Here's a science question that I can't seem to get a good answer to. Is there a difference between a cold and a warm crosswind in the way it acts on a landing airplane. It seems to me air density makes a difference in everything else it should also make a difference here.

The short question is if the OAT is 70 degrees F with a 10 knot x-wind does it feel different if than if the OAT 10 degrees F with the same wind?

Maybe this is a dumb question

Brad

I think this is a good question. However, since the chart that calculates a crosswind component does not use temperature as a factor, then I would conclude that while there may be a difference it is so slight that it should not be considered material.

I think the short answer is yes. You're basically asking a density altitude question that's been turned on its side.

At higher altitudes, true airspeed (and therefore groundspeed) is higher. Your aircraft's momentum is a function of its groundspeed. So my guess is that in gusty conditions, the colder crosswind would be more noticeable.

Now a 10 kt wind is a 10 kt wind, and has the same force exerted regardless of temperature, because this velocity is basically another indicated airspeed. A colder wind has more molecules moving slower, but the pressure is the same, so in a constant crosswind there shouldn't be much difference.

But if you notice a difference in your airplane's handling in crosswinds of different temperature, the larger difference might be in the pilot. Numb feet in large boots and frozen hands in thick gloves might desensitize to a point where the minute adjustments needed for a smooth landing aren't feasible.

I'm going to venture out here and say no, that air density doesn't have an isolated effect on crosswind intensity from the pilot's perspective. There are several other factors that come into play in a season when the air is cold, like wind and weather intensity, and maybe even different terrain-induced dynamics, but from an analysis of just the air density, I think it's all relative.

Sure, with high density altitude, groundspeed is going to be greater, so the second your wheels touch down, you're just moving faster down the runway, and to get to comfortable braking speeds (groundspeed) you have to transition though lower airspeeds that with hot air just aren't going to be as controllable. But prior to touchdown, it's all airspeed, so it's all relative.

If an entire air mass is the same temperature, let's say at a fixed altitude, then the wind blowing across the runway is the same temp and density as the air you're flying through, the same dense air that's keeping you aligned and flying straight down the strip. It's not a big dense mass blowing into a less dense mass, unless the front is right on the airport. So, I think it's a wash.

If it's just a nasty winter wind and it's more intense than the last time you flew in early fall, then it will probably seem more challenging.

I would think summer crosswinds would be more challenging for the reasons listed above regarding hot thin air and faster groundspeeds, as well as thermals lifting off on short final and kicking you around.

Just be careful not to stall in that downwind turn at low airspeeds....

Theoretically it will, but from a practical point of view it won't. We can calculate the relative difference:

The force a wind makes against an object is reasonably modeled with the equation F=KDV^2. Where F is the force of the wind, D is density of the air, V is the velocity of the wind, and K is a constant that takes into account the various units of measure. Since we are concerned with relative changes, we don't need to know what K is.

What, you don't see temperature in there? It is incorporated into the density variable, D. D=kP/T. Where P is pressure, T is absolute temperature, and k is another constant.

So the effect boils down to one of the effect of temperature on density. In this case, it would be the ratio of 70+460/10+460=1.13, or about 13% Note that you need to add 460 to degrees F to get the absolute temperature.

So the effect on a warm day will be lower, but I bet most pilots would not be able to distinguish the difference.

The effect would be similar to compensating for a cross wind in long range shooting. Any of you gun nuts doing 1000yd targets care to comment?

tom

Nice explanation, Tom. I think though that what the question comes down to is not necessarily will the wind have more force per volume, but will it displace the aircraft more than any a higher temperature air? Especially given that the force needed to displace the aircraft laterally is theoretically greater.

I think no, because the fluid that the fuselage and vertical stab are flying through is also more dense, providing more resistance to lateral deflection than warm air would. Whether the dynamics that achieve this balance of forces diverge or not, I don't know. Is more aileron input needed to compensate for a cooler 10 kt wind than a warmer one?

