Jfkaua From United States of America, joined Aug 2004, 1000 posts, RR: 3 Posted (11 years 4 months 1 week 4 days 7 hours ago) and read 2720 times:

Hey I was just wondering after reading some threads, how strongly does air push against flight surfaces. For example if a 777 is going 250 and uses spoilers to slow down.. How much force is pushing on those spoilers.. I always imagined how much force they must be under. You can tell from inside the cabin because when ever the spoilers are used to slow down you can immediately feel the power of the air hitting them from the cabin shaking...

2H4 From United States of America, joined Oct 2004, 8957 posts, RR: 56
Reply 1, posted (11 years 4 months 1 week 4 days 7 hours ago) and read 2695 times:

Couldn't find any info on spoilers, but the maximum output force of an EMB-135 horizontal stabilizer actuator is 5600 lbs under extension and 4750 lbs under retraction. I also found out that the wing flap actuator for the P-3C is rated to 3600 lbs.

Keep in mind this isn't necessarily an indication of how hard the air is pushing on the control surfaces...the numbers I found only indicate the maximum output of the actuators. None of this actually answers your question, but I thought it was interesting nonetheless.

FredT From United Kingdom, joined Feb 2002, 2185 posts, RR: 26
Reply 2, posted (11 years 4 months 1 week 4 days ago) and read 2607 times:

The dynamic pressure is half the density of the air times the velocity of the air squared. If you use that on the component of the velocity of the air perpendicular to the control surface times the area of the control surface, it should give you a ballpark figure.

Regards,
Fred

I thought I was doing good trying to avoid those airport hotels... and look at me now.

Jfkaua From United States of America, joined Aug 2004, 1000 posts, RR: 3
Reply 3, posted (11 years 4 months 1 week 3 days 12 hours ago) and read 2558 times:

OK hold on I think I did something wrong...
I found the density of air is .088 lb/ft and I figured 250 mi/hr which converts to 366 ft/sec

So [½(.088)]·[366²]

This comes to 5894 lb/sec . Now if this was multiplied by the size of the spoiler that would be alot! Say the spoiler is 1ft x 4ft that would be 4ft² so that would come to a grand total of over 23 thousand pounds! So where did I go wrong?

Vikkyvik From United States of America, joined Jul 2003, 12147 posts, RR: 24
Reply 5, posted (11 years 4 months 1 week 3 days 10 hours ago) and read 2533 times:

One problem might be that in English units, the standard unit for mass is slugs, not pounds (there is a pound-mass, but for these calculations, you should use slugs). The reason for this is that a pound, as it is commonly used, is a unit of force, not mass (same as a Newton is force, and a kilogram is mass). The density at sea level is 0.00238 slugs/ft^3. Now I personally find English units to be an unwarranted pain in the ass, so I'm going to use metric.

Let's assume V = 100 m/s (approx. 224 mi/h)
The density at 5,000 meters (approx. 16,400 ft) is 0.736 kg/m^3

Let's assume the spoiler is sticking up into the flow at a 30 degree angle (not too unreasonable I hope), which means the flow perpendicular to the spoiler will be 1/2 the total flow speed (so V = 50 m/s for our purposes).

So:

q = 0.5*0.736*50^2 = 920 kg/(m*s^2) = 920 N/m^2

That equates to 19.2 lb/ft^2

Doing it with English units, I got 19.1 lb/ft^2

So for a 4'x1' spoiler, your total force would be 76.8 lbs.

I seriously hope this is right, and I welcome corrections - it's been awhile.

~Vik

I'm watching Jeopardy. The category is worst Madonna songs. "This one from 1987 is terrible".

if you're suitably enthusiastic. Wind tunnel tests on a scale model.

Abstract....

The influence of Reynolds number on the performance of outboard spoilers and ailerons was investigated on a generic subsonic transport configuration in the National Transonic Facility over a chord Reynolds number range from 3 x 10 super 6 to 30 x 10 super 6 and a Mach number range from 0.50 to 0.94. Spoiler deflection angles of 0 degrees, 10 degrees, 15 degrees, and 20 degrees and aileron deflection angles of -10 degrees, 0 degrees, and 10 degrees were tested. Aeroelastic effects were minimized by testing at constant normalized dynamic pressure conditions over intermediate Reynolds number ranges. Results indicated that the increment in rolling moment due to spoiler deflection generally becomes more negative as the Reynolds number increases from 3 x 10 super 6 to 22 x 10 super 6 with only small changes between Reynolds numbers of 22 x 10 super 6 and 30 x 10 super 6. The change in the increment in rolling moment coefficient with Reynolds number for the aileron deflected configuration is generally small with a general trend of increasing magnitude with increasing Reynolds number.

Jfkaua From United States of America, joined Aug 2004, 1000 posts, RR: 3
Reply 7, posted (11 years 4 months 1 week 3 days 6 hours ago) and read 2496 times:

hhmm intresting only 78 pounds.. I always thought it would be so much more

Jetlagged From United Kingdom, joined Jan 2005, 2620 posts, RR: 25
Reply 8, posted (11 years 4 months 1 week 3 days 1 hour ago) and read 2474 times:

Vikkyvik, your calculations are a bit on the low side. Firstly, to compute dynamic pressure you need to square the aircraft true airspeed, not square half the velocity. Secondly 100 m/s is a bit low for 5,000 m altitude. If the IAS is 250 knots (288 mph), the true airspeed would be around 163 m/s. You then take half the panel area to approximate the force for 30 deg deflection (not half the speed).

q = 0.5 * 0.736 * (163^2) = 9777.4 N/m^2

So for this 4 sq ft (0.37 m^2) panel, the force is:

9777.4 * (0.37 * 0.5) = 1808.8 N (or 406 lbf)

I've not measured a 777 spoiler panel, but my guess is that they are much bigger than 4 ft by 1 ft (maybe 8 ft by 2 ft). Also maximum in-flight spoiler deflection is more like 45 degrees. Finally, a 777 descending at 16,400 ft needing speedbrake deployment would probably be doing 300 kts or thereabouts. All these factors mean the spoiler panel is taking quite a beating.

