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Power Off - 1-G vs. "Controllable" stall

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BJC

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Descent, or deceleration. I belieev that a deceleration rate is suggested somewhere.
Using deceleration mkaes it a dynamic test, which isn't the easiest the measure speed in, especially if you are also trying to fly a stalling aircraft.
IIRC, Part 23 suggests (requires?) a deceleration rate of no greater than 1 knot per second when determining stall speed.


BJC
 

pictsidhe

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Why not do a "stall turn" then? (Hammerhead)The speed will be zero.
Why do you care when AC103-7 is available to use?
Or why not follow what EAA recommends (and the popular LE) and simply disregard the speed rules somewhat. (like 90% of Americans on the highways). The FAA doesn't want to enforce the speed rules. Only the 5 gallon and one seat rule.
Lets say that 'Harry halfwit' decides to build his own 103
A bit of rummaging and he finds that the LE has a CLmax of 3.00 Woohoo, no need for an silly oversized wing on a 103, then!
He then sees that the FAA doesn't care if a 103 is a few kts over. Heh, he'll be fine at 30kts.
Harrys also 300lb, since the FAA isn't too worried about weight either, 50lb of extra material to carry his bulk seems like a good idea.

His 'almost legal' 103 is actually going to stall at over 50kts. I have a suspicion that if some FAA suit who's having bad day sees a '103' flaring at twice the speed that it should be, Harry may soon be having a bad day, too. If there are too many Harrys, the rest of us will be very upset too.

Guys, please don't extract the urine, I like having 103.

I'd suggest keeping actual stall speed under 30kts. 10% extra because you aren't 170lb, 10% because you want to cut wing size...
 

BBerson

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On page 12 of AC103-7 it says the pilots weight will be considered 170 pounds for the purpose of stall speed calculations.
 

REVAN

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A power off stall speed will be descending. It isn't the lift alone acting on the airframe. The drag is also relevant, and ultralights usually have a lot of it. A round parachute has no lift, only drag, yet it will still bring you down at speeds below 25 knots. High drag and a steep glide slope makes this increasingly important to account for its contribution to the power off stall speed.
 

Dana

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But if you put it in a deep stall like a Kasperwing, the forward speed could be nil, with greater vertical rate. So is the legal stall speed a measure of horizontal rate or vertical rate or combination of both?
Depends on whether the plane would be reusable after hitting the ground in that condition...

But seriously, that's post stall behavior. The defined stall point (silly humans, always want to draw lines) occurred at Clmax.
 

BBerson

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No requirement to fly it to the ground in a stall. Disregarding my deep stall comment, gliding an unpowered UL at CLmax might be a 45° glide angle (as Revan mentioned because of high drag). So a GPS would measure horizontal ground speed and get a significantly lower reading than actual airspeed. And even if actual airspeed was measured in the 45° glide it might be lower than the lift formula because of excess drag component. As Revan said, a round parachute has no Cl but still has a low speed unpowered.
 

12notes

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No requirement to fly it to the ground in a stall. Disregarding my deep stall comment, gliding an unpowered UL at CLmax might be a 45° glide angle (as Revan mentioned because of high drag). So a GPS would measure horizontal ground speed and get a significantly lower reading than actual airspeed. And even if actual airspeed was measured in the 45° glide it might be lower than the lift formula because of excess drag component. As Revan said, a round parachute has no Cl but still has a low speed unpowered.
Would putting the pilot on a swiveling angle of attack type vane give an accurate airspeed at steep angles? Something like this:
 

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BBerson

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Putting the pitot (not pilot :eek: ) on a swivel would help get closer to calibrated airspeed. Also needs to be out two feet in front of the wing. I think the FAA offered AC103-7 to avoid all this complication with a flight test. It also gives heavy pilots a free pass.
 
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Aesquire

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Impact type, pitot tube, airspeed sensors are typically wildly inaccurate at high AOA.

We found with hang gliders that a trailing bomb type airspeed indicator on a 20 foot line was needed to get really accurate information. The flow field extends further than you'd expect.