1SeventyZ wrote:I think no, because the fluid that the fuselage and vertical stab are flying through is also more dense, providing more resistance to lateral deflection than warm air would. Whether the dynamics that achieve this balance of forces diverge or not, I don't know. Is more aileron input needed to compensate for a cooler 10 kt wind than a warmer one?

If your argument were correct, a paddle would be just as hard to move through air as it is through water.

I think the reason folks don't worry about it is that it is too small to worry about. Most stuff to do with flying has so much variability in it that these small effects are not noticeable. Every time I make a cross wind landing, the wind is gusting way more than 13%. Going back to the first equation, you can see that the force is velocity squared. That means that relatively small changes in wind speed will be magnified in effect. A 16 knot wind will only have to increase by 1 knot to increase it's effect by 13%.

tom

Savannah-Tom wrote:If your argument were correct, a paddle would be just as hard to move through air as it is through water.

I don't think that's an appropriate analogy. Perhaps more relevant would be to compare the amount of deflection that a fluid of equivalent velocity could exert on the broadside of a boat keel.

Sustained 10 kt of air against the side of a keel aligned with a 60 kt velocity wind in air

vs

Sustained 10 kts of water against the side of a keel in water, aligned with a 60kt velocity current.

You're saying if we put a newtonometer on the keel, the underwater one will register a higher static force?

Tom, if you can convince me of this, there's a Local Boyz lunch in it for you.

Durango Rebel wrote:The short question is if the OAT is 70 degrees F with a 10 knot x-wind does it feel different if than if the OAT 10 degrees F with the same wind?

Maybe this is a dumb question

It's not a dumb question, but maybe it's not quite the right question.

Perhaps it would be better to ask if a 10 knot crosswind at 10 degrees will have more or less of an effect on your touchdown and rollout than a 10 knot crosswind at 70 degrees.

Here's my take on it.

An anemometer doesn't know air density, it only knows how fast it's spinning, so even though the air molecules may have to move faster to spin it the same RPM at higher temperatures and lower densities, it will still FEEL the same to any FIXED object. If it said 50, you would have to lean into the wind the same amount at 10 degrees as at 100 degrees.

However, your aircraft is NOT a fixed object until it is completely stopped! Especially if it's a taildragger, what you are most concerned about in any crosswind landing is controllability versus the effect of the wind. If it's quite hot and you get slow, the controls will lose their effectiveness at a higher groundspeed, and since the likelyhood of a groundloop goes up as the square of the groundspeed, this is a bad situation. Also, as other folks have mentioned, hot air is just more squirelly, up, down or sideways.

On the other hand, the required amount of crosswind correction is a function of airspeed, so a jet landing at 140 knots may handle a 40 knot crosswind, but an ultralight landing at 25 might only handle an 8 knot one. This means that your required correction in degrees of crab will be less for 10 knots at 80 degrees than for 10 knots at 10 degrees. These two variables may somewhat cancel each other out during the approach and flare part of a landing, but once you are solidly on the ground the higher groundspeed at the higher temperatures takes precedence.

What this all means to me is that I will have higher chances of a poor landing in the same velocity of crosswind at lower air densities than at higher densities. This has been amply borne out in 39 years of experience...I have landed the 180 at sea level and 55 degrees with a solid 25 knots being reported, and although the crab angle was impressive, the touchdown and rollout were not too scary . I have also landed at 85 degrees at 8000 feet in a 10 knot crosswind, and the slight crab angle lulled me into a false sense of security, so that I almost lost the airplane during the rollout, before the tail came down. To me, the critical phase is always the rollout...if I'm going to lose it, it will be between a second or two after touching and before slowing to a taxi speed. Thus I am much more concerned with variables that affect that phase(such as groundspeed) rather than crab angles. Rudder and aileron effectiveness most definitely affect the rollout, and my criteria for a go-around are whether or not I can maintain the wing low enough to stop the drift while having enough rudder to keep it straight after transitioning from the crab, not what the crab angle actually is, although that is a real good indicator of what will happen while you are trying to plant the upwind wheel. If I'm working too hard to keep straight and on centerline, I'd rather go-around before touchdown than after! I also carry a little more power and speed and plan on a much longer landing than normal.

