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Message: <blockquote data-quote="arscott" data-source="post: 1976389" data-attributes="member: 17969">The mistake we're making here is that we normally think of falls in terms of time, but d&d handles them in terms of distance.  falling acceleration in earth-normal gravity is about 32 feet per second squared*.  so if you fall for one second, then you'll fall 16 feet, but if you fall for twice that time, then you'll fall four  times that distance, or 64 feet.But if you fall for ten feet, then you've fallen for ten feet.  and if you fall for twice that distance, then youve only fallen twenty feet.Here's how it works:t is the time, in seconds, that you've been falling, starting a t = 0m is the mass of the object in question, in pounds masss(t) is you the distance you've fallen with respect to time, in feet, starting at s(0)=0v(t) is your velocity, in feet per second.  v(0)=0k(t) is your kinetic energy in feet squared per second squared times pounds massvelocity is the derivitive of positon with respect to timekinetic energy is velocity squared times mass over twoone of the major laws governing physics is the conservation of energy.  energy never increases or decreases, it just changes forms.at the starting point, all of your energy is potential.  That basically means that the energy used to raise the you above the floor is stored inside you, ready to be released when you begin to fall.Next is kinetic energy.  Thats the energy in your motion.After that it gets messy.  When you hit the ground, your kinetic energy is transformed into all kinds of energy.  The effects of that energy are represented by damage in the D&D world.So the amount of damage you take is proportional to your kinetic energy when you hit the ground.  And since we want to find out how your damage varies with respect to the distance you fall, we're trying to find the derivative of your kinetic energy with respect to your position.you position the distance you've fallen, sos(t) = 16t^2your velocity is the derivitive of that with respect to time, sov(t) = d/dt 16t^2v(t) = 32tyour kinetic energy is the square of your velocity times mass, sok(t) = (32t)^2*m/2k(t) = 32m * 16t^2but there's a 16t^2 in there.  that's the position. so we can just sayk(t) = 32m*s(t)we're trying to the derivitive of kinetic energy with respect to position, so we can just differentiate both sides with respect to s(t)dk(t)/ds(t) = d/ds(t) 32m*s(t)dk(t)/ds(t) = 32mthe derivitive is a constant, which means the relationship is linear.  If you fall twice the distance, you take twice the damage.  which is exactly how it's represented in d&d.What's interesting to note is that the mass doesn't cancel out.  Thus, the larger you are, the more kinetic energy you have.  But given that big creatures take no extra damage from Area effects despite taking up a greater portion of the effected area, you could argue that hit-points are already adjusted for size.Of course you could also argue that your gnome should only take 1/6th damage from the fall 'cause he's 1/6th the mass of everyone else, but I'll leave that up to you and your DM.*pardon my use of imperial units in a discussion of physics, but it's a hell of a lot easier this way 'cause the important figures are a power of two.  plus, that's what d&d uses.Edit:  Aaagh.  I spent three hours on this thing.  Now I remember why I switched majors to Theatre Arts.Edit two: I initially made an error regarding the equation for position, which is s(t) = 16t^2, not s(t) = 32t^2.  madsen this out.  It is now fixed.</blockquote>

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