Kiwiineurope From New Zealand, joined Sep 2004, 5 posts, RR: 0 Posted (9 years 7 months 4 days 5 hours ago) and read 24709 times:
I hope this is the right place to post this, since I don't see many threads on private-pilot type questions, but I did a good search through the archive for an answer...
I am trying to fathom out the practical difference between this text book's explanation of rate of climb versus angle of climb. I have no problem with understanding the practical difference, i.e. one you gain more height in a given time,t, and the other you trade that off for more horizontal distance, i.e. a greater TAS.
What I don't understand is why rate of climb is defined has having the greatest excess power over the power required and maximum angle of climb is defined as having the greatest excess of thrust over the drag. Aren't they both pragmatically speaking a matter of excess thrust over the minimum required to counteract drag? Power would only come if to it when you're talking about climb performance in general, not highlighting the difference between angle and rate.
FredT From United Kingdom, joined Feb 2002, 2185 posts, RR: 26
Reply 1, posted (9 years 7 months 4 days 5 hours ago) and read 24651 times:
If your text book is any good, it derives the equations.
If it's not, either do a web search or do it yourself. It's not complex. Just solve for thrust, weight, drag and lift for an aircraft in climb. Assume the thrust to be parallell to the direction of flight and remember that you can work in either the aircraft reference system or in the world reference system and the formulae will pop out.
If you run into difficulties, just come back here and ask.
I think I posted the derivations here a while back... or if it was somewhere else. I'd redo them but I haven't got the time at the moment.
I thought I was doing good trying to avoid those airport hotels... and look at me now.
777236ER From , joined Dec 1969, posts, RR:
Reply 2, posted (9 years 7 months 4 days 4 hours ago) and read 24620 times:
Vertical speed = (Thrust - Drag) x TAS / Weight
(Thrust - Drag) x TAS is just the power, so to have the maximum vertical speed (rate of climb) you need a maximum power.
Sine of the climb angle is (thrust - drag) / weight, so to get the biggest angle you need the highest value of (thrust - drag). Because the TAS doesn't come into this equation, we're talking about a force, not power.
Timz From United States of America, joined Sep 1999, 6836 posts, RR: 7
Reply 3, posted (9 years 7 months 4 days 2 hours ago) and read 24586 times:
Just do a simple example. Aircraft A has a net thrust (actual thrust minus drag) of 400 units (Newtons, pounds, whatever) at speed 100 units (m/sec, miles/hr, whatever). Aircraft B has a net thrust of 200 at speed 300; the two aircraft weigh the same.
Aircraft A will climb steeper, right? And Aircraft B will gain altitude faster?
Flyer737sw From United States of America, joined Oct 2004, 135 posts, RR: 2
Reply 4, posted (9 years 7 months 4 days 1 hour ago) and read 24585 times:
Its as simple as:
Best Rate of climb is over time...
Best Angle of climb is over lateral distance...
There are many factors in climb performance in an airplane...Density altitude, Weight of the aircraft, and terrain obstructions...
Max angle of climb would be used more for terrain obstructions...For example I flew into KRNO airport that has a field elevation of about 4412 feet...There are very steep mountains(10,000 feet) to the west side of the airport which create a problem...In calculating climb performance I opted a greater angle of climb becuase of this...
Best rate of climb would be used more for economy climb...
Another thing to look at as well, is if you were to have an engine failure during climb out...I prefer to use best angle since i will gain the most altitude over lateral distance...In the event of a engine failure that extra altitude might be enough to get back to the airport or an alternate...Just food for thought...
727EMflyer From United States of America, joined Mar 2005, 547 posts, RR: 0
Reply 5, posted (9 years 6 months 2 weeks 5 days 18 hours ago) and read 24333 times:
Visualize (or look at) a diagram of how Vx and Vy change with altitude. The point where they are equal is the aircraft's ceiling, right? Ok, now analyze why that is... to hell with the numbers. At Vy your extra thrust is producing extra lift, thus making you climb. At Vx, the lower airspeed is producing just enough lift to keep the airplane up, it is actually thrust pulling the airplane higher. When you finally get to the ceiling, Vx=Vy, you can't go higher. You will find Vx is very close to stalling speed, so if you were to raise the nose to get more vertical thrust, the increased drag would cause you to loose horizontal speed (actually airspeed, but visualize it as horizontal), thus loosing lift, thus losing any altitude you might have gained. Now, if you lower the nose to gain airspeed and climb with lift you will actually be driving the plane down. Air mass flowing over the wings creates the lift, and the mass at altitude is so low that you need maximum lift to maintain altitude.... lift increases at a higher angle of attack, so if you lower the nose, thus lowering the angle of attack, you lose the lift your higher speed would have generated.
Make sense? I know it is hard to visualize four dimensions interacting with each other, and the math only makes sense in numbers. I hate numbers. If you are a pilot you can demonstrate why Vy works to yourself using a soft field take off, or going around on landing. You will push the nose down to build the speed and it will feel like you should fly into the ground, but actually you climb. To demonstrate why Vx works, try a power-on stall. It is hard to do because the thrust is keeping you in the air. If you are not a pilot, get in touch with one and go for a ride in the cockpit and have him demonstrate. Better yet, make that pilot a CFI and start learning!