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Max crosswind for takeoff and landing a 1965 E model?


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2 hours ago, jaylw314 said:

Yikes, was there any poo? :unsure:

I'd also add that ground effect slows down winds in the last 50-100 feet, so if you run out of rudder above, you may still have enough below, but it'd still be pretty sketchy...

FWIW 15 kts is okay in a J, but I stopped there because I figured that was enough...

No poo because I was expecting it to be rough to some degree. What I did not expect was the wind tunnel effect between two sets of hangars. It was like a rolling into a horizontal wind shear. Ground friction certainly diminishes wind speed. Thank god for small mercies. Nevertheless I once took off in wet, windy conditions in a C172 where The plane was not yet ready to fly nor had sufficient traction to track runway heading. Spring steel Cessna gear chatter worse than a Mooney when side loaded. As I slid closer to the DW side of the runway, I ruddered it into the wind and lifted off about 50° Right of runway heading.

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7 hours ago, 201er said:

Depends on runway width. You don't want to mess with too strong a crosswind on a narrow runway. Once it pops off the ground, it will drift significantly sideways until you become aware of the intensity and turn into a crab. You don't want to dip a wing into the runway, so you're limited how quickly you can bank into that crab while sliding across the runway in the air.

HUH.  Is that how you were taught?

You NEED some time in a tailwheel aircraft.

I start my take off roll with full aileron into the wind.  Rudder to maintain straight down the runway.  Aileron is removed as needed, but in a strong cross wind, after I raise the nose, the down wind main lifts off, then finally the up wind one.  So as soon as I break ground, I am in a slip, that transitions to a crab.

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3 hours ago, Pinecone said:

HUH.  Is that how you were taught?

You NEED some time in a tailwheel aircraft.

I start my take off roll with full aileron into the wind.  Rudder to maintain straight down the runway.  Aileron is removed as needed, but in a strong cross wind, after I raise the nose, the down wind main lifts off, then finally the up wind one.  So as soon as I break ground, I am in a slip, that transitions to a crab.

And YOU need to keep your assumptions in your own ass. I learned in a conventional gear trainer before I’ve ever flown tricycle gear. The ground clearance in a low wing Mooney is nothing like in those high wing conventional gear trainers!

 

Sorry I don’t have the obvious markings of a taildragger pilot by going around telling everyone they need to fly tailwheel or they’re not worth a dime.

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And here I thought my own 20G30 with an 18 Kt direct crosswind was something to crow about.  My POH says 11 mph is the max demonstrated crosswind.  I guess Kerrville was not as windy as Tucumcari that day. 

I knew I was a test pilot that day on a 3800’ X 150’ runway.  My out was our local international airport with 4 runways so it was going to be a relatively safe learning experience.  Worked out good. Upwind wheel down first. Then the other side followed very quickly by the nose wheel.  This 12 years ago.  Never had to try it since.  
 

Crosswind landings in my 65 E model are pretty non eventful.

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18 minutes ago, bluehighwayflyer said:

Same here.  I think there is a relatively common misperception that “short rudder” M20Cs and Es are somehow yaw control limited when in fact this was addressed in 1962 when the M20C was first developed from the M20B and rudder travel was increased and the aileron interconnect springs were added. The long rudder was developed for the F in 1966 and I think it was only added to the Cs and Es in 1969 (I think) as a cost savings measure to standardize production across the entire model line.  Not because the C and E needed more rudder.  There were a lot of cost cutting measures implemented in the late 60s, but not many advancements.  Times were getting tough for Mooney by then.   

You've done a lot of J/C time. How does the short rudder feel by comparison? How much less authority do you actually get?

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18 hours ago, 201er said:

And YOU need to keep your assumptions in your own ass. I learned in a conventional gear trainer before I’ve ever flown tricycle gear. The ground clearance in a low wing Mooney is nothing like in those high wing conventional gear trainers!

 

Sorry I don’t have the obvious markings of a taildragger pilot by going around telling everyone they need to fly tailwheel or they’re not worth a dime.

Interesting.  Because my experience with taking off the way you stated in a tailwheel aircraft will be a bit exciting.

