I prefer to think of it in this way, which I find more intuitive than thinking about "negative lift" created by different angles of attack. If ((RPM * Prop Pitch)/1056) > Current airspeed, then the propeller will cause an increase in airspeed, inversely if ((RPM * Prop Pitch)/1056) < Current airspeed, then the prop will act as a "brake" because the work it is doing is less than required to maintain airspeed. Now, if the propeller is simply free-wheeling rather than the engine running at idle, the propeller will act as a break. Simple physics will tell us that you can't get more out than you put in. If the propeller is free-wheeling, the only energy input into the situation would be the force of the wind (no, I'm not forgetting gravity, but that really factors its way into the wind). Since the spinning of the propeller is not completely friction-free, there will be energy lost to heat, and the propeller will not be 100% efficient in its use of the energy. The problem most R/C pilots have is dealing with the difference between airspeed and groundspeed, when we think braking we think in terms of ground speed, but while the airplane is in the air, all we can control directly is airspeed. By the way, the 1056 is the factor required to convert distance travelled (in/min) into miles/hr. (60/(12*5280)). Now, to address davidej's question as to whether the spinning, idling prop has more drag than a stationary prop: it depends. If it is a situation where the airspeed/RPM comparison above is in equilibrium, I would say the drag is virtually equal to that of the stationary prop, but if the RPM side is less than the current airspeed the "braking" effect of the prop (for lack of a better term, drag) will be greater than that of a stationary prop because as stated in a previous post, the propeller while spinning has almost the same effect as having a flat plate mounted on the front of the airplane almost equal in diameter to the prop.