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enter image description here

In the above image lift is the same during a climb and during a descent. Is lift force not supposed to be less than weight during a descent?

Secondly, does descending at an airspeed much lower than the dive velocity (Vd) of the airplane affect the rate of descent of plane?

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  • $\begingroup$ This question is misleading. What reason is there to believe that the red arrow points to a climb and the blue arrow points to a descent? $\endgroup$ – quiet flyer Nov 2 '18 at 14:59
  • $\begingroup$ Really the question should not be answered at all, but I did anyway. $\endgroup$ – quiet flyer Nov 2 '18 at 15:08
  • $\begingroup$ @quietflyer There must be a chart or drawing that is not shown here. Point A probably indicates climb while point D indicates descent. Without the chart, though, you're right. A and D are not defined $\endgroup$ – TomMcW Nov 2 '18 at 18:39
  • $\begingroup$ I bet point A indicates maneuvering speed and point D indicates Vne-- I bet it's a V-n diagram $\endgroup$ – quiet flyer Nov 2 '18 at 18:45
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The content reproduced in the question is basically exploring what happens if we suddenly bring the wing to the angle-of-attack required to produce some given G-load, while flying at some given airspeed. The value of lift given in both cases -- 4166 pounds -- is undoubtedly NOT the weight of the microlight aircraft. It is undoubtedly much higher than the weight of the microlight aircraft. We are NOT looking at a situation here where net force and net acceleration are zero, so this content really doesn't shed light on the questions being asked.

On the content reproduced in the question, it's not clear why the asker believes that that the red arrow points to a climbing situation and the blue arrow points to a diving situation. At any rate, the content reproduced in the question has nothing to do with the forces in a steady-state climb or descent.

As to the questions asked--

"Is lift force not supposed to be lesser than weight during a descent?"

Yes, both in a steady-state climb, and a steady-state descent, lift is less than weight. To understand this better, keep in mind that in the context of winged flight, lift is defined to act perpendicular to the aircraft's flight path through the airmass, not vertically. Similarly, the drag vector is defined to act parallel to the aircraft's flight path through the airmass, and to a first approximation, the thrust vector can often be considered to also act parallel to the aircraft's flight path through the airmass. As a result, thrust supports part of the aircraft weight in a steady-state climb, and drag supports part of the aircraft weight in a steady-state descent, and in both cases the wing's lift vector is reduced. To see the force vector diagrams illustrating these situations, visit these related answers to related questions:

What produces Thrust along the line of flight in a glider?

Does lift equal weight in a climb?

(Note-- to apply the vector diagrams shown in the first link above to the case of a powered glide or powered dive, simply replace the label "drag" with the label "drag minus thrust" –- much as the label "thrust minus drag" appears in the diagrams in the second link above.)

"Is descending at an airspeed much lower than the dive velocity (Vd) of the airplane affect the rate of descent of plane?"

Yes, for a given throttle position or a given thrust force or a given power output, if we change the airspeed, we'll change the rate of descent. In general the slowest descent rate is obtained flying just a little faster than stall speed.

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  • $\begingroup$ Descending will also reduce the rate of descent due to the air density increase along the way. $\endgroup$ – Peter Kämpf Nov 15 '18 at 19:33
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For a constant (vertical-) speed descent the lift is equal to the weight of the craft.

Vertical acceleration only happens when the net vertical force is not equal to 0.

So the time when lift is less than weight is when the plane starts or speeds up the descent or stops or slows down a climb.

Similarly the lift force is only greater than the weight of the plane when the plane stops or slows down a descent or starts or speeds up a climb.

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  • $\begingroup$ Re "For a constant (vertical-) speed descent the lift is equal to the weight of the craft. Vertical acceleration only happens when the net vertical force is not equal to 0."-- The problem with this answer is that lift is NOT the same thing as net vertical force. See my answer, and the associated links, for more. $\endgroup$ – quiet flyer Nov 16 '18 at 15:12
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First to compliment @quiet flyer for deciphering the data provided with the question. There is nothing wrong with going to a flying school and taking a "ground lesson" if there is anything in the POH that needs explanation.