You really have to creep up on a stall to get an accurate, not accelerated number. Change, stabilize, change, stabilize, repeat.

It's also possible to fly with the center section of a wing stalled, but the outer sections still lifting. ( twist,cuffs, etc. )

But my opinion is some folk just reverse engineer stall numbers. You do the math & make up numbers until the equation looks not too fake and call it a day.
 

pictsidhe

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No requirement to fly it to the ground in a stall. Disregarding my deep stall comment, gliding an unpowered UL at CLmax might be a 45° glide angle (as Revan mentioned because of high drag). So a GPS would measure horizontal ground speed and get a significantly lower reading than actual airspeed. And even if actual airspeed was measured in the 45° glide it might be lower than the lift formula because of excess drag component. As Revan said, a round parachute has no Cl but still has a low speed unpowered.
Didn't consider that most people would use a commercial GPS rather than build their own datalogger that will GPS absolute velocity...
Hmmm, maybe this would be good product. Hardware parts would be perhaps $100.
 

BJC

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Didn't consider that most people would use a commercial GPS rather than build their own datalogger that will GPS absolute velocity...
Hmmm, maybe this would be good product. Hardware parts would be perhaps $100.
Not velocity, but my AFS EFIS shows where the airplane is going with an indication on the ground, horizon, or sky.


BJC
 

autopilot

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V = √( 2 m g / ( ρ S Clmax ) Where mg = MTOW, p=air preasure, S=wing area.
[/QUOTE]
It is my understanding that CLmax in this equation is given at a specified AoA (no high-lift devices). Therefore (and I'm happy to be wrong as I am trying to work through this issue now) this would indicate that the ascertained Vso is at that AoA., and that is (or should be) still controllable or at the point of no control (I understand this is based on a continuous wing). I would further postulate that it doesn't matter weather or not you are accelerating, de-accel' or power of (not gliding as this would entail vert mntm) but your vert momentum is 0 and your fwd V is the (theoretical) V arrived at using this equation.
However, when I do the math I do get a very low Vs0 then I refer to the above. It is not V a level or near level AoA (or cruise attitude). I would expect that for a given aircraft, this cannot be ascertained until testing, As in when you do a 'home-build' and then must test and fill-in the blanks.
For my aircraft given the characteristics below I get a figure of 8.68 m/s (31 km/h, ?m/h). But if I put the figures into Vcalc I get a very different figure (over 3 times). If I read the site very carefully it may be giving me the Take-of speed, but it isn't (100%) clear and doesn't say by what factor it is being multiplied by or why.
MTOW 1200kg , WA 15 sq mtr, Clmax 1.67 (NACA 23012), p 1.27, (Ar 6 )
Pls keep in mind I'm a little slow and I need any response to be understood by a 6th grader.
 

Dana

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ρ is air density, not pressure. Weight needs to be in newtons, not kg. The speed calculated is using the Clmax in whatever configuration you're looking at, with or without any high lift devices.
 

12notes

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ρ is air density, not pressure. Weight needs to be in newtons, not kg. The speed calculated is using the Clmax in whatever configuration you're looking at, with or without any high lift devices.
You are correct about rho being air density, but since the equation above has the term m for mass and not W for weight, you need to use kg for the mass, as that is what the equation calls for. The term immediately after mass is g for the acceleration of gravity, which, when multiplied together, gives weight in Newtons (kgm/s^2).
 

radfordc

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His 'almost legal' 103 is actually going to stall at over 50kts. I have a suspicion that if some FAA suit who's having bad day sees a '103' flaring at twice the speed that it should be, Harry may soon be having a bad day, too. If there are too many Harrys, the rest of us will be very upset too.
I'm always amazed when so many people have that "suspicion" without any basis in fact to support it. Has there ever been an instance where an inspector questioned the speed of an ultralight (either stall or top speed)? I have never heard of a verified case. In fact every bit of anecdotal evidence I've seen points to the FAA not giving a fig about ULs in general. In fact, if a FAA guy wanted to bust a UL around here he would have to search diligently to even find one.

