3.1 Revised lift equation

In this section, we compute the lift generated by a helicopter rotor blade with the modified equation

$$ \begin{align*} L_{\text{mod}}(u, \theta) = \int_{0}^{R} \frac{\rho c C_{L}}{2}(\Omega r + u\cos(\theta))|\Omega r + u\cos(\theta)| \,dr. \end{align*} $$

Similar expressions can be found in [Reference Johnson2, Reference Leishman5]. Expressing the velocity term as $\nu |\nu |$ rather than $\nu ^{2}$ allows the term to become negative, as experienced by retreating blades at high velocity. Upon explicit calculation,

$$ \begin{align*} L_{\text{mod}}(u, \theta) &= \frac{\rho c C_{L}}{6\Omega}[(\Omega R + u\cos(\theta))^{2} |\Omega R + u\cos(\theta)| - u^{3}\cos^{2}(\theta)|\!\cos(\theta)|]. \end{align*} $$

3.1.1 Retreating blade special case

For retreating blades ( $\theta = 180^{\circ }$ ), the lift generated simplifies to

$$ \begin{align*} L_{\text{mod}, \, R}(u) = \frac{\rho c C_{L}}{6\Omega}[(\Omega R - u)^{2}|\Omega R - u| - u^{3}]. \end{align*} $$

Given $\Omega =(\nu - u)/R$ to ensure maximum RPM, the lift becomes

$$ \begin{align*} L_{\text{mod}, \, R}(u) = \frac{R\rho c C_{L}}{6(\nu - u)}[(\nu - 2u)^{2}|\nu - 2u| -u^{3}]. \end{align*} $$

To find u such that $L_{R}(u) = 0$ , we solve

$$ \begin{align*} [(\nu - 2u)^{2}|\nu - 2u| - u^{3}] = 0. \end{align*} $$

Given $u < \nu /2$ ,

$$ \begin{align*} u = \frac{\nu}{3}. \end{align*} $$

Given $\nu \kern1.3pt{=}\kern1.3pt 343\,\text {m}\text {s}^{-1}$ , we see that traditional helicopters cannot exceed ${\nu /3 \kern1.3pt{\approx}\kern1.3pt 114.3\,\text {m}\text {s}^{-1}}$ . This limit is slightly higher than the known world speed record of $111.35\,\text {m}\text {s}^{-1}$ by the Westland Lynx set in 1986 [Reference Fay1].

3.1.2 Negative lift region

If $u> \nu /3$ , a region of the rotor disk about the tip of the retreating blade will produce negative lift when integrated. To find this region, we solve $L(u, \theta _{c}) = 0$ for $\theta _{c}$ given $u> \nu /3$ . That is, we solve

$$ \begin{align*} \frac{\rho c C_{L}}{6\Omega}[(\Omega R + u\cos(\theta_{c}))^{2}|\Omega R + u\cos(\theta_{c})| - u^{3}\cos^{2}(\theta_{c})|\!\cos(\theta_{c})|] = 0. \end{align*} $$

Letting $\epsilon = \cos (\theta _{c})$ and using $\Omega = (\nu - u)/R$ ,

$$ \begin{align*} (\nu - u(1 - \epsilon))^{2}|\nu - u(1 - \epsilon)| - u^{3}\epsilon^{2}|\epsilon| = 0. \end{align*} $$

Since $\epsilon < 0$ , by expansion and factoring,

$$ \begin{align*} \bigg(\frac{\nu - u}{u}\bigg)(2\epsilon + \nu)\bigg(\epsilon - 2\frac{\nu - u}{u}\bigg) = 0. \end{align*} $$

Therefore, we have $\epsilon = (\nu - u)/2u$ . Upon substitution,

$$ \begin{align*} \theta_{c} = \arccos\bigg(\frac{\nu - u}{2u}\bigg). \end{align*} $$

Therefore, the helicopter blades will produce negative lift for azimuth angles, satisfying

$$ \begin{align*} \theta \in \bigg[\pi - \arccos\bigg(\frac{\nu - u}{2u}\bigg), \, \pi + \arccos\bigg(\frac{\nu - u}{2u}\bigg)\bigg]. \end{align*} $$

