0
$\begingroup$

It is often said that the lift vector helps to propel a glider forwards. The lift vector has no component acting parallel to the glider's trajectory through the airmass, but in many cases the lift vector does have a component acting parallel to the glider's trajectory as seen from the ground. Note that for a given airspeed, the direction of the glider's trajectory through the airmass-- i.e. the direction that is exactly opposite to the "relative wind" that is "felt" by the glider-- is not affected by sustained updrafts, downdrafts, headwind, or tailwinds. But this is not true of the glider's trajectory as seen from the ground. A glider in a powerful mountain wave updraft may be rising straight up as seen from the ground, while the trajectory through the airmass will still be the normal glide path that would be obtained at whatever airspeed the glider is flying at.

Under what circumstances does the lift vector of a glider contribute a force component along the glider's path of travel (trajectory) as seen from the ground?

Under what circumstances does the drag vector of a glider contribute a force component along the glider's path of travel (trajectory) as seen from the ground?

Likewise for a powered airplane.

Note -- to avoid any possible ambiguity, please be aware that this question is using the word "lift" exclusively to mean an aerodynamic force generated by the aircraft, not to mean rising air. These are two completely different things.

Note -- the scope of this question is meant to be confined to linear (but not necessarily horizontal) straight-line flight with constant airspeed and groundspeed, at least over the short term. We're assuming the glider (or airplane) is flying in an airmass that is locally uniform. In other words, if the glider has entered a thermal updraft or penetrated through an abrupt wind shear or flown from sinking air into a ridge updraft or wave updraft, we're assuming that it has been in the new airmass long enough to come into equilibrium, so that net force is zero for the time being. This is not a question about "dynamic soaring" as practiced by the albatross over the open ocean, r.c. glider pilots flying in loops on the lee side of the hill, someone trying to put into practice Taras Kiceniuk's ideas about exploiting the boundaries between still air and downdrafts as a source of energy, etc. Those subjects are worthy of an ASE question, or many questions, but this is not that question.

$\endgroup$
  • $\begingroup$ Comments are not for extended discussion; this conversation has been moved to chat. Also note that comments' section is not a place to jot down your future edits or thoughts. Please refrain from doing that. $\endgroup$ – Farhan Apr 9 at 16:29
2
$\begingroup$

In thermals.

More precisely: Every time when the updraft strength increases. The glider will continue on its original path due to inertia while the airflow will have a positive angle when referenced to the path of travel. Now the aerodynamic force will point slightly forward and accelerate the glider.

Experienced pilots use this by pulling more than 1 g when flying into a thermal and less when updraft strength decreases. This helps them to gain some more energy from the rising air.

Drag is per definition the part of aerodynamic forces parallel to the flow direction, so for drag the answer is: Always. With changes in vertical airspeed, the drag component parallel to the direction of movement will merely change with the cosine of the angle between the aerodynamic and the kinetic coordinate systems.

If you insist on excluding changes in vertical air speed, your question will have no answer regarding lift.

| improve this answer | |
$\endgroup$
  • $\begingroup$ "If you insist on excluding changes in vertical air speed, your question will have no answer regarding lift."-- why not? $\endgroup$ – quiet flyer Apr 7 at 13:06
  • $\begingroup$ @quietflyer Because in the absence of wind the aerodynamic and the kinetic coordinate systems are identical. $\endgroup$ – Peter Kämpf Apr 7 at 14:51
  • $\begingroup$ But there can be a CONSTANT updraft or downdraft. Locally constant I mean, over the time and distance scale needed for the glider to come into equilibrium (just a few seconds). Also even in the presence of a tailwind but no updraft, the lift vector has a component acting along the glider's path of travel (trajectory) as seen from the ground, as noted in the (currently) downvoted answer. Maybe it will become more clear if I add a diagram. $\endgroup$ – quiet flyer Apr 7 at 14:56
0
$\begingroup$

Any glider is powered in part by its lift vector at all times when in normal flight (as opposed to a loop, roll, or other aerobatics, where a reserve of kinetic energy is used and often partially converted to potential). The lift vector is always angled slightly forward when in gliding flight.

