25 August 2013 Beard and eye update

[RedwoodElf]

@warrl: The problem is the range of the featherfall spell:
it's too short ranged to allow are braking to safe speeds if cast by someone trying to catch you from the ground.
Cast on yourself no problem.
But in any case, you'd need to use the spell early enough to have quite a way to fall for the aerobraking to work.
And it might never be sufficient.

@Redwoodelf:
double posting is considered to be unnecessary on most forums. But I should be quiet about it....

Ah, first edition. Who still plays with that? So, no. The push spell in later editions will do nothing for featherfalled objects in regard to accelerating.
And push as in "and is pushed from the end it entered the tube toward the other end." could just be an arm pushing. Didn't register that in the next instance you added push spell.

Googled a bit more: http://mouseferatu.com/index.php/blog/w ... -spelling/ tells us, that in 2^nd edition mass was reduced, but speed too.
more of it:
1^st edition rule, as on this page:

Feather Fall (Alteration)
Level: 1 Components: V, M Range: 1"/level Casting Time: 1/10 segment Duration: 1 segment/level Saving Throw: None Area of Effect: Special
Explanation/Description: When this spell is cast, the creature(s) or object(s) affected immediately assumes the mass of a feathery piece of down. Rate of falling is thus instantly changed to a mere constant 2' per second or 12' per segment, and no damage is incurred when landing when the spell is in effect. However, when the spell duration ceases, normal rate of fall occurs. The spell can be cast upon the magic-user or some other creature or object up to the maximum range of 1" per level of experience of the spell caster. it lasts for 1 segment for each level of the magic-user. The Feather Fall affects an area of 1 cubic inch, and the maximum weight of creatures and/or objects cannot exceed a combined total equal to a base 2,000 gold pieces weight plus 2,000 gold pieces weight per level of the spell caster. Example: a 2nd level magic-user has a range of 2", a duration of 2 segments, a weight maximum of 6,000 gold pieces (600 pounds) when employing the spell. The spell works only upon free-falling or propelled objects. It will not affect a sword blow or a charging creature, but it will affect a missile. The material component is a small feather or a piece of down somewhere on the person of the spell caster.

So, even in first edition speed is instantly reduced.

Only until the push spell is applied, at which point acceleration takes over...since the two spells are continuous within the tube and coterminous (end at precisely the same time) the final velocity would be maintained (Believe me, when we designed the thing we checked EVERYTHING in first edition rules to make sure it would work as advertised.) It's certainly more realistic than the "Peasant Railgun" as far as actual (Fantasy) physics is concerned. Realistically, a peasant could only hand an item at a rate of about one "pass" per segment. Since there are 10 segments in a round, the projectile would only go through the hands of 10 peasants in that round.

Here is the part of the math I can remember (using Minmax as a convenient guinea pig):

First Edition Minmax is falling down a very very deep pit...He has been falling for 10 seconds, when a Featherfall spell is cast on him.

For this calculation, let's assume Minmax weighs 250 LBS with all his weapons and gear.

After 10 seconds, Minmax's Velocity would be 320 Feet per second (neglecting air resistance for simplicity's sake), so his momentum at the time of casting , using the standard formula were momentum P equals Mass Times Velocity ( (P = MV) - to simplify, treating pounds as mass, though it is actually weight) is 80,000 Foot Pounds.

For the purposes of calculation, let's assume Featherfall reduces Minmax's mass to 1 LB (Pretty heavy for a feather, but it makes the calculation simpler)

Solving the momentum formula for Velocity, we get V = Momentum/Mass which, since we chose the convenient number of 1 for mass post-featherfall, means that after the Featherfall is cast on him, IF MOMENTUM is A CONSTANT, would mean that his velocity would be 80,000 feet per second. Since this is not what is observed (his velocity doesn't change until air resistance slows him down) we can conclude that velocity, not momentum, is a constant when a Featherfall spell is activated or deactivated.

Thus it is proved.



And "Double Posting" is posting the exact same thing twice, NOT responding to two different posts with two different replies. Learn your terms before typing, please, it makes the whole forum go a lot more smoothly.