109 advice needed (climb)

To translate the comparison of Ivan's chart:

For ease of use I take the example of point P: TAS = 250 mph, turn at 3g.

Me109 (written on bottom line):

Phi = 71deg
time for 360deg turn = 25.5sec
turn radius = 1480 ft
angle of descent = 5.2deg (= 2000ft/min)

That means that for a 360deg turn at sustained speed the 109 would have lost about 850 ft.

Spit (type? some eye measure here so plus minus a tat):

Phi = 70.6deg
time for 360deg turn = ~26seg (attention: logarithmic scale!)
turn radius = ~1450-1500ft (I would have to measure it but I don't have a printer)
angle of descent = ~0.5deg (= ~ 190 ft/min)

That means that for a 360deg turn at sustained speed the spit would have lost about 80 ft.

This is albeit for 3g turns only. For a TAS of 250 mph a 3g turn is hence most energy preserving for the Spit. It would loose more energy in a tighter turn at this speed.

For 250 mph the 109 optimal turn load energy-wise would be about 2.3g but then it would also have increased its turn radius and increased its turn radius. Its 360deg turn time would then be about 35-38sec (+40% wrt Spit) and its turn radius 2000 ft (+30% wrt Spit).

Obviously the numbers for energy loss seem significantly different. However in a dogfight one usually never pulls a 360deg turn nor constantly at the same load. In a short duration turn the 109 should be able to turn with the Spit w/o loosing too much energy but of course not continuously.

The Spit also should loose energy if the combination velocity-turn load is above the energy-optimal line (the "Angle of Straight Climb" line). If it retains always energy then something is wrong.