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Today I was reading about the aerospike nozzle and found this image:

image1

(from section 10.3.1 of this lecture)

This image tells us how to measure the thrust of this kind of nozzle. I can understand all of it except this part:

image2

Where does this integral come from? How do they derive it?

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    $\begingroup$ Where is the image from? Besides citing your source being a common courtesy, it might make answering the question easier. $\endgroup$
    – Jan Hudec
    Commented May 1, 2017 at 16:06
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    $\begingroup$ @JanHudec here is a link to the source of the image: web.csulb.edu/colleges/coe/ae/engr370i/ch10/sect_3-1 $\endgroup$
    – Roh
    Commented May 1, 2017 at 16:31
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    $\begingroup$ On this site question (and answers) are editable and additional information should be edited in. I've added the citation for you now. $\endgroup$
    – Jan Hudec
    Commented May 1, 2017 at 16:44

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I expect you know how a conventional rocket nozzle works: It expands the exhaust gas, and the pressure inside helps to increase thrust, pushing the nozzle forward.

Now turn the conventional nozzle inside out. The expansion happens on its outside, but there is still a pressure acting on a backward-facing surface. The graph doesn't explain it, but the ramp area $\text{A}_{\text{Ramp}}$ is the projection of the nozzle area in flight direction. The ramp pressure $\text{p}_{\text{Ramp}}$ is the pressure acting along the nozzle surface, and it must be reduced by ambient pressure $\text{p}_{\infty}$ in order to arrive at the correct thrust. The integration is needed because $\text{p}_{\text{Ramp}}$ varies over the nozzle area.

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  • $\begingroup$ But the integration is proportion to A not Pramp.why? $\endgroup$
    – Roh
    Commented May 2, 2017 at 4:16
  • $\begingroup$ @Roh: The pressure is integrated over the area. Both influence the result. How else should this be done? $\endgroup$
    – Peter Kämpf
    Commented May 2, 2017 at 10:07

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