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THERE is, however, one method of considering the problem theoretically which is given tentatively here. Assuming the helicopter has its blade angle reduced so that it will auto‐rotate, the power required to sustain the screw is given by equation (10). This power must be precisely the same as that expended in dropping at its normal velocity Vv because, considering an interval of time t, there is a certain difference of potential energy between the two machines—one hovering, the other descending. In the same time t, therefore, there must be an equivalent input of energy to sustain the helicopter to balance the difference in potential energy. We are considering now only airscrew horse‐power, not engine horse‐power. The power expended in dropping at Vv is WVv. Therefore we have or Put in another form by means of the previous equations:

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