1
$\begingroup$

Let imagine that we put wind indicator on rotating merry-go-around. Will wind indicator stay prepedicular to the radius of circle,showing tangetinal airflow velocity?

(Wind indicator has equal mass and distance from pivot point so moment about pivot point is zero.It desing like this to stop centrifugal force to move one end out of turn..)

I think wind indicator will be prependicular to radius of rotation.. (I ask this because of glider yaw string...)

enter image description here

enter image description here

enter image description here

$\endgroup$
  • $\begingroup$ It's going to depend on the speed of the natural wind, as well as the speed of the merry go round... $\endgroup$ – Michael Hall Aug 1 at 16:17
0
$\begingroup$

If there is no moment around the pivot point, then you do not have a wind indicator at all. One prominent writer here spoke of "curved airflow" in a turn, and this is a good example of it.

In order to be a "wind indicator" one must have greater Area aft of the pivot point. Because it is fixed, mass distribution is not a factor in indication of airflow direction, but may affect rapidity of response to changing relative wind.

The solution to understanding is to exaggerate the point of contact between the circle and the pivot. One will see that the back end has air flow from the outside and the front end from the inside. Because the back end has more area, the front end will point out of the circle.

It really depends how large a circle you are working with. For an aircraft with a turn radius of 1000 meters, the "weather vane" works fine.

But now comes an interesting twist. Because the weathervane is accelerating in a circular motion, center of mass must be at the pivot point, or error with be introduced to the direction of "point". Interestingly, for a known radius and rpm, placing the CG slightly behind of the pivot point will make the arrow point tangent to the circle!

Circular physics can be a bit different.

| improve this answer | |
$\endgroup$
  • $\begingroup$ @Roberto Digiovanni,for example if nose is very heavy it will point to much outward because of centrifugal force so it will not show correct relative wind.If radius of curve is much bigger than length of indicator then indicator will stay almost prependicular to the radius? $\endgroup$ – user50657 Jul 31 at 21:58
  • $\begingroup$ @Noah Prandtl Just got that too! Yes, the "error" decreases with increasing radius (or decreasing indicator length). $\endgroup$ – Robert DiGiovanni Jul 31 at 22:02
  • $\begingroup$ @RobertoDiGiovani,front end feel airflow from inside,back end feel airflow from outside,so if we put identical surface (flate plate for example) at both end, then indicator will point even more outward because "error-torque" is now even larger end then indicator will start rotating in clockwise direction in may case,because tangential velocity will produce greater drag force at flat surface which is at greater radius.Isnt it? $\endgroup$ – user50657 Jul 31 at 22:44
  • $\begingroup$ Then it depends on where CG is, now I search for application. RPM and radius give velocity, and direction is known, so why measure? But yes, velocity will be greater for outside plate. Given equal area, it still will have front plate point out until ... yes, it may rotate. Theoretically, would have to check interferences to rotation, it might. $\endgroup$ – Robert DiGiovanni Jul 31 at 22:51
  • $\begingroup$ @RobertoDiGiovani,In my case cg is at pivot point,indicator is "neutral"So if both ends of indicator have same surface it will rotate in clock wise direction in my case.Isnt it? $\endgroup$ – user50657 Jul 31 at 22:55
4
$\begingroup$

Let imagine that we put wind indicator on rotating merry-go-around. Will wind indicator stay perpendicular to the radius of circle,showing tangential airflow velocity?

Assuming that the indicator has a mass distribution such that the apparent "centrifugal force" from the rotation has no effect on it, and also assuming that the physical length of the wind indicator is trivially small compared to the radius of the circle, then yes, the wind indicator will remain perpendicular to the radius of the circle at the point where it is located, showing that the "relative wind" is tangent to the circle.

A very lightweight piece of yarn would do essentially the same thing, though there would be a very slight tendency for the apparent "centrifugal force" to displace the free end of the yarn toward the outside of the circle.

However the problem gets more tricky if the length of the wind indicator is not trivially small compared to the radius of the circle. If the length of the wind indicator is not trivial, this means that the "vane" or fin is located well aft of the weathervane's pivot point. The relative wind at every point along the circumference of the circle is aligned with the line that is tangent to the circle at that point. So when the vane streamlines itself to to the relative wind which is tangent to the circle at a point well behind the pivot point of the weathervane, this means that the weathervane is not parallel to a line drawn tangent to the circle at the pivot point of the weathervane. Rather, the weathervane is skewed with its nose outboard, and its tail inboard, of the line drawn tangent to the circle at the pivot point of the weathervane.

Your intuition is correct that this is why tend to see some sideslip in circling flight, especially in slow-flying aircraft whose linear dimensions are not trivial compared to the radius of turn. The vertical fin tends to streamline itself to the relative wind, which means that the fuselage tends to be parallel to a line drawn tangent to the circular flight path at a point well aft of the CG.

This effect is illustrated in section 8.10 of the "See How It Flies" website.

In short, in turning flight, the "relative wind" is curved, rather than linear. It curves to follow the curvature of the flight path.

| improve this answer | |
$\endgroup$
  • $\begingroup$ This is the correct answer. It can be shown by considering an exaggerated case. Attach a small very lightweight silk cloth to a long silk thread, say at least as long as the radius of the spinning platform. For his cloth to trail the point on the platform at a 90 degree angle, it would have to be moving on a circle significantly larger (with a greater radius), than the point the thread is attached to. That point, (outside the radius where the thread is attached) is experiencing a relative wind with both a radial and tangential component, not only tangential. But the thread is only tangential. $\endgroup$ – Charles Bretana Aug 1 at 2:51
1
$\begingroup$

If the vertical surface at the nose and at the tail of the wind indicator (usually called a weather vane) are both flat plates, have the same surface area and are equi-distant from the pivot point then the wind indicator will remain perpendicular to the radius because the force generated by the nose and tail will be equal. If, however, it is designed like a normal weather vane and has a larger surface area at the tail then the nose will point slightly away from the merry-go-round, in your diagram above it would rotate slightly clockwise.

| improve this answer | |
$\endgroup$
  • $\begingroup$ ,You are wrong ,nose can point slightly outward only if has bigger moment (mass x lever arm) then flate plate moment.In my case both moment is same,So I just need to know does wind "blow" 100% in tangetial direction.. $\endgroup$ – user50657 Jul 31 at 17:56
  • $\begingroup$ @Noah Prandtl: No. I'm afraid you are wrong. Mass and arm are only two considerations. The shape of the mass has a significant impact as well. What David is saying is that your hypothetical wind indicator can have equal mass on either side of the pivot point, but if the mass is shaped differently, it will interact with the airflow in different ways. A flat plate will deflect more airflow that a rounded dart, even if they have the same mass. $\endgroup$ – Aaron Holmes Jul 31 at 18:38
  • $\begingroup$ @NoahPrandtl If you (thought you) knew the answer, why ask the question? $\endgroup$ – HiddenWindshield Jul 31 at 18:45
  • $\begingroup$ @NoahPrandtl Because your little vane has a finite length. When it has length, the relative airspeed experienced by one end is different from the other. $\endgroup$ – JZYL Jul 31 at 21:34
  • $\begingroup$ @JZYL,@AaronHolmes, David is wrong when he say that indicator will point in tanget to the cirlce if both ends have same surface.If both ends have same surface, indicator will rotating in clockkwise direction for my case.Because tangential velocity is greater at larger radius so it will create more drag at outward surface and make indicator rotating all the time $\endgroup$ – user50657 Jul 31 at 23:18

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy