I'm having trouble understanding the derivation of the integral for Coefficient of Friction in Example 2.2 of John Anderson's Fundamentals of Aerodynamics 5e. I'm not sure how the author cancels $$ V_\infty^2c $$ and places the freestream velocity inside the integral. $$ \frac{u_2}{V_\infty}(\frac{u_1}{V_\infty}-\frac{u_2}{V_\infty}) $$ Equations below:

From Equation $(2.84)$ $$C_f=\frac{D'}{q_{\infty} c}=\frac{\rho_{\infty}}{\frac{1}{2}\rho_{\infty} V^2_{\infty}c}\int_0^{\delta}u_2(u_1-u_2)dy$$ where the integral is evaluated at the trailing edge of the plate. Hence, $$C_f=1\int_0^{\delta/2}\frac{u_2}{V_{\infty}}\left(\frac{u_1}{V_{\infty}}-\frac{u_2}{V_{\infty}}\right)d\left(\frac yc\right)$$

  • $\begingroup$ Hello! It is preferable to type out screenshots; for formulae, one can use MathJax. I have edited your question accordingly. Thanks! $\endgroup$
    – jng224
    Commented May 28, 2021 at 18:20

1 Answer 1


This is quite simple. Since $\dfrac{1}{V_\infty^2c}$ is a constant, it can be inserted into the integral. Thus, one $\dfrac{1}{V_\infty}$ goes into the factor $$\frac{u_2}{V_\infty},$$ the other $\dfrac{1}{V_\infty}$ goes into the factor $$\left(\frac{u_1}{V_\infty} - \frac{u_2}{V_\infty}\right),$$ and the $\dfrac{1}{c}$ goes into $$d\left(\frac{y}{c}\right).$$ Notice that the upper limit of the integral becomes $\delta/c$ and not $\delta/2$.

  • $\begingroup$ Of course! Brilliant, thank you so much! @Tofi $\endgroup$
    – Tom
    Commented May 28, 2021 at 19:00
  • $\begingroup$ @Tom Glad to help 👍 $\endgroup$
    – Tofi
    Commented May 28, 2021 at 19:17

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.