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→Best Glide
# Increasing the aspect ratio of the wing is an effective (and, in fact, easiest) way to improve the best performance, as the best glide ratio scales with \( \sqrt{A} \). This is the reason why high performance gliders have slender wings.
# Improving aerodynamic design, such that \( C_{D0} \) or \( k \) is reduced, can improve the best glide ratio as we would intuitively expect. However, modern advancement in aerodynamics has been agonisingly slow and you realise that there is not much potential to be released by comparing a fibre glass glider built in the 1980s with a modern one. What differences do you spot?
From a geometric point of view, the above solution process is equivalent to finding a ray from the origin that is tangent to the polar curve. You should ask an instructor to demonstrate this to you to reinforce the understanding. This geometric method is useful when more factors are taken into account, such that an analytical solution cannot be obtained easily.
== Adjustments to the Analytical Polar: Headwind and Sinking Air ==
You may have been instructed that you need to fly faster in headwind or sinking air to cover ground efficiently. This section briefly demonstrates the underlying mathematics, but you are encouraged to use the geometric method to prove to yourself that it is indeed the case.
=== Effects of Headwind ===
We firstly approximate true airspeed with indicated airspeed. By doing this, the following construction is possible:
\[ V = V_g + V_w \]
Where \( V_g \) is ground speed and \( V_w \) is the headwind speed. This expression is to be substituted into the analytical polar. We should also bear in mind that it is the most efficient covering of '''ground''' distance that is of interest, therefore:
\[ \frac{V_S}{V_g} = \frac{\rho C_{D0}}{2 \omega} \frac{V^3}{V-V_w} + \frac{2k \omega}{\pi A \rho}\frac{1}{V(V-V_w)} \]
This quantity must be differentiated with respect to \( V \) to find the optimum. Performing the differentiation and, after considerable simplifications:
\[ 2V^5 - 3V^4 V_w - 2CV + CV_w = 0 \]
Where:
\[ C = V_{BG}^4 \]
This is a fifth-order polynomial. It is a fact established by Abel in 1820 that a fifth-order polynomial cannot, in general, be solved by radicals. However, we can 'prove' that the solution lies in \( V > V_{BG} \) by considering the following:
# Substitute \( V=V_{BG} \) into the equation above, and show that the value is negative.
# Differentiate the expression and substitute \( V=V_{BG} \) into the differentiated expression, and show that the value is positive.
# Hence, we have a function that is presently negative, but it is increasing, so we would expect a root (the required solution) at a larger \( V \) than the present \( V \) which is \( V_{BG} \)
The above arguments are far from watertight: The differentiated expression only gives a positive value if \( V_w \leq \frac{2}{3} V \). While this is usually the case, the above cannot constitute a proof. A more rigorous analysis on the locations of the maxima and minima is required.
The problem is much simpler if the geometric method is used: to use the geometric method, imagine setting up a ground speed zero which is different from the airspeed zero. The polar is plotted with respect to the airspeed zero but the tangent ray needs to start from the ground speed zero. Because the ground zero is located in \( V>0\), the tangent is steeper and intersects the polar at a larger \( V \).
