Spir4,
First off,
power does
not equal
thrust multiplied by
time. Below are correct equations.
Work = force x distance
Power = work / time
Power = torque x angular velocity (e.g., RPM)
Torque = force x length of lever arm
Power = force x distance / time
Velocity = distance / time
therefore;
Power = force x velocity
therefore;
Power = thrust x TAS
You first need to understand all those equations. If you're unsure about one of them ask for clarification. This page might help:
http://en.wikipedia.org/wiki/Power_(physics)
Regarding piston engines:
The expansion of gas inside the cylinder creates a
force on the piston and moves it a certain
distance. Force x distance = work, meaning the burning of the mixture inside the cylinder does work on the piston. Since power = work / time, the shorter the time period (i.e., faster) the work is done in, the more power is required.
That power gets transferred into
torque as the connecting rod applies a force a certain distance from the center of the crankshaft. So torque is now being applied to the crankshaft. Torque overcomes the friction and begins moving the crankshaft at a certain speed (angular velocity [RPM]).
To calculate the
power created at the crankshaft you take the
torque applied to it and multiply it by the
RPM of the crankshaft.
The propeller then converts that power into thrust. The power from the engine is required to overcome the drag due to lift (thrust in this case), parasitic drag, and internal friction. When RPM is constant - the propeller-drag torque, plus internal friction, is equal to the torque applied to the crankshaft by the expanding gasses in the cylinder. Basically, the moments (torques) clockwise about the crankshaft center equal the moments (torques) counterclockwise about the crankshaft center.
Regarding climbs:
Forget about Vx and Vy for a moment -
every climb has a certain climb angle and climb rate! When you're thinking about excess thrust and excess power just think of any sort of climb - not Vx or Vy. What it's saying is that the angle of climb (regardless of what it is) is directly related to the excess thrust, and rate of climb (regardless of what it is) is directly related to excess power.
These two images are from Aerodynamics for Naval Aviators:
Climbing - Imgur
For the angle of climb bit, it's a simple analysis of the 4 forces in flight. At the top of Figure 2.21 you'll see how the forces are broken up for analysis. Essentially the more excess thrust you have, the more horizontal component of weight you're able to balance, meaning the steeper you can climb. Reading those pages I've attached should explain it quite clearly.
As for the rate of climb bit, you'll see at the top of Figure 2.21, beside the airplane, a triangle where rate of climb, velocity, and angle of climb are plotted. Essentially, for the same angle of climb, increasing the TAS (velocity) will increase the rate of climb directly. Since thrust is directly related to angle of climb, power = thrust x TAS. That's explained in more detail on the second image above.
In a combustion
engine, the power is related to the speed of the
engine (RPM). In the case of the
airplane, power is related to the speed of the
airplane. There is no difference in this relationship when comparing piston to jet performance. In the case of a propeller engine, the power is called "brake or shaft horsepower"; for an airplane, the power is called "thrust power" or typically "thrust horsepower" (THP) as it's generally displayed in units of horsepower. Piston, turboprop, and jet
airplanes all have THP and in all 3 cases it's the same - thrust x TAS.
The only difference happens because there is a difference in how a prop engine's thrust varies with TAS to how a jet's thrust varies with TAS. Since a jet's thrust is relatively constant throughout the range of TAS, the power (THP) will increase linearly with an increase in TAS. Notice that
ALL THP vs TAS graphs will start at 0,0. That is because at zero TAS, the THP is zero! Even though you're at the runway with the engine at full BHP/thrust, the velocity of the airplane relative to the air is zero, therefore, the THP is zero. It's that simple! I'm ignoring wind. As for the prop engines, the thrust will decrease with increasing TAS because prop efficiency goes down. Initially the decrease in prop efficiency (thrust) is not enough to overcome the increasing TAS, so THP will increase. Eventually, the decrease in thrust overcomes the increase in TAS and you'll notice the THP curve will taper to zero and then decrease.
Now going back to Vx and Vy:
If you want the
maximum angle of climb, you need the
maximum excess thrust. If you want the
maximum rate of climb, you need the
maximum excess power.
Additional reading:
US Navy - Thrust and Power
https://www.box.com/s/3302d62b8d269bfc96cb
Aerodynamics for Naval Aviators - Propeller Efficiency
https://www.box.com/s/86499db811c2ca977456