ths@lanl.ARPA (09/11/85)
I have received a number of correct answers to the question "During a constant airspeed climb what is the relationship between the four primary forces of flight - lift, drag, thrust and weight". They are equal. The "trick" is why does the airplane climb? Ted Spitzmiller
john@gcc-bill.ARPA (John Allred) (09/17/85)
In article <30512@lanl.ARPA> ths@lanl.ARPA writes: >I have received a number of correct answers to the question "During a >constant airspeed climb what is the relationship between the four primary >forces of flight - lift, drag, thrust and weight". They are equal. The >"trick" is why does the airplane climb? > >Ted Spitzmiller In general, the nose is above the horizon in a climb, so the plane "drives" up. -- John Allred General Computer Company uucp: seismo!harvard!gcc-bill!john
oscar@oakhill.UUCP (Oscar Strohacker) (09/17/85)
In article <30512@lanl.ARPA> ths@lanl.ARPA writes: >I have received a number of correct answers to the question "During a >constant airspeed climb what is the relationship between the four primary >forces of flight - lift, drag, thrust and weight". They are equal. The >"trick" is why does the airplane climb? > >Ted Spitzmiller False, Ted. Before you set yourself up as an aerodynamics teacher you ought to at least understand the definitions and the physics. These are vectors. L L L T L T >> D W D W W W W Weight - the force exerted by gravity on the a/c. (In the vertical direction, obviously. This is the only one which is constant in magnitude). The OTHER three vectors are defined with respect to the flightpath of the aircraft. Thrust - the force produced by the powerplant in the forward direction along the flightpath. Drag - the component of the total aerodynamic forces on the aircraft in the reverse direction along the flightpath. Lift - the component of the total aerodynamic forces on the aircraft perpendicular to the flightpath. In any condition of unaccelerated flight the vector sum of forces on the aircraft is zero. This includes a constant rate climb. However, in a constant rate climb, because of the fact that we have an angle between the flight path and the horizontal, we have a vertical component of thrust, a horizontal component of lift, and a vertical component of drag. Generally speaking, no two are equal in magnitude in a constant rate climb. Perhaps the area of confusion implicit in the question and answer is a little sloppy thinking about mechanics: it doesn't take any more force to raise an object vertically than to hold it level (except the initial vertical acceleration). But to exert the same force through a distance requires that work be done. ___________ Oscar Strohacker Commercial,Instrument,ASMEL
ron@brl-tgr.ARPA (Ron Natalie <ron>) (09/18/85)
> I have received a number of correct answers to the question "During a > constant airspeed climb what is the relationship between the four primary > forces of flight - lift, drag, thrust and weight". They are equal. The > "trick" is why does the airplane climb? > Inertia.
foster@nsc.UUCP (Jerry Foster) (09/18/85)
>I have received a number of correct answers to the question "During a >constant airspeed climb what is the relationship between the four primary >forces of flight - lift, drag, thrust and weight". They are equal. The >"trick" is why does the airplane climb? Antigravity? "Beam me up Scotty, my vectors are all equal!"
rick@apple.UUCP (Rick Auricchio) (09/19/85)
-------- [] >In article <30512@lanl.ARPA> ths@lanl.ARPA writes: >I have received a number of correct answers to the question "During a >constant airspeed climb what is the relationship between the four primary >forces of flight - lift, drag, thrust and weight". They are equal. The >"trick" is why does the airplane climb? > I seem to recall it being an "excess of horsepower". -- {decwrl,hplabs,ihnp4}!nsc!apple!rick N1150G (408) 973-4227 Reality is an illusion caused by alcohol deficiency. Opinions expressed are not necessarily those of my employer.
notes@ucf-cs.UUCP (10/01/85)
the 4 forces are NOT equal in the same way as in level flight. L=W*Cos(c) T=D + W*SIn(c) where c is the climb angle, hence more the engine thrust/power the better the climb. elementary my dear physicists. .