Taking the larger area, higher speed and larger deflection the force works out at approx 3,200 lbf (eight times as much load).

The glass isn't half empty, or half full, it's twice as big as it needs to be.

Vikkyvik From United States of America, joined Jul 2003, 12147 posts, RR: 24
Reply 9, posted (11 years 4 months 1 week 2 days 20 hours ago) and read 2448 times:

Jetlagged, thanks for that. However, in terms of IAS versus TAS: I just meant the velocity of the flow as it passed the wing; I wasn't referring to IAS. Now that I think about it, that is low for that altitude. Oh well.

The one question I do have is why do you use half the area instead of half the velocity? If you're concerned with the component of velocity that is perpendicular to the spoiler, couldn't you just set up a velocity vector diagram and use the perpendicular part? If you do use half the velocity instead of half the area with your numbers, you get half the force (203 lbf on the 4 sq ft area).

Doing it your way with my numbers, I got 153 lbf.

Thanks again - always appreciate the corrections.

~Vik

I'm watching Jeopardy. The category is worst Madonna songs. "This one from 1987 is terrible".

FredT From United Kingdom, joined Feb 2002, 2185 posts, RR: 26
Reply 10, posted (11 years 4 months 1 week 2 days 20 hours ago) and read 2451 times:

It should be the velocity component, not the pressure component. The pressure component comes in later, in that part of the force excerted goes towards retarding the aircraft while part of it counteracts lift... in excess of the lift that was already spoiled by extending the spoiler in the first place.

Regards,
Fred

I thought I was doing good trying to avoid those airport hotels... and look at me now.

Vikkyvik From United States of America, joined Jul 2003, 12147 posts, RR: 24
Reply 11, posted (11 years 4 months 1 week 2 days 13 hours ago) and read 2411 times:

FredT,

Sorry, I'm not exactly sure what you're saying. Can you phrase it in the context of our calculations in the previous few posts? Thanks.

~Vik

I'm watching Jeopardy. The category is worst Madonna songs. "This one from 1987 is terrible".

FredT From United Kingdom, joined Feb 2002, 2185 posts, RR: 26
Reply 12, posted (11 years 4 months 1 week 2 days 12 hours ago) and read 2400 times:

The assumption made, which is not at all exact, is that the velocity component of the onrushing air which is perpendicular to the surface of the spoiler is reduced to zero while the parallell component remains untouched. This is far from correct, so you have to remember we are talking about a ballpark figure.

If the velocity is V and the angle of the spoiler is a (measured from the velocity of the aircraft, the perpendicular velocity will be V*sin(a).

The dynamic pressure q = 1/2*rho*(V*sin(a))^2

The force on the spoiler, perpendicular to the surface: F = q*A

The retarding component of the force: Fr = F*sin(a)

I reiterate that this is far from scientific but will give you an idea about the magnitude of the forces involved.

Assuming M.85@FL330 ISA, V will be roughly 250 m/s and rho 0.4 kg/m^3. Assume a to be 70 degrees.

q = 11e3 N/m^2.

Fr=A*q*sin(a)

With a weight of, say, 100 metric tons, this means you get

100/11*sin(70) m/s of deceleration per square meter of spoiler, a little less than 1G.

Hint: Look into the change of momentum of the air deflected. That would probably be a more correct approach!

Regards,
Fred

P.S. Way too late to be doing calculus... so be kind if I messed something up.

I thought I was doing good trying to avoid those airport hotels... and look at me now.

Jetlagged From United Kingdom, joined Jan 2005, 2620 posts, RR: 25
Reply 13, posted (11 years 4 months 6 days 4 hours ago) and read 2280 times:

My calculation had an error, in that I was not allowing for the angle of the spoiler, just it's projected area. This overestimated the force on the panel by a factor of two. FredT and Vikkyvik's approach of resolving velocity first is equally valid.

FredT's estimate of 1g deceleration sounds too high though. That's about 19 knots TAS per second. I would guess it would be roughly a tenth of that.

Airliner speedbrakes are not that effective, although it varies. They work mainly be reducing lift (as well as adding some drag). Angle of attack has to be increased to restore lift, thus creating additional drag. Users of Microsoft Flight Simulator's default airliners may be surprised how little effect speedbrakes have compared to the rapid speed loss experienced in the game.

Interestingly the force on the spoiler panel has a limit, the hydraulic pressure in the actuator. If the airload gets too high, it will force the spoiler angle to decrease, reducing the airload until it balances the actuator force again. This is known as blowdown. Therefore the maximum load on the actuator is the hydraulic pressure times the piston area. It's easy when you think about it.

The glass isn't half empty, or half full, it's twice as big as it needs to be.

2H4 From United States of America, joined Oct 2004, 8957 posts, RR: 56
Reply 14, posted (11 years 4 months 6 days 4 hours ago) and read 2277 times:

Interestingly the force on the spoiler panel has a limit, the hydraulic pressure in the actuator.

Jetlagged...is it possible that on certain aircraft the spoiler panel limitation is structural, rather than hydraulic? In other words, such that the hardware would deform or damage before blowdown occurs?

I don't mean to doubt you...I honestly don't know the answer.