 

And let's see.  The tip is 13.5 feet outside the main gear.  The tip is over 3.5 feet off the ground (I will actually measure next time I am at the hangar).  So that angle where the wing tip hits is over 14 degrees.  which is a pretty hefty bank angle for still on the runway.

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6 hours ago, bluehighwayflyer said:

Same here.  I think there is a relatively common misperception that “short rudder” M20Cs and Es are somehow yaw control limited when in fact this was addressed in 1962 when the M20C was first developed from the M20B and rudder travel was increased and the aileron interconnect springs were added. The long rudder was developed for the F in 1966 and I think it was only added to the Cs and Es in 1969 (I think) as a cost savings measure to standardize production across the entire model line.  Not because the C and E needed more rudder.  There were a lot of cost cutting measures implemented in the late 60s, but not many advancements.  Times were getting tough for Mooney by then.   

Generally speaking, I think the long rudder design was easier and less expensive to produce than short rudder/tail cone. I’ve never put a lot of stock in the differences nor have I noticed the short rudder birds to lack yaw authority.

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3 hours ago, Pinecone said:

So that angle where the wing tip hits is over 14 degrees.

With 5.5° of dihedral, I would think a lot more than then 14° For wingtip contact with the gear is down. Geometry is not my strong point, perhaps someone can workout the bank angle required to drag a tip with the gear down.

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Whenever I takeoff with a stiff cross wind, I use full aileron into the wind. The downwind wing raises first, but doesn’t keep rising, as the airspeed increases the upwind wing will rise. I don’t remember it ever going more than level. As soon as the upwind wheel is airborne, I level the wings.

I would think if dragging a wing was possible, I would have heard of someone doing it, but I have never heard of it.

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5 hours ago, Shadrach said:

With 5.5° of dihedral, I would think a lot more than then 14° For wingtip contact with the gear is down. Geometry is not my strong point, perhaps someone can workout the bank angle required to drag a tip with the gear down.

Yeah, I was using a flat wing to get an idea.  So it would be even more than 15 degrees.  But the idea is, it is really hard to drag a wing tip.

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In the Kc-135 they put CFM-56 engines on it and at max weight you had 16 inches of ground clearance on the inboards before dragging an engine pod. This came out to a little more than just 4 degrees in the 3 point attitude and once the nose was rotated off the ground with the mains still on the ground the tilt limit went up to 8 degrees but ironically it now would be the outboard narcelles that would touch first. We were limited to 25 max crosswind but could be waived up to 33kts max. I got to experience a 33 kt crosswind takeoff as a typhoon was coming and we needed the airplane out of it’s path and the winds were already being reported by tower at 33. We got the approval and as we took off it took full rudder and most aileron to maintain runway centerline. Had we lost a upwind engine after V1 we would’ve not been able to stay on the runway.

in the simulator they would show us 4 degrees of tilt to see how far over that looks like and the good news is it looks much more like 6 degrees than 4 so when you think you were at 4 it was more like 2.5 to 3 degrees in reality. Boeing told us the majority of engine pod strikes did not occur at 25 degrees but at 15 degrees as less than that wasn’t enough for when you land in a crab and straighten out, the upwind wing would not rise high enough to cause the low wind  wing engine to strike the ground. And at 20 to 25 kts engine stikes were on every pilots mind to avoid. But the 15-20 kts pilots would let their guard down and relax their control after touch down or put in rudder then try to stop the drift with aileron and over bank instead of putting in aileron first then add rudder to align the nose with the runway and a strike would happen. Most of the time the crew would not know they had a strike unless tower saw sparks and told them or on post flight inspection there would be scrape marks. So if 4 degrees allowed a 25 kts cross wind i would think the crosswind limitation in a mooney would be something else before the tilt to scrape a wing tip would be the limiting factor. 

Edited by Will.iam
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16 hours ago, Pinecone said:

Yeah, I was using a flat wing to get an idea.  So it would be even more than 15 degrees.  But the idea is, it is really hard to drag a wing tip.

