Originally Posted by dangermouse
as a first order approximation.
kinetic energy required to keep the rotor going round needs to come from somewhere when the motors stop (this value is pretty fixed regardless of aircraft mass or collective pitch)
KE =1/2xMxVxV where M is the rotor mass & V is rotor speed
an aircraft at height has potential energy
PE=MxGxH where M is aircraft mass, G is a fixed constant, H is height (so a heavy helicopter has at the same height more potential energy than a light one)
therefore converting aircraft potential energy (variable with mass) into rotational kinetic energy (fixed by rotor mass) requires a loss of height.
if the rotational power is fixed (ie rate of energy use) the rate of potential energy change is also fixed, so an aircaft with more mass must lose less height in a fixed time than a light one for the same energy conversion rate, ie a heavy aircraft will lose less height in the same time than a light one when in autorotation as can be demonstrated
This is the same explanation that I use when someone asks me about this, but it skips one important point (highlighted). Energy use must actually be higher in the heavier machine, because it requires more lift to control its rate of descent - or in other words, more pitch = more drag = more power consumed by the rotor.
For some reason, the additional energy of the extra mass outweighs the higher power consumption of the rotor. I suspect there is no easy way to show why that is the case, and a mathematical explanation might be necessary to really get to the bottom of this.
I also suspect that the relationship between weight and ROD is not consistent, because the efficiency of the rotor system will change at different AOA. Higher weight will only reduce ROD up to a certain point, and there must be a "best weight" for autorotation for any given helicopter.