speed with flaps as in situation here and weight 6400kg)=135.7kts
Increase load factor for turn increases stall speed by 15%, therefore stall speed is 156.6kts.
Allowing for the fact that we are not given wing area so would expect a slight error, I'd go for 146 as the nearest. I think your 18% is a bit off - note that CLmax doubles.
Join Date: Sep 2002
Location: La Belle Province
Posts: 2,179
Likes: 0
Received 0 Likes
on
0 Posts
REFERENCE wing area does not (should not!) change with flap deployment - or anything else. It's purely a convenience ofr the purpose of non-dimensionalising forces.
So, using the equation for stall speed:
Vs = sqrt[ (2*nz*mass) / (rho * S * CLmax)]
for the first case (m=10000, nz=1, Vs=240) we get:
240 = sqrt[ (2*1*10000)/(rho*S*CLmax)
which means that
rho*S*CLmax = (2*1*10000)/240*240 = 0.3472
Now, for the second cased we're told that CLmax has doubled in the landing config, so now we'll use rho*S*CLmax=0.6944
We also have mass=6400, and nz for a 30 degree bank is 1.155'g'
Therefore, we have:
Vs=sqrt [ (2*1.155*6400)/0.6944] = sqrt [21290] = 145.9
So the answer they want is 146
EDIT: I think the problem the last two posters have is that they are increasing stall speed by 15% for load factor.
In fact, the load factor for a 30 deg bank is 1.155'g' (1/cos30). That means the load factor is 15.5% above 1.0, so the stall speed increase for load factor is half that amount - 7.75%. If Airbus38 increases his 135.5 number by 7.75% he'll get 145.7. Which is, allowing for various roundings, the same as my answer.
So, using the equation for stall speed:
Vs = sqrt[ (2*nz*mass) / (rho * S * CLmax)]
for the first case (m=10000, nz=1, Vs=240) we get:
240 = sqrt[ (2*1*10000)/(rho*S*CLmax)
which means that
rho*S*CLmax = (2*1*10000)/240*240 = 0.3472
Now, for the second cased we're told that CLmax has doubled in the landing config, so now we'll use rho*S*CLmax=0.6944
We also have mass=6400, and nz for a 30 degree bank is 1.155'g'
Therefore, we have:
Vs=sqrt [ (2*1.155*6400)/0.6944] = sqrt [21290] = 145.9
So the answer they want is 146
EDIT: I think the problem the last two posters have is that they are increasing stall speed by 15% for load factor.
In fact, the load factor for a 30 deg bank is 1.155'g' (1/cos30). That means the load factor is 15.5% above 1.0, so the stall speed increase for load factor is half that amount - 7.75%. If Airbus38 increases his 135.5 number by 7.75% he'll get 145.7. Which is, allowing for various roundings, the same as my answer.
Join Date: Sep 2002
Location: La Belle Province
Posts: 2,179
Likes: 0
Received 0 Likes
on
0 Posts
I wouldn't worry too much. The first pass I did at the full calc gave a stall speed of 50knots, because I had the calculation of rho*S*CLmax "upside down" (i.e. I mucked up swapping the terms around) and it was only the "there's no $#%^ way its that low" factor that spared my blushes. 