3.2 Modified lift equation

To recapture the loss of lift on the retreating rotor blades, we modify the lift equation to be

(3.1) $$ \begin{align} \ell_{\text{mod}} = \tfrac{1}{2}\rho c v|v| C_{L}, \end{align} $$

replacing $v^{2}$ with $v|v|$ . This allows the velocity of the blade to take negative values, which will occur on the retreating side for sufficiently high relative wind speeds. The lift distribution in static hover matches Figure 4 as $v = \Omega r$ . Given a relative wind u, we set $v = \Omega r + u\cos (\theta )$ as before. Figure 6 shows that the relative wind greatly affects the lift produced near the rotor shaft on the retreating side.

Figure 6 Modified lift distribution of the rotor disk given a relative wind $u = 50\,\text {m}\text {s}^{-1}$ (left) and $u = 120\,\text {m}\text {s}^{-1}$ (right).

Figure 7 Plot of the lift generated over time for several forward velocities.

3.2.1 Modified integrated lift equation

Using the same techniques as the standard lift equation, we may calculate the total lift produced by a helicopter rotor blade by integrating the lift over the blade. That is, we compute

$$ \begin{align*} L(\alpha, u, \theta) = \int_{0}^{R} \frac{1}{2}\rho c (\Omega r + u\cos(\theta))|\Omega r + u\cos(\theta)|C_{L}(\alpha) \,dr. \end{align*} $$

Upon computation,

$$ \begin{align*} L(\alpha, u, \theta) = \frac{\rho c C_{L}(\alpha)}{6\Omega}[(\Omega R + u\cos(\theta))^{2}|\Omega R + u\cos(\theta)| - u^{3}\cos^{2}(\theta)|\!\cos(\theta)|]. \end{align*} $$

With $\Omega = (\nu - u)/R$ , the lift becomes

$$ \begin{align*} L(\alpha, u, \theta) = \frac{R\rho c C_{L}(\alpha)}{6(\nu - u)}[(\nu - u + u\cos(\theta))^{2}|\nu - u + u\cos(\theta)| - u^{3}\cos^{2}(\theta)|\!\cos(\theta)|]. \end{align*} $$

Figure 7 plots the lift generated by the blade over time using the usual azimuth angle $\theta (t) = \Omega t$ . The lift oscillates as the blade transitions between advancing and retreating. As the forward velocity increases, the oscillation amplitude increases and the centre decreases. As the velocity may take negative values with the expression $\nu |\nu |$ , we see the special case $u = 120\,\text {m}\text {s}^{-1}$ produces negative lift at the retreating side.

To ensure $L = mg/n$ for vertical hover, the coefficient of lift must satisfy

$$ \begin{align*} C_{L}(\alpha) = \frac{6\Omega m g}{\rho c n}[(\Omega R + u\cos(\theta))^{2}|\Omega R + u\cos(\theta)| - u^{3}\cos^{2}(\theta)|\!\cos(\theta)|]^{-1}. \end{align*} $$

With $\Omega = (\nu - u)/R$ , the coefficient of lift becomes

$$ \begin{align*} C_{L}(\alpha) &= \frac{6\eta(\nu - u) m g}{\rho R c n}\\ &\quad\times [(\eta(\nu - u) + u\cos(\theta))^{2}|\eta(\nu - u) + u\cos(\theta)| - u^{3}\cos^{2}(\theta)|\!\cos(\theta)|]^{-1}. \end{align*} $$

Figure 8 is a plot of the coefficient of lift over time given the standard azimuth angle $\theta (t) = \Omega t$ . For static hover and the relatively low forward velocity $50\,\text {m}\text {s}^{-1}$ , we see the coefficient of lift take reasonable values between $0$ and $1$ . For the forward velocity $120\,\text {m}\text {s}^{-1}$ , we see that the negative lift produced on the retreating side leads to discontinuities the coefficient of lift cannot reconcile. The angle of attack is, therefore, unable to balance the lift on the retreating side for forward velocities $u> \nu /3$ .

Figure 8 Plot of the lift coefficient $C_{L}$ required for vertical hover against time.

