Originally Posted by
Jim59
Precisely. That's what keeps both aeroplanes and gliders in the air. Jongster was considering the thrust vector when climbing under engine power and the thrust is greater than the drag and is perfectly correct.
Umm...no it isn't, because whilst in a climb there is a vertical component of thrust which opposes weight, there is also a vertical component of drag which adds to the weight. Thrust and drag are equal and opposite, so these two vertical components cancel out. I'm aware that the "lift is less in a stable climb" myth is taught to PPLs, but it's by no means the only bit of basic aeronautical theory taught to PPLs doesn't really stand scrutiny.
But even this simplistic view doesn't really cover it. The point becomes obvious if you draw out the force diagram. Weight always acts vertically downwards. Lift always acts at right angles to the airflow (it has to - it's a hydrostatic pressure effect) and drag always acts parallel to the airflow (because it's a dynamic pressure effect) so lift and drag always act at rightangles to eachother. Thrust acts in the direction which it is pointed - for simplicity lets assume that the designer made a decision to bolt the thrust-maker so that it's alighed with the direction of the drag vector at cruise speed. So in stable straight&level flight (with respect to the surrounding air mass) at cruise speed we find lift,drag, weight and thrust all at right-angles to eachother. This is the picture you'll find in all the PPL textbooks.
What happens if we slow down, adjusting power as required to remain straight& level? Obviously lift, drag and weight will still be mutually at right angles, but to maintain height we need to trim back, so the thrust vector angled upwards. So thrust now has a horizontal component equal to drag PLUS a vertical component. So in slow S&L flight "wing lift" will be less than it is a cruise speed, because total lift must equal weight.
OK, accelerate back to cruise and retrim for S&L. Now add power and climb at 10 degrees. The "air velocity vector" is now angle 10 degrees upwards, so the drag vector points 10 degrees downwards, adding to the weight, but this is cancelled out by the trust vector being pointed upwards. Weight is still acting vertically, but that is now angled 10 degrees backwards compared to the velocity vector so thrust and weight are no longer at right angles. Therefore you will need to add thrust to allow the engine to do work against gravity (or you could view this as the horizontal component of lift in the climb - same thing) as well as just against drag to restore the equilibrium. Last of all we have lift - lift acts at right angles to the airflow velocity vector, so it is angled 10 degrees backwards. That means that it is no longer parallel to weight, so only a COMPONENT of the lift is opposing gravity. Therefore the total "wing-lift" must be GREATER so that (lift*cos(climb_angle)) is equal to weight.
NALOPKT(&EFGAS),
PDR