Regardless of the physics of it, most of the variables at high temps and low air densities conspire against you, and most of the variables at low temperatures and high air densities conspire for you in the landing and rollout.

Watch out for all crosswinds when it's high, hot, and humid!


Rocky

Personally, I think you're trying reeeeaaaallllllyyyyyy hard to over-think this.

A little too much free time lately?

MTV

Blue&Yellow Luscombe wrote:....

Hey there B&Y Luscmbe, your bird wouldn't happen to be "Little Buster" would it? If so, I saw your airplane over at Stehekin a couple years ago, also talked to you (or the at-the-time owner) at Auburn once. Nice airplane....

Eric

My two cents say yes the colder air will have a greater effect than the warmer because........ The upwind wing will want to fly longer on the denser air at lower air speed. My understanding is that denser air provides us better wing performance. However the increased perfomance would also have to apply to the ailerons allowing you to correct the wing lift w/ your normal amount of aileron input thus making it the same adjustment as in the warm air.......

I just confused myself, You guys are to smart Just do some pilot shit and keep the plane going straight down the strip!

1SeventyZ wrote:

Savannah-Tom wrote:If your argument were correct, a paddle would be just as hard to move through air as it is through water.

I don't think that's an appropriate analogy. Perhaps more relevant would be to compare the amount of deflection that a fluid of equivalent velocity could exert on the broadside of a boat keel.

Sustained 10 kt of air against the side of a keel aligned with a 60 kt velocity wind in air

vs

Sustained 10 kts of water against the side of a keel in water, aligned with a 60kt velocity current.

You're saying if we put a newtonometer on the keel, the underwater one will register a higher static force?

Tom, if you can convince me of this, there's a Local Boyz lunch in it for you.

My whole argument centers on the assumption that the side force on a landing airplane is equivalent to the force the wind would make on an airplane statically suspended in the air. And that this side force is calculated the same as forces on fixed structures like flagpoles and buildings. These widely different shapes all use different constants, K, but again if you are looking at relative differences, the absolute doesn't matter. You just want to look at how the various variables effect the result.

The only thing wrong with the paddle analogy is that it did not take into effect the viscosity of the two fluids. Viscosity equals friction. In typical "thin" objects like buildings, flagpoles, and airplanes, the viscosity is not nearly as important as the simple equation I proposed. Long pipes and perhaps trains are examples where viscosity friction can't be ignored.

Pick a decent day, and I'll fly up to Aurora and we can fly to someplace nearby. Win, lose, or draw, I'd enjoy hauling you to lunch.

tom

mtv wrote:A little too much free time lately?

Yup! My plane is 70 miles away in a freezing hangar with a leaky prop 3 hours after an overhaul and it's snowing and 8 degrees.

Besides, hangar flying is almost as much fun as the real thing! I've gotten a lot of enjoyment discussing all manner of abstruse/obtuse ideas on this forum. It's got the most varied pilot types of any forum I've ever seen!

Rocky

Savannah-Tom wrote:My whole argument centers on the assumption that the side force on a landing airplane is equivalent to the force the wind would make on an airplane statically suspended in the air.

Then I have no argument for that.

I would respectfully point out that the largest single reason that a tailwheel airplane, landing at high density altitude is a bit more of a handful has to do with the fact that you are touching down at a significantly higher ground speed.

Indeed, crosswinds may be more of a challenge in that environment, but it's largely due to the speed at which you touch, and the consequent longer period of time and distance to decelerate and stop, NOT so much due to side load differences, etc, as discussed here.

There may indeed be a little difference in side loads with changes in air density, but I doubt that effect is noticeable.

MTV