3.3 Average modified lift

To better understand the lift performance of the helicopter blade over the entire rotor disk, we also compute the average lift produced over time $\lambda $ with the expression

$$ \begin{align*} L_{\text{mod}, \lambda} = \frac{1}{\lambda}\int_{0}^{\lambda} \int_{0}^{R} \frac{1}{2} \rho c (\Omega r + u\cos(\Omega t))|\Omega r + u\cos(\Omega t) |C_{L}(\alpha) \,dr \,dt. \end{align*} $$

We compute this integral numerically with a standard midpoint Riemann sum with 300 evenly spaced intervals in Figure 9. As the forward velocity increases, the negative lift produced by the retreating side lowers the average lift across the disk. A higher forward velocity results in a more significant difference.

Figure 9 Plot of the average lift $L_{\text {mod}, T}$ over time interval $\lambda $ .

3.4 Average lift across rotor disk

Since the helicopter rotor speed $\Omega $ is dependent on u, the time interval required to consider one revolution is also dependent on u. Therefore, for the standard lift equation, we compute the average value over one particular revolution with the alternative integral

$$ \begin{align*} L_{\text{std}, T}(u) = \frac{1}{2\pi}\int_{0}^{2\pi} \int_{0}^{R} \frac{1}{2}\rho c \bigg(\frac{\nu - u}{R}r + u\cos(\theta)\bigg)^{2}C_{L} \,dr\, d\theta. \end{align*} $$

Upon simplification,

$$ \begin{align*} L_{\text{std}, T}(u) = \frac{\rho c R C_{L}(\alpha)}{12}[5u^2 -4\nu u + 2\nu^2]. \end{align*} $$

For the modified lift, we compute

$$ \begin{align*} L_{\text{mod}, T}(u) = \frac{1}{2\pi}\int_{0}^{2\pi} \int_{0}^{R} \frac{1}{2}\rho c \bigg(\frac{\nu - u}{R}r + u\cos(\theta)\bigg)\bigg| \frac{\nu - u}{R}r + u\cos(\theta) \bigg|C_{L} \,dr\, d\theta. \end{align*} $$

Finally, to reconcile the negative lift from the modified velocity model $\nu |\nu |$ , we also compute the integral

$$ \begin{align*} L_{\max, T} = \frac{1}{2\pi}\int_{0}^{2\pi} \int_{0}^{R} \max\bigg(\frac{1}{2}\rho c \bigg(\frac{\nu - u}{R}r + u\cos(\theta)\bigg) \bigg| \frac{\nu - u}{R}r + u\cos(\theta) \bigg|C_{L}, 0\bigg) \,dr\, d\theta. \end{align*} $$

To better understand the lift differences, we scale the result by the lift obtained in static hover. From Figure 10, we find the models produce similar average lifts for low forward velocity as the effect of dissymmetry of lift is at its lowest influence.

Figure 10 Plot of the normalized average lift over forward velocity for each model.

As the forward velocity increases, the lift produced by the blade decreases as the rotor speed $\Omega $ must decrease. Once the negative lift is produced by the retreating blade at $u = \nu /3$ , we find the lift from the standard and modified model diverge. The standard model increases lift as the forward velocity remains nonnegative. For the modified model, the overall lift decreases as the negative lift on the retreating blades counteracts the positive lift produced by the advancing sides. If the angle of attack can be set to $0^{\circ }$ , the negative effect of the retreating blade diminishes and the lost lift begins to recover.

Since the retreating side of the rotor disk cannot produce any positive lift for velocities above $114.3\,\text {ms}^{-1}$ , we find that only coaxial helicopters are able to rectify the issue. We also note the lift produced in these velocities varies between $50\%\ \mathrm{and}\ 75\%$ of the lift generated in static hover. This is primarily a consequence of the decreased rotor speed $\Omega $ as u increases.