 Nevertheless, my dad watched a brand new F model drag a wing at our airport in the late 60s. There was a “penny a pound” airplane ride fundraiser with many local pilots participating. No one knows what caused the uncommanded roll but it happened fast at about 10agl. Didn’t cartwheel but I’m sure the passengers were terrified.

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1 hour ago, Shadrach said:

 Nevertheless, my dad watched a brand new F model drag a wing at our airport in the late 60s. There was a “penny a pound” airplane ride fundraiser with many local pilots participating. No one knows what caused the uncommanded roll but it happened fast at about 10agl. Didn’t cartwheel but I’m sure the passengers were terrified.

you can calculate the angle for your F model with a trigonometry formula by taking the "height of the wingtip above the ground" divided by "span MLG to wingtip" and then multiplying that figure by the inverse tangent function on your smartphone by - press "X", press "2nd", press "tan-1". That should be a close enough estimate & I wouldn't sweat the dihedral. Using the 3.5 & 13.5 figures @Pinecone provided produces an angle of 14.53 degrees. Assuming @Pinecone used 252/Encore figures your numbers for a F may differ slightly. Hope this helps.

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1 hour ago, rbp said:

now we just need a formula to convert that to crosswind component! 

I don't think my circa 1974 High School trigonometry skills will be near enough to figure that out. I'll ask my college buddy who has a Bachelor's in Aero Eng and worked as a Flt Control Specialist in the Space Shuttle program in Houston. He likes a challenge. 

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10 hours ago, C.J. said:

you can calculate the angle for your F model with a trigonometry formula by taking the "height of the wingtip above the ground" divided by "span MLG to wingtip" and then multiplying that figure by the inverse tangent function on your smartphone by - press "X", press "2nd", press "tan-1". That should be a close enough estimate & I wouldn't sweat the dihedral. Using the 3.5 & 13.5 figures @Pinecone provided produces an angle of 14.53 degrees. Assuming @Pinecone used 252/Encore figures your numbers for a F may differ slightly. Hope this helps.

FYI, I measured the ground to wingtip and it is 42 inches or 3.5 feet.

So if you scrape the wingtip with the main wheel on the ground you are at 14.5 degrees.  But the other wingtip will have risen about 6 more feet off the ground.

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1 hour ago, McMooney said:

reread my poh, unless i missed it, there is 0 mention of crosswind take offs or landings

"Mooneys were originally certified under the old CAR 3 (Civil Aviation Regulations) dating back to the 1930’s. There is very little guidance in CAR 3 pertaining to crosswinds. Thus you find little information about them in most of the pre-M20J Owners Manuals."

*** copied & pasted from the download available on MS referred to in my previous post that was written by Mooney factory test pilot Bob Kromer

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Max Demonstrated Crosswind is one of the most misunderstood parts of the POH. MDC has minimal bearing on the actual ability of the pilot and aircraft to make a crosswind landing under a specific set of conditions. It is simply a set standard that had to be met under CAR 3 (and now the FARs) for the aircraft to be certified. Here is one of many explanations you can find on the Internet. https://www.boldmethod.com/learn-to-fly/maneuvers/how-maximum-demonstrated-crosswind-is-calculated-aircraft/ Basically, the standard requires that, to be certified, the manufacturer must demonstrate that the aircraft can be landed in a 90 degree crosswind with a velocity of 0.2 Vso . That’s it. Doesn’t matter that the plane is, let’s say a DC3 with the biggest paddle rudder in the world and can handle twice that. The POH only needs to state that the aircraft is certified to that level. 

Mooney’s vary from model to model in the amount of rudder authority. In some instances it is necessary to fly the aircraft down onto the runway at speeds much greater than normal landing speed in order to get the necessary rudder authority. I used to go out at least once a summer and make a bunch of landings with a 20-30 kts 90degree crosswind. Sometimes the landing speed was 90 kts. The thrill has worn off and I try to avoid those landings, for one thing, there will always be a point during the rollout when the aircraft is still light on its feet and the crosswind wants to skitter its across the runway, which is hard on the tires. Often the rudder offers plenty on its own. But Max Demonstrated is not a measure of how much rudder authority the aircraft offers. It is a set standard for certification and that’s it.

Edited by jlunseth
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