Second, to shed light on how important it is to get vector diagrams correct and account for "that which produces velocity in what direction". This is why using gravity vector as the universal reference for direction and magnitude literally allows one to draw force vectors to scale, break them down into vertical and horizontal components, and understand what is involved with direction and speed. Unfortunately, so many people are missing one very obvious fact, a heavier than air craft must be moving in flight, and, in an aerogravimetric enviornment, to be in steady state 0 acceleration, the thrust vector is balanced in the direction of velocity by drag.

Yes, all heavier that air craft in steady state flight need "push" to keep them moving. DRAW IN THE THRUST VECTOR DIRECTLY OPPOSED TO DRAG VECTOR, and let's see what happens:

Parachute straight down. Direction straight down. Power source gravity. Lift source drag only. Horizontal velocity none. Drag and Gravity vectors are equal and cancel at steady state at a given velocity.

Glider descending steady state. Direction angled towards earth. Power source gravity. Lift source drag and airfoil. Horizontal and vertical velocity. Vertical drag component contributes to vertical lift (resists gravity). Horizontal lift component provides forward motion, balances horizontal drag component at steady state. To the "purist" lift is less than gravity, but this is only considering the vertical component of lift!

Powered aircraft ascending. Direction angled away from earth. Power source aviation fuel. Lift source propeller and air foil. Horizontal and vertical velocity. Vertical drag component OPPOSES vertical lift component. Horizontal lift and drag components OPPOSE horizontal direction of flight. What MUST compensate to maintain steady state flight: the thrust vector. This contributes to vertical lift component to balance vertical drag, and to the "forward" motion component, balancing horizontal drag and horizontal lift components.

Anyone who has done near stall powered slow flight with the engine screaming high RPMs knows what I mean.

With gravity vector constant WRT to earth, and drag, lift, and thrust expressed as vertical and horizontal components, the olde four forces of flight can be used to describe steady state flight regardless of direction.

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    $\begingroup$ And why not be a purist? $\endgroup$ – quiet flyer Nov 4 '18 at 13:51
  • $\begingroup$ Btw, the case of "Anyone who has done near stall powered slow flight with the engine screaming high RPMs knows what I mean." appears to introduce the complication of the thrust line being significantly different than the direction of the flight path through the airmass-- due to very high angle-of-attack-- consider for example the case where altitude is constant (flight path horizontal) during the slow-flight maneuver-- would be interesting to know how much of the motor's "supporting" effect is due to enhanced airflow over wings and how much is due to "thrust vectoring" effect, percentage-wise. $\endgroup$ – quiet flyer Nov 4 '18 at 13:52
  • $\begingroup$ For example how much would the minimum sustainable airspeed jump up if we took away the enhanced airflow over the wings but left the "thrust vectoring effect" in the picture, and how much would the minimum sustainable airspeed jump up if we took away the thrust vectoring effect but kept the enhanced airflow over the wings? Would be interesting to know-- video noting extreme performance of over-powered bush plane Wilga (now Draco) w/ radial engine swapped for turboprop is fascinating re extreme nose-high attitude in power-on flight near stall youtube.com/watch?v=PqhI4MeCn1c $\endgroup$ – quiet flyer Nov 4 '18 at 14:49
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    $\begingroup$ "To the "purist" lift is less than gravity, but this is only considering the vertical component of lift!"-- actually this is not a true statement. Sorry to keep locking horns with you Robert but if you look at the gliding-case diagram linked in my answer you'll see that the entire lift vector, not just its vertical component, is less than weight $\endgroup$ – quiet flyer Nov 4 '18 at 18:03
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    $\begingroup$ No need to. See you next question, and BTW, I did like the way you figured what was in that POH. $\endgroup$ – Robert DiGiovanni Nov 4 '18 at 18:35

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