The energy for this comes from conversion of potential energy (in the form of altitude) into kinetic energy (in the form of forward speed). Potential energy, in turn, for a glider, is in general supplied by rising air (initially for each flight, some is given by the launch assist -- whether aero tow, ground tow, winch/bungee, or by rolling/running down a slope into ridge lift).

Drag, on the other hand, by definition is always opposite in direction to the velocity vector, so will never have a component that adds to current velocity (though if the glider is, for instance, in a deep stall with tail down, there may be a drag component "forward" in the direction the glider's nose is pointed).

For powered airplanes, there will be times when there's a lift component that acts to increase forward speed (for instance, when descending more or less steeply at low throttle), and times (most steady-state flight) when lift contributes to drag -- which still, by definition, acts against current velocity vector.

| improve this answer | |
$\endgroup$
  • $\begingroup$ I think I want to confine the question to steady-state linear flight. Would that invalidate anything in your answer? I don't think so. $\endgroup$ – quiet flyer Apr 6 at 18:00
  • $\begingroup$ Not directly -- but don't forget that in still air, a glider can't maintain level flight for long (it's converting kinetic energy to replace drag losses if it quits giving up altitude). $\endgroup$ – Zeiss Ikon Apr 6 at 18:02
  • $\begingroup$ @Zeiss Ikon your answer is a plus, but the wing is only the lifter/propeller. The "power" comes from potential energy converted to kinetic energy "falling". Notice birds have lifter/propellers too, but the "power" is from the muscles, and bird seed is the "fuel". $\endgroup$ – Robert DiGiovanni Apr 7 at 20:01
  • $\begingroup$ @RobertDiGiovanni You've almost quoted my second paragraph. "The energy for this comes from conversion of potential energy (in the form of altitude) into kinetic energy (in the form of forward speed)." $\endgroup$ – Zeiss Ikon Apr 8 at 11:05
0
$\begingroup$

"As seen from the ground" will be the net trajectory of the gliders vertical and horizontal velocity components added or subtracted from the vertical and horizontal velocity components of the air (discounting the third axis for now).

We consider the theoretical uniformly moving air mass and the relative wind the glider creates in flight. As discussed many times on this site, a perfect circle in the air will look a lot different as a ground track if there is wind. A glider with an airspeed of 40 knots may be motionless "as seen from the ground".

Logicly, the answer is always, with the caveat that the air mass movement velocity and direction must be considered, and yes, along all 3 axes.

Finally, one of the keys is to consider that weight is subtracted from any vertical drag or lift vector components, the remainder is balanced in the glider velocity and direction with respect to the airmass.

| improve this answer | |
$\endgroup$
  • $\begingroup$ I found this easier to work backward from the aircraft's velocity and direction (starting with no wind case). The horizontal component of the lift vector will match the horizontal drag component. The vertical lift component plus the vertical drag component equal the weight component. Net forces are 0 at a given equilibrium velocity. $\endgroup$ – Robert DiGiovanni Apr 8 at 23:27
  • $\begingroup$ "Finally, one of the keys is to consider that weight is subtracted from any vertical drag or lift vector components,"-- that is the purpose of drawing the vector triangle, as noted in the downvoted answer. $\endgroup$ – quiet flyer Apr 10 at 11:20
  • $\begingroup$ @quiet flyer something about a Billy Joel record being "cut down to 3:05". Some people just need it to be short. Your question is very good and still not completely answered. Vertical drag may indeed contribute a horizontal component in the direction of flight. A falling object tilted forward gets pushed forward by the "up draft". Re-read Kampf, an amazing and detailed source of knowledge. $\endgroup$ – Robert DiGiovanni Apr 10 at 13:52
-1
$\begingroup$

Assuming steady-state linear flight and the other constraints stated in the question--

The idea that a component of the Lift vector is helping to pull the glider forward along the glider's trajectory as viewed from the ground is only true when the glider's achieved glide ratio relative to the ground is better than the L/D ratio (which is also the still-air glide ratio), or when the glider is climbing in an updraft (unless the glider is drifting backwards over the ground along an achieved climb path that is flatter than the direction of the drag vector/ relative wind). These are the only times the that the Lift vector has a forward component relative to the glider's trajectory as viewed from the ground.