3.5 Minimum rotor speed requirements

To obtain the maximum lift attainable from helicopter blades, we prescribe $\Omega $ with equation (1.1) to the lift equation (3.1). To find the minimum requirements for $\Omega $ to hold a given lift L, we estimate (to enable us to give an approximate analytic formula) the average lift by

$$ \begin{align*} L_{\text{avg}}(u, \Omega) = \frac{1}{4}\bigg[L(u, 0, \Omega) + 2L\bigg(u, \frac{\pi}{2}, \Omega \bigg) + L(u, \pi, \Omega)\bigg], \end{align*} $$

which simplifies to

$$ \begin{align*} L_{\text{avg}}(u, \Omega) = \frac{\rho c}{24\Omega}[(\Omega R + u)^{3} + (\Omega R - u)^{2}|\Omega R - u| + 2\Omega^{3}R^{3} - 2u^{3}]. \end{align*} $$

Solving for $\Omega $ ,

(3.2) $$ \begin{align} \Omega(u, L) = \left\{ \begin{aligned} &2^{-2/3}\dfrac{2^{7/3}L - 2^{1/3}(Rc\rho u^{2}) + ( (Rc\rho u^{2})^{3/2} + K)^{2/3}}{R\sqrt{Rc\rho}((Rc\rho u^{2})^{3/2} + K )^{1/3}}, & u < \sqrt{\frac{3L}{c\rho R}} , \\[.1cm] & -\dfrac{3u}{2R} + \dfrac{\sqrt{3c\rho R(16L + 3c\rho Ru^{2})}}{2c\rho R^{2}}, & u \geq \sqrt{\frac{3L}{c\rho R}}, \end{aligned} \right. \end{align} $$

where $K = \sqrt {-128L^{3} + 96(Rc\rho u^{2})L^{2} - 24(Rc\rho u^{2})^{2}L + 3(Rc\rho u^{2})^{3}}$ .

From Figure 11, we see that the minimum threshold for $\Omega $ decreases as forward velocity u increases. While this will allow the rotor disk to spin at a lower RPM, the maximum threshold imposed by equation (1.1) decreases at a faster rate. Consequently, the interval between the maximum and minimum values for $\Omega $ decreases in length until they coincide at some $u = u_{M}(L)$ dependent on the lift L required by the minimum threshold.

Figure 11 Plot of minimum $\Omega $ for various weights with the theoretical maximum against forward velocity.

By solving $\Omega (u_{M}, L) = (\nu - u_{M})/R$ , we are able to find the maximum allowable speed limit as imposed by the minimum lift L that the rotor disk(s) must generate. Assuming $u_{M}> 3L/c\rho R$ , we find the maximum speed limit is given by

(3.3) $$ \begin{align} u_{M}(L) = \frac{\nu}{4} + \frac{\sqrt{3Rc\rho(3Rc\rho\nu^{2} - 32L)}}{4Rc\rho}. \end{align} $$

In Figure 12, we plot the maximum speed limit as a function of lift required for a variety of helicopter blade lengths. If the forward velocity of a helicopter should exceed this limit (if $u> u_{M}$ ), the rotor angular velocity $\Omega $ must decrease, and the lift produced will not be sufficient to maintain altitude. To compensate for this, other methods for producing lift will be necessary, such as wings or a modified fuselage shape.

Figure 12 Plot of maximum speed $u_{M}$ against lift L for several blade lengths.

We note in equation (3.3) that $u_{M}$ attains its minimum at $L = 3Rc\rho \nu ^{2}/32$ , where $u_{M} = \nu /4$ .

3.6 Lift on the advancing side of the rotor disk

To study the effect of the relative wind on the advancing side of the rotor disk, we calculate the integral

$$ \begin{align*} L_{\text{adv}}(u) = \int_{-\pi/2}^{\pi/2} \int_{0}^{R} \frac{1}{2\pi}\rho c (\Omega r + u\cos(\theta))|\Omega r + u\cos(\theta)| \,dr \,d\theta. \end{align*} $$

Since $\Omega r + u\cos (\theta )> 0$ for advancing rotor blades and the integral is even, we may simplify the expression to

$$ \begin{align*} L_{\text{adv}}(u) = \frac{2}{\pi} \int_{0}^{\pi/2} \int_{0}^{R} \frac{1}{2}\rho c (\Omega r + u\cos(\theta))^{2} \,dr \,d\theta. \end{align*} $$

Upon simplification,

$$ \begin{align*} L_{\text{adv}}(u) = \frac{R\rho c}{12\pi}[3\pi u^{2} + 12\Omega R u + 2\pi(\Omega R)^{2}]. \end{align*} $$

In Figure 13, we plot $L_{\text {adv}}$ against u for the maximum theoretical $\Omega $ prescribed by equation (1.1) and for the minimum values for $\Omega $ required to maintain altitude for $1000$ kg, $2000$ kg and $3000$ kg helicopters. As the forward velocity increases, it reaches the maximum allowable limit $u_{M}$ , at which point the lift generated coincides with the lift obtained by the maximal rotor angular velocity $\Omega $ .

