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I've worded the question in two ways, before and after context, both bolded text below.

Let's say you have some fly speed with average maneuverability. Speed is not important. You have a minimum forward speed (MFS) of half, an up angle of 60 degrees, and an up speed of half. Imagine you begin ascending. Aside from beginning your ascent (where you have to fly steady for half maximum distance before ascending), does MFS still apply if you continue ascending the next round? If minimum forward speed applies when ascending (moving at your "up speed"), then implicitly you need a full double move to ascend at the full 60 degrees, and in general you cannot ascend at 60 degrees (math below).

For 60 degrees, the standard trig triangle has a hypotenuse of 2 and a base of 1. In D&D terms, however far you fly at 60 degrees, you fly half that distance forward. Therefore, if your up speed is half, flying half your speed at 60 degrees means you fly only one-quarter your speed forward, which doesn't meet the minimum forward speed of half requirement.

It's not hard to see that flying at 60 degrees under this logic requires a double move, and you must use all your movement to achieve the minimum forward speed (a double move at half speed is full speed along 60 degrees, which means you moved half speed forward).

Otherwise for a single move action, you're capped at much smaller angles of ascension, if that's even technically possible within the rules. For example, with a fly speed of 60 and average maneuverability, you could still move foward 30 and up 15. Again, it doesn't really say whether or not you can ascend at lower angles, but I wouldn't see why not.

So, is this implicit logic correct, or is ascension supposed to be a different mode of movement whereby minimum forward speed shouldn't apply?

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Yes, minimum forward speed applies to all kinds of aerial movement. But minimum forward speed does not mean "minimal horizontal distance covered", it simply refers to the minimum movement you have to make in whatever direction - may it be up, down, left, right, ... ALso, if your up speed is half and your minimum forward speed is half, you only need to move a quarter of your normal fly speed to stay airborne (half of your already reduced speed). For instance: if your maneuverability is average and your normal flight speed is 60ft, your up speed will be 30ft and your minimal flight speed 15ft (as long as you keep moving upwards).

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  • \$\begingroup\$ Thanks, this answer seems a bit more sensible... Would this then imply that a mid-flight turn has to occur halfway during movement, so that you get half speed in both directions? \$\endgroup\$ – Erik Feb 21 at 15:19
  • \$\begingroup\$ I´m not quite sure, whether I understand you correctly, but: No. The movement you make (provided your maneuverability is average, poor or clumsy) is always forward, even if you fly in circles - a turn to the left or the right makes no difference. Only if you change between up, horizontal and downward movement during your turn, calculating minimum forward speed will become a bit more complicated. \$\endgroup\$ – Peregrin Took Feb 21 at 17:31
  • \$\begingroup\$ Looks like you've got my idea. It seems like they want to say "In horizontal, upward or downward flying, you must travel half your allowable distance"... if that's the case then it sounds like transitioning from horizontal to up/down movement requires a full move action (half in horizontal, half in up/down) \$\endgroup\$ – Erik Feb 21 at 19:51
  • \$\begingroup\$ I think, what you have to do is count the different parts of your flying movement against each other. E.g. let´s say you have a fly speed of 60 ft with average maneuverability. If you fly 20ft horizontal, you have spent 1/2 of your normal fly speed and have covered 2/3 of the 30ft normal (horizontal) minimum forward speed). If you then accend for at least another 10 ft (= 1/3 of your up speed (which is half your normal fly speed (=30ft)) you have maintained your minimum forward speed – all within one move action. \$\endgroup\$ – Peregrin Took Feb 22 at 12:38
  • \$\begingroup\$ Hmm, so take distance moved in each mode as a fraction of its maximum, add the fractions up. If the sum is less than 1/2, you fall. Seems a bit complicated, not sure if this one is really resolved \$\endgroup\$ – Erik Feb 22 at 18:45

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