Peregrinefalcon’s NASA quote explains it better;
The with zero excess power flies in Thrust = drag condition. If thrust is higher than drag (or lower), there will be a force (kg*m/s^2) on a moving object (m/s), there will be a rate of increase in energy in Joule/s, or kg*m^2/s^2 to write it in a more open form.
Thermodynamics law: Total energy of the system = Kinetic energy of system + Potential energy of system + internal energy of the system.
This law can apply to everything, and if we consider whole aircraft as this thermodynamic system, internal energy won’t change (ie, aircraft won’t heat up etc). So equation becomes E = KE+PE.
As a performance parameter, we are looking on rate of kinetic energy change or potential energy change, so we take the differential of both sides;
rate of energy change(dE) = dKE (rate of KE change) + dPE (rate of PE change).
Power = mass*Velocity*dVelocity + mass*G*dHeight.
dVelocity = rate of change in velocity, which means acceleration
dHeight = rate of change in height, which means climb rate.
divide both sides by mass and there is;
Specific Power = velocity*acceleration + G*climb rate.
Now as I’ve written above specific power has no meaning for the pilot; Some 1000 watt/kg is not a quantifiable measurement for him. So instead of giving such performance parameter, we divide both sides with G so we have
Climb-rate/potential (as performance parameter) = velocity*acceleration/G + climb rate (as in aircrafts actual change in height).
Now if you are flying at 300 m/s, and you have 200 m/s climb rate as written in the manual, this means you can climb at 200m/s without losing speed. You can surely go 90deg vertical and climb at 300m/s for an instant, but accordingly to this -scalar- energy equation you would be trading KE (slowing down) for this rate of increase in PE. Likewise, you can climb at 100m/s and accelerate at the same time (both KE and PE increase). Or you can stick at 0m/s climb rate (dPE = 0) and climb rate will mathematically give the level flight acceleration of the aircraft. On descent rate of PE change is also added to the rate of KE change.
Even if you are flying at 309 m/s, an aircraft can surely have 345m/s climb rate. A SAM recently launched and flying 20-30 m/s can even have 500+ m/s climb rate. This climb rate is a mere representation of energy change of aircraft, which pilot can use in any combination of KE and PE change.
If climb rate is equal to aircraft’s speed, this means aircraft can climb at 90 degree angle and won’t speed up or slow down. If its higher, than this excess climb-potential will result in KE change, which means acceleration on vertical climb.
Despite it says “climb rate” graph its solely related with energy and should be read with caution: if a MiG-29 wants to make a quick climb from 11 km altitude, actual climb rate won’t be highest around 180m/s @M1,7. It will be highest at M2,3 @ 678 m/s when MiG-29 makes an exact 90 degree vertical climb (assuming no direction change delays). But since MiG-29 has only 60m/s climb rate at that speed, it will see a quick deceleration equivalent to (60-678) m/s climb rate; 60-678 = v*a/g = -8,94 m/s^2 deceleration. Assuming constant deceleration, it will slow down to below M2,0 just within 10 seconds, but will have 633 m/s average climb rate during that time, to ascend it to 17km altitude.
For aerodynamics 101 this does not matter, as people are teached without this background, examplified by something with puny climb rates like B-737 or something, so certain know-it-all people can easily calculate angle from simple trigonometry. Its perfectly ok, but gets unamusing when they draw conclusions and making assumptions about all aircraft should climb at 60 degrees so climb rate will give an imaginary speed and draw even more conclusions from that.
Climb rate/potential is also variable with dynamic thrust, L/D and how much G aircraft pulls (which is also independent of aircraft weight), and its shown as lines for Energy-maneuverability diagram F-16 manual for example. In a sense, that diagram doesn’t only show ITR and STR, it also shows the vertical play and acceleration capability of the F-16 as well. As energy is a scalar quantity (instead of vectoral), one an use this graph to say F-16 at Mach A pulling B amount of Gs can climb/descent C amount of meters and accelerate/decelerate by D amount.
Nope, up to the bit in bold you were doing so well.
You are trying to conflate SEP and Climb Rate.
rate of energy change(dE) = dKE (rate of KE change) + dPE (rate of PE change).
I.e.
It is not a measure of potential to climb but potential to climb + potential to accelerate.
You can’t turn it all to climb rate unless you satisfy V*sin(alpha) = hdot.
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