Figure 13 Plot of the average lift over the advancing half of the helicopter rotor disk against forward velocity.

To elucidate the effect of the angle of attack on the advancing side of the blade, we multiply the lift by the coefficient of lift using the maximal values for $\Omega $ . In Figure 14, we see lower angles of attack predictably produce less lift up to the stall threshold $\alpha _{c} \approx 10^{\circ }$ . When the angle of attack exceeds this limit, the lift begins to decrease, as expected with retreating blade stall.

Figure 14 Plot of the average lift over the advancing half of the helicopter rotor disk against forward velocity for several angles of attack.

To additionally incorporate the effect of drag, we alter the integral on the advancing blades with the proportion $(C_L(\alpha ) - C_D(\alpha ))$ . Given that drag increases sharply with the angle of attack, we expect an optimal angle of attack that allows the greatest lift coefficient under the drag coefficient, as shown in Figure 15.

Figure 15 Plot of the lift coefficient and the difference between lift and drag against the angle of attack.

Figure 16 shows the lift on the advancing side with the coefficient of drag included in the computation. Though the effect of the drag is slight, the increased angle of attack magnifies the gap between the lift at $\alpha = 9^{\circ }$ and $\alpha = 12^{\circ }$ .

Figure 16 Plot of the average lift over the advancing half of the helicopter rotor disk against forward velocity for several angles of attack.

3.7 Lift on the retreating side of the rotor disk

To obtain the lift generated on the retreating side of the rotor disk, we calculate

$$ \begin{align*} L_{\text{ret}}(u) = \frac{1}{\pi}\int_{{\pi}/{2}}^{{3\pi}/{2}} \int_{0}^{R} \frac{1}{2}\rho c (\Omega r + u\cos(\theta))|\Omega r + u\cos(\theta)| \,dr \,d\theta. \end{align*} $$

Given that $\Omega r + u\cos (\theta )$ is not necessarily nonnegative nor nonpositive, there is no simplification. To calculate this integral, we use MAPLE’s built-in numerical integration with a partition of $100$ points. In Figure 17, the retreating lift can be seen to rapidly decrease. Without measures to nullify negative lift, the retreating side continues to build negative lift as forward velocity increases. For coaxial helicopters, Figure 13 shows us that the lift increase from the advancing side is more than capable of balancing the lift loss on the retreating side.

Figure 17 Plot of the average lift over the retreating half of the helicopter rotor disk against forward velocity.

3.8 Angle of attack considerations for coaxial helicopters

In our previous mathematical derivations for the lift across the rotor disk (or particular regions of the rotor disk), we have assumed that the coefficient of lift constantly takes its maximum value $C_{L} = 1$ . In this section, we relax this assumption to derive conditions on the coefficient of lift that will enable the lift to be generated equally across the rotor disk.

For low forward velocities, we are able to apply the simple formula

$$ \begin{align*} C_{L}(\theta) = \frac{L}{L_{\text{std}}(u, \theta)}, \end{align*} $$

where $L_{\text {std}}(u, \theta )$ is the lift generated by the blade at forward velocity u and azimuth angle $\theta $ , given by

$$ \begin{align*} L_{\text{std}}(u, \theta) = \frac{\rho c}{6\Omega}[(\Omega R + u\cos(\theta))^{2}|\Omega R + u\cos(\theta)| - u^{3}\cos^{2}(\theta)|\!\cos(\theta)|]. \end{align*} $$

This expression remains valid until the lift generated on the retreating blade is no longer capable of producing sufficient lift. That is, the first threshold is the forward velocity $u_{1}$ such that

$$ \begin{align*} L_{\text{std}}(u_{1}, \pi) = L. \end{align*} $$

Assuming the rotor angular velocity is at its maximum, as shown by equation (1.1), we may simplify the above equation to show $u_{1}$ satisfies

$$ \begin{align*} u_{1}^{3} - \frac{4}{3}\nu u_{1}^{2} - \frac{2}{3}\bigg(\nu^{2} + \frac{L}{\rho c R}\bigg)u_{1} - \frac{\nu}{9}\bigg(\nu^{2} - \frac{6L}{\rho c R}\bigg) = 0. \end{align*} $$

For velocities $u> u_{1}$ , we set the angle of attack on the retreating blades to maximum and increase the angle of attack on the counterpart advancing blade to compensate for the insufficient lift generation.

