4
$\begingroup$

Consider the 2D airfoil below.

2D Airfoil Pressure
(source: gsu.edu)

In engineering (and maybe physics) you will often see something like the following as an expression for the pressure force acting on a surface (in this case a curve but imagine it having unit depth into the screen).

$$ \mathrm{d} \mathbf{F} = p \, \mathrm{d} \mathbf{s} \\ \text{where} \, \mathrm{d} \mathbf{s} = \mathbf{\hat{n}} \, \mathrm{d} s $$

If you attempt to integrate this over a curve C to find the force you get;

$$ \int_?^? \mathrm{d} \mathbf{F} = \int_C p \, \mathrm{d} \mathbf{s} $$

where there doesn't seem to be obvious corresponding limits for integrating on the LHS. Is it fine to consider the limits as from 0 to $\mathbf{F}$ or is this some kind of engineering "shorthand" that you often see which makes no sense mathematically. I am trying to interpret it as a "change in force" but it doesn't really make sense to me.

Improve this question
$\endgroup$

2 Answers 2

Reset to default
3
$\begingroup$

You are summing "small amounts of force" over "all points on the surface".

This is the tricky thing about integrals - they are not always nice one dimensional constructs with limits in the units of the quantity shown as $dx$.

The integral symbol is actually a very stylized letter S: once you realize that, you see that you are "summing something" and the limits just describe the region over which the summing happens.

In your example, the "region" is the surface of the air foil; and you cannot give "neat" numerical limits unless you find a way to parameterize the location on the foil so that you get a separation of variables. But for the concept of integration, that doesn't matter...

Note on separation of variables: if you have a rectangle parallel to X and Y, you can describe each part of the surface with x, y coordinates and area as dx dydx dy. Then the limits become $$\int_{x_1}^{x_2}\int_{y_1}^{y_2}\vec{F}(x,y)\cdot \vec n ~dx ~dy$$

I don't know if that made things clearer, or just confused you more. Comments...

Improve this answer
$\endgroup$
6
  • $\begingroup$ Could you clarify what you mean by separation of variables? $\endgroup$
    – user56658
    Aug 5, 2014 at 2:48
  • $\begingroup$ I think perhaps the way to do it would be to define $\mathbf{F}$ as a function of the length (or other parameterisation perhaps) of the curve you are integrating over, starting from an arbitrary point (which would be the same point that the parameterisation of the curve starts at). If it is 0 at the initial point, then $\mathrm{d} \mathbf{F}$ would represent a change in the force as you go along the curve. Extending this to a surface though is a bit trickier. $\endgroup$
    – user56658
    Aug 5, 2014 at 2:55
  • $\begingroup$ @user56658 All that's happening is taking $dF/ds = p$ and "multiplying" the $ds$ to the other side and then integrating. That's the separation of variables and it's completely a "trick" in this instance. You don't need limits on the left hand side -- the total force comes from integrating $p$ over the surface. $\endgroup$
    – tpg2114
    Aug 5, 2014 at 2:58
  • 1
    $\begingroup$ Separation of variables: if you have a rectangle parallel to X and Y, you can describe each part of the surface with X, Y coordinates and area as $dx\ dy$. Then the limits become $\int_{x1}^{x2}\int_{y1}^{y2}\vec F(x,y)\cdot \vec n\ dx\ dy$ $\endgroup$
    – Floris
    Aug 5, 2014 at 3:20
  • 1
    $\begingroup$ @Floris: It would be nice if you converted the last comment to an edit to the answer. $\endgroup$
    – Jan Hudec
    Aug 5, 2014 at 11:37
-2
$\begingroup$

Limits can be fuselage to the tip of the wings :) !! As some user mentioned its just a force in total, nothing to worry on the limits over the integration. If one take it as a line just summing all lines will give the area and of course the force. For example if xxxx Newtons force in a single line and summing it over the sampled area (each sample is considered as thick line) will give the total force. As sampling rate increases(sampling thickness --->0) error will come down. Integration eliminate this complexity but it depends the governing equation. Am I making it more complex :(

Improve this answer
$\endgroup$
2
  • $\begingroup$ So the answer is $\int_{fuse}^{wing}{\rm d}\mathbf{F}$? How does that work? More specifically, what are $\mathbf{F}_{fuse}$ and $\mathbf{F}_{wing}$? Shouldn't it really just be the total force, $\mathbf{F}$? $\endgroup$
    – Kyle Kanos
    Jul 3, 2015 at 13:23
  • $\begingroup$ Am sorry. I meant to say this integration result is force over the wing. Am sorry if I understood wrongly and clarify me. What I thought to integrate the governing equation of the pressure variation across the wing area will give the total force on the wing. I am just a graduate :( $\endgroup$
    – jai
    Jul 6, 2015 at 5:39

Your Answer

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

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