Dear Friends,
I'm in need of your assistance, if you'd be so kind. I'm currently casting my gaze upon the flight model of the Me 163 "Comet". In this endeavour, I've delved into numerous books to gain a more precise understanding of the rocket engine's burn duration and to better comprehend its exact operational profile.
The latter is more or less identically described in all the documents, so for simplicity's sake, I'll quote
Wikipedia, where the launch and subsequent climb are described as follows:
At a speed of over 320 km/h (200 mph) the aircraft would take off, in a so-called "scharfer Start" ("sharp start", with "Start" being the German word for "take-off") from the ground, from its two-wheeled dolly. The aircraft would be kept at level flight at low altitude until the best climbing speed of around 676 km/h (420 mph) was reached, at which point it would jettison the dolly, retract its extendable skid using a knob-topped release lever just forward of the throttle48 (as both levers were located atop the cockpit's portside 120-litre T-Stoff oxidizer tank) that engaged the aforementioned pneumatic cylinder,31 and then pull up into a 70° angle of climb, to a bomber's altitude. It could go higher if required, reaching 12,000 m (39,000 ft) in an unheard-of three minutes.
This raises some questions for me. The Me 163, when fully fuelled and armed, has a take-off weight of 4309 kg. The fuel weighs a total of 2018 kg and according to the majority of sources, the burn duration is 7 1/2 minutes, or 450 seconds. This means that after 450 seconds, the Me 163 still weighs 2291 kg, or in other words, it loses about 4 1/2 kg of weight per second at full thrust. The engine has a thrust of 14.71 kN. At a climb angle of 70°, the (decreasing over time) weight force of the Me 163 counteracts this with the factor sin(70°). We thus obtain an accelerated motion with a starting speed of 676 km/h, a climb angle of 70°, and a linearly decreasing mass by about 4 1/2 kg with a starting weight of 4309 kg.
I'll spare you the detailed calculations (function for acceleration, speed (integral of acceleration), and distance (integral of speed)). The result is that at a climb angle of 70°, the speed reaches 0 after 32 1/2 seconds, and the Me 163 has climbed to just under 2900m during this time. And that's without even considering air resistance, which would actually further reduce the climb performance!
How on earth do all the documents on the Me 163 come up with the idea that the climb could be carried out at a constant 70°?
In order not to lose speed during the climb - again without considering air resistance, which would also significantly worsen the values here - the Me 163 would in reality only be allowed to climb at an angle of 20° at the beginning and 40° at the end of the burn duration (thanks to the then lower weight). An optimal climb to combat altitude would thus, taking into account air resistance, start at about 15° climb angle and slowly increase to about 30°.
This is because the engine has a thrust of 14.71 kN, which corresponds to a thrust force of 1500 kg. This 1500 kg must not be exceeded by the current mass of the Me 163 during the climb at the respective angle. This doesn't even work out with the empty weight of the Me 163 at a climb angle of 70°...
Question to the physicists among you: Have I overlooked something or have the documents on the Me 163 simply all copied the same nonsense from each other?
Mike