3.8.1 Lift compensation for retreating blades

As the forward velocity u increases above the first threshold $u_{1}$ defined above, we must produce excess lift on the advancing blades to compensate for the lack of lift produced on the retreating blades. Suppose each blade must generate L lift. If $u < u_{1}$ , $L_{\text {std}}(u, \pi ) < L$ . Therefore, the excess lift E that the advancing blades must produce is given by

$$ \begin{align*} E = L - L_{\text{std}}(u, \pi). \end{align*} $$

In Figure 18, we see the lift compensation E that the advancing blade must produce for its counterpart retreating blade. The curves in Figure 18 are proportional to the maximum lift produced in hover that the highest weight helicopters may attain. Negative values indicate the retreating blade is producing sufficient lift, and compensation is not necessary. As the forward velocity increases, the critical values $E(u) = 0$ coincident with $u_{1}(L)$ are found. The lift compensation required also increases up to the critical value $\nu /3$ .

Figure 18 Plot of the average lift over the retreating half of the helicopter rotor disk against forward velocity.

Therefore, in the case that $u> u_{1}(L)$ , the coefficient of lift will take the form

$$ \begin{align*} C_{L}(\theta) = \begin{cases} \dfrac{2L - \min\bigg(\dfrac{L}{L_{\text{std}}(u, \theta + \pi)}, 1\bigg)L_{\text{std}}(u, \theta + \pi)}{L_{\text{std}}(u, \theta)}, & \theta \in \bigg[0, \dfrac{\pi}{2}\bigg]\cup\bigg[\dfrac{3\pi}{2}, 2\pi\bigg], \\ \min\bigg(\dfrac{L}{L_{\text{std}}(u, \theta)}, 1\bigg), & \theta \in \bigg[\dfrac{\pi}{2}, \dfrac{3\pi}{2}\bigg]. \end{cases} \end{align*} $$

3.8.2 Speed limit imposed on transitioning blades

Consider the lift produced by the blades as they transition between the advancing and retreating halves of the rotor disk (at azimuth angles $90^{\circ }$ and $270^{\circ }$ ). The lift produced on these blades is equal and unaffected by the forward velocity. That is,

$$ \begin{align*} L_{\text{std}}\bigg(u, \frac{\pi}{2}\bigg) = L_{\text{std}}\bigg(u, \frac{3\pi}{2}\bigg) = \frac{\rho c R^{3} \Omega^{2}}{6}. \end{align*} $$

Since the rotor angular velocity $\Omega $ decreases as forward velocity u increases, the lift produced on the transitioning blades decreases. For a given lift L, with the maximal rotor angular velocity $\Omega $ given in equation (1.1),

$$ \begin{align*} \frac{\rho c R}{6}(\nu - u)^{2} = L. \end{align*} $$

Solving for u,

(3.4) $$ \begin{align} u = \nu - \sqrt{\frac{6L}{\rho c R}}. \end{align} $$

We plot the maximum speed limit as a function of lift in Figure 19. As we increase the amount of lift the blade must produce, the maximum forward velocity at which the blades are able to maintain the lift decreases. Therefore, coaxial helicopters are subject to an additional speed limit in equation (3.4) as imposed by the reduced lift on the transitioning blades. Though using four blades per disk reduces the lift demand for each blade, the problem of reduced lift on transitioning blades remains, as seen in Figure 20.

Figure 19 Plot of the maximum speed limit imposed by equation (3.4) against lift.

Figure 20 Plot of the maximum lift produced by the blades at $\theta = 90^{\circ }$ and $\theta = 270^{\circ }$ .