Second though... Suppose you saw a UL flairing at what you consider to be too high a speed (say 50 mph for instance), is that prima facia evidence of an illegal UL? Could there be valid reasons for the speed....hot/high DA, heavy pilot, etc.
 

BJC

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The whole reason for Part 103 was that the FAA wanted nothing to do with ultralights .... or near ultralights.


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autopilot

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OK, yes, I did mean air density (my bad). The book I'm reading from (Airplane Design Manual Frederick Teichmann 1939) clearly says Gross Weight. I accept this may be wrong as his calculation for Aspect Ratio has the WA divided by the span squared?

So, if I use Nm I get the answer I want BUT is this the correct answer? I would like some 3rd party checking here.
I am referencing NACA 20312, the table I have says Clmax 1.67 @ 16 degrees
Wing area 15sq mtr
p 1.275
MTOW 1200kg

Further, if I multiply my answer by a factor of x1.4 I get close to the answer in vcalc.com (airplane speed-lift, which I have decided is Take-Of speed). Is x1.4 the correct factor?
Can someone give me a Cl @ 2 degrees for the airfoil above please, working this out is way beyond me but I may be able to work through the finished equation.
V = √( 2 m g / ( ρ S Clmax ) Where mg = MTOW, p=air preasure, S=wing area.
............. It is not V a level or near level AoA (or cruise attitude). I would expect that for a given aircraft, this cannot be ascertained until testing, As in when you do a 'home-build' and then must test and fill-in the blanks.

The above should read .......... Is the published Vs0 not measured at level or near level AoA..........?

(I cannot access Aifoiltools.com from China)
 

BJC

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Can someone give me a Cl @ 2 degrees for the airfoil above please, working this out is way beyond me but I may be able to work through the finished equation.
It will be around 0.25, depending on Reynolds number.


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BJC

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The above should read .......... Is the published Vs0 not measured at level or near level AoA..........?
That terminology can be confusing. AoA is the angle between the wing chord and the relative wind, and is not defined by the word “level”.

The FAA definition of Vso is:
VS0Stall speed or minimum flight speed in landing configuration.[7][8][9]

That is at maximum unaccelerated sustained Cl, which also is at maximum unaccelerated AoA.


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Dana

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OK, yes, I did mean air density (my bad). The book I'm reading from (Airplane Design Manual Frederick Teichmann 1939) clearly says Gross Weight. I accept this may be wrong as his calculation for Aspect Ratio has the WA divided by the span squared?
Aspect ratio is span squared divided by wing area. For a rectangular wing this works out to span divided by chord.

So, if I use Nm I get the answer I want BUT is this the correct answer? I would like some 3rd party checking here.
I am referencing NACA 20312, the table I have says Clmax 1.67 @ 16 degrees
Wing area 15sq mtr
p 1.275
MTOW 1200kg

Further, if I multiply my answer by a factor of x1.4 I get close to the answer in vcalc.com (airplane speed-lift, which I have decided is Take-Of speed). Is x1.4 the correct factor?
Can someone give me a Cl @ 2 degrees for the airfoil above please, working this out is way beyond me but I may be able to work through the finished equation.
I've never heard of vcalc before, but the person who put the aerodynamic calculations up explains it as "The algorithm uses a wind velocity starting with 0.1 m/s and increases the value until the lifting force exceeds the weight of the aircraft. The formula returns that velocity." Huh? You don't use an "algorithm" to calculate the stall speed of an aircraft, you plug the numbers into a single equation and get a single result. He also says, "(W) This is the weight or mass of the aircraft". Weight and mass are not the same thing! That makes anything else there suspect.

But anyway, as stated previously:

Vs = √( 2 W / ( ρ S Clmax )

Using your numbers, and W = 11768 since 1200kg = 11768N, Vs is 27.2 m/s (98 kph, 60 mph).

That is hardly an ultralight.
 
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