This would only happen in one the following cases -- the glider must be flying in:

1) Air that is neither rising or sinking, but is moving with tailwind component relative to the glider's actual heading (not relative to the "course" or achieved ground track-- the distinction comes into play when there is a strong crosswind component.)

2) Air that has no horizontal motion, but is rising.

3) Air that is rising fast enough to offset the degradation in achieved glide ratio caused by a headwind component.

4) Air that is moving with a tailwind component that is strong enough to offset the degradation in achieved glide ratio caused by a sinking (downdraft) component.

5) Air that is rising and also is moving with a tailwind component

To a first approximation, the idea that a component of the Drag vector is helping to pull the glider along the glider's trajectory as viewed from the ground is only true when the glider is moving backwards over the ground-- i.e. when a component of its "course" or achieved ground track vector points in the same direction that the tail of the glider is pointing, and opposite the direction that the nose is pointing. This can easily happen in strong wind-- the glider can easily drift backwards over the ground.

Strictly speaking this approximation is only true if the L/D ratio is infinite. To be more precise, we have to recognize that the true criteria is that the Drag vector will help pull the glider along the glider's trajectory as viewed from the ground whenever a component of the actual three-dimensional flight path as viewed from the ground-- not just the "course" vector or achieved ground track-- points in the same direction that the Drag vector is pointing. Since the Drag vector is parallel to the flight path through the airmass or "relative wind", it is inclined relative to the horizon. Therefore for non-infinite glide ratios this criteria can be achieved when the aircraft is rising straight up vertically relative to the ground with zero horizontal groundspeed in a strong updraft, or even when the aircraft is creeping forward very slowly over the ground as it rises rapidly upwards. Also, there can be cases where the criteria is not met even if the aircraft is drifting backwards very slowly over the ground, if it is also sinking very rapidly due to a strong downdraft. But for reasonably high glide ratios, it's a fairly good approximation to say that the Drag vector is helping pull the aircraft along its trajectory as viewed from the ground only when the aircraft is actually drifting backwards over the ground.

What is really going on here is that we have a closed vector triangle of Lift, Drag, and Weight, with Lift smaller than Weight and inclined slightly forward relative to weight. Drag is perpendicular to Lift and acting at right angles to Lift, as illustrated in this related answer. There is no question that net force is zero, and net acceleration is zero, from any inertial reference frame, including the airmass reference frame and the ground reference frame. But the velocity vector is different in different reference frames-- for example consider the case where the glider is slowly rising straight up as seen from the ground reference frame-- so the issue of which forces are contributing a component along the glider's direction of travel depends on which reference frame we choose.

Note that it is often said that gliders are powered by gravity. This is always true as seen from the airmass reference frame-- of the three forces Lift, Drag, and Weight, Weight is the only one with any component parallel to the glider's path of travel through the airmass. But as seen from the ground-- i.e. in relation to the path of travel relative to the ground-- gravity is exerting a force component parallel to the trajectory only in cases where the glider is losing altitude. If the glider is climbing, then the Weight vector is exerting a force component against the direction of glider's trajectory as seen from the ground, not in the same direction as the glider's trajectory as seen from the ground. If the glider is flying horizontally, then from the viewpoint of the ground-based reference frame, the Weight vector is exerting no force component along the direction of the glider's trajectory, and is doing no work on the glider.

