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The Art of the Helicopter John Watkinson - Karatunov.net

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38 <strong>The</strong> <strong>Art</strong> <strong>of</strong> <strong>the</strong> <strong>Helicopter</strong><br />

Fig. 2.18 A body following a circular path is at constant acceleration and if this is calculated, it can be used<br />

to derive <strong>the</strong> forces acting. (a) shows that after a short time δT, <strong>the</strong> velocity has changed direction. (b) <strong>The</strong> rate<br />

<strong>of</strong> change <strong>of</strong> velocity is <strong>the</strong> acceleration. (c) shows that if δT is allowed to fall to zero, <strong>the</strong> acceleration points<br />

to <strong>the</strong> centre <strong>of</strong> rotation.<br />

becomes vanishingly small this vector will be seen to point to <strong>the</strong> axis <strong>of</strong> rotation as<br />

can be seen in (c).<br />

As <strong>the</strong> RPM increases, <strong>the</strong> length <strong>of</strong> <strong>the</strong> velocity vectors in Figure 2.18 increases in<br />

proportion to RPM, as does <strong>the</strong> angle between <strong>the</strong>m. As <strong>the</strong> resultant for small angles<br />

is <strong>the</strong> product <strong>of</strong> <strong>the</strong> vector length multiplied by <strong>the</strong> angle between <strong>the</strong>m, it is easy to<br />

see that <strong>the</strong> acceleration or rate <strong>of</strong> change <strong>of</strong> velocity is proportional to <strong>the</strong> square <strong>of</strong><br />

<strong>the</strong> RPM. <strong>The</strong> advantage <strong>of</strong> <strong>the</strong> use <strong>of</strong> ω to measure angular velocity is that if this is<br />

done <strong>the</strong> velocity at any radius r is simply ω × r and <strong>the</strong> acceleration is simply ω 2 × r.<br />

<strong>The</strong>re are no constants or conversion factors to remember which is a major advantage<br />

<strong>of</strong> <strong>the</strong> MKS metric system.<br />

It was shown above that F = m × a, it can be seen that <strong>the</strong> inward or centripetal<br />

(Latin: centre seeking) force needed to accelerate an object in a circular path is simply:<br />

F = m × ω 2 × r as ω = v/r <strong>the</strong>n F = m × v 2 /r<br />

When a mass moves along, or translates, it has ki<strong>net</strong>ic energy. When an object rotates,<br />

it also has ki<strong>net</strong>ic energy, but <strong>the</strong> amount is less easy to calculate. This is because all<br />

elements <strong>of</strong> a translating mass move at <strong>the</strong> same velocity, whereas in a rotating mass<br />

different elements move at different velocities according to how far from <strong>the</strong> axis <strong>the</strong>y<br />

are. By adding up all <strong>of</strong> <strong>the</strong> ki<strong>net</strong>ic energies <strong>of</strong> blade elements at different velocities, <strong>the</strong><br />

ki<strong>net</strong>ic energy <strong>of</strong> <strong>the</strong> rotor can be found. If an elemental mass δm is rotating at radius r<br />

with angular velocity ω, its velocity is ω × r and so its ki<strong>net</strong>ic energy must be:<br />

δm × ω 2 × r 2<br />

2<br />

An arbitrary body can be treated as a collection <strong>of</strong> such masses at various radii and<br />

<strong>the</strong> integral <strong>of</strong> <strong>the</strong> ki<strong>net</strong>ic energies <strong>of</strong> all <strong>of</strong> <strong>the</strong>se gives <strong>the</strong> total ki<strong>net</strong>ic energy. If this<br />

is divided by <strong>the</strong> square <strong>of</strong> <strong>the</strong> angular velocity <strong>the</strong> result is <strong>the</strong> moment <strong>of</strong> inertia;<br />

<strong>the</strong> rotational equivalent <strong>of</strong> mass. Any rotating body could have <strong>the</strong> same moment<br />

<strong>of</strong> inertia if all <strong>of</strong> its mass instead were concentrated at one radius from <strong>the</strong> axis <strong>of</strong><br />

rotation. This is known as <strong>the</strong> radius <strong>of</strong> gyration.<br />

<strong>The</strong> rotor blade requires an inward or centripetal force to accelerate it into a circular<br />

path. Each element <strong>of</strong> <strong>the</strong> blade is at a different radius and so calculating <strong>the</strong> overall

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