This means that whenever a glider's velocity vector as seen from the ground is constant and is purely horizontal, the forces exerted along the direction of travel by the Lift and Drag vectors must be exactly equal and opposite. Whenever a glider is travelling horizontally and the glider's velocity vector as seen from the ground has a component toward the nose of the glider rather than toward the tail of the glider, the Lift vector is helping to pull the glider along its path of travel, and the Drag vector is resisting that pull in a way that exactly cancels it out. Conversely, if the glider is travelling horizontally and backwards as seen from the ground, then Drag vector is helping to pull the glider along its path of travel, and the Lift vector is resisting that pull in a way that exactly cancels it out.

The Lift and Drag vectors are fixed in their geometric relationship to the relative wind-- Lift is perpendicular to the relative wind, and Drag is parallel to the relative wind. Furthermore, for any given L/D ratio and Weight value, the Lift and Drag vectors are fixed in magnitude. The fact that, from the ground reference frame, the force component exerted in the glider's direction of travel by the Lift and Drag vectors varies according to the wind direction and updraft or downdraft velocity, is really just an artifact of the fact that direction of the relative wind is not fixed in relationship to the direction, and magnitude, of the glider's velocity vector as seen from the ground.

The work done by a force is defined as distance traveled times the force component exerted in the direction of travel, and it takes energy to do work on a glider or any other body. It doesn't really make sense to ask what energy source is responsible for any one particular aerodynamic force component, such as the component of the Lift or Drag vector that is acting parallel to the flight path as viewed from the ground. Ultimately, as seen from the ground, any NET energy transferred from the atmosphere to the glider, via the combined action of the Lift and Drag vectors, can only come from updrafts. Only in the presence of an updraft can a glider gain altitude while flying at a constant velocity, regardless of the horizontal wind speed and direction. Of course, as the glider extracts Kinetic Energy from the updraft and converts it into its own gravitational Potential Energy, the updraft must be slowed by an infinitesimal degree. The only instance where the glider is neither increasing nor decreasing the Kinetic Energy of the surrounding local airmass is when the glider is travelling purely horizontally, rather than sinking or climbing.

And the ultimate energy source that powers updrafts-- whether they are thermal updrafts or "ridge lift" updrafts or mountain wave-- is the sun. This power may be conferred either via direct solar heating of the ground directly below, or by complex meteorological processes involving the entire global weather system.

The question also asked about powered aircraft.

Assume for simplicity that Thrust acts exactly opposite in direction to Drag.

In the case where Thrust is less than Drag, the aircraft is exactly like a glider, except we substitute "(Drag minus Thrust)" wherever "Thrust" appears in the explanation.

In the case where the Thrust and Drag are equal, the explanation regarding the Drag vector is exactly the same as the "first approximation" explanation (assuming infinite L/D ratio) above. And in this case, the Lift vector is adding helping to pull the aircraft forward along its trajectory relative to the ground anytime the aircraft is in rising air, without exception, and without regard to headwind or tailwind.

In the case where Thrust is greater than Drag, then the idea that a component of the Lift vector is helping to pull the aircraft forward along the aircraft's trajectory as viewed from the ground is only true when the aircraft's achieved climb ratio relative to the ground is better (i.e. a lower number than) than the L/(T-D) ratio. This will require an updraft, or a headwind, or both, and a weak updraft may not be enough to offset a strong headwind, and a weak headwind may not be enough to offset a strong downdraft.

In the case where Thrust is greater than Drag, then the idea that the Drag vector will help to pull the aircraft along the aircraft's trajectory as viewed from the ground can only be true when the aircraft's trajectory as viewed from the ground has a component pointing in the same direction as the drag vector. For ratios of L/(T-D) that are not too high (i.e. the still-air climb angle is not extremely good), to a first approximation this means that the aircraft is drifting backwards over the ground, but this approximation will not apply well to extremely high still-air climb angles. An aircraft that could climb straight up at a high rate of climb in still air could, in that same configuration, only move (as seen from the ground) in the direction of the Drag vector in the presence of an incredibly strong downdraft.

| improve this answer | |
$\endgroup$

Your Answer

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

Not the answer you're looking for? Browse other questions tagged or ask your own question.