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I read that when you want to double the max speed of e.g. a boat you would have to increase the power eightfold! I could understand that you have to square it (= fourfold) because you are working against "a wall" of water which makes it an area (= power of two)... but why to the power of three?

Could you please give a physical/mathematical explanation but also an intuition on why this is the case.

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  • $\begingroup$ "I read that..." please provide a link to the text or page which makes this statement. $\endgroup$ – Carl Witthoft Oct 2 '17 at 14:54
  • $\begingroup$ @CarlWitthoft: I think it wouldn't be of much help because it is in German (a book about the Titanic), but if you think it worthwhile I will edit my question. $\endgroup$ – vonjd Oct 3 '17 at 18:08
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Think of it as the volume of water you're displacing per second - the area multiplied by the distance moved.

So intuitive I hope I can get away without an equation, but there's some good stuff in the "Power" paragraph here :https://en.wikipedia.org/wiki/Drag_%28physics%29#Drag_at_high_velocity

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  • $\begingroup$ Good link, and there are some explanations for "how wind turbines work" that say the same... $\endgroup$ – Solar Mike Sep 30 '17 at 18:14
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    $\begingroup$ to make this a high quality answer you really should elaborate here and not make folks follow a link. The answer is simple enough to be self contained here, $\endgroup$ – agentp Oct 1 '17 at 19:24
  • $\begingroup$ I hadn't intended to make anyone do anything. The question asked for a physical explanation and something intuitive. Done. If anyone wanted additional information, the link gave that. I disagree that copying a bunch of equations would have added any quality to the answer, but perhaps I'm missing something. Can you show me what you had in mind in a further answer below? $\endgroup$ – ItWasLikeThatWhenIGotHere Oct 2 '17 at 7:06
  • $\begingroup$ The risk is that the link will disappear (hard to believe, but that happens now and then to links on the Intartubezz :-) ). By posting some text & a picture, your answer is at less risk of becoming useless. $\endgroup$ – Carl Witthoft Oct 2 '17 at 14:56
  • $\begingroup$ The volume of water you are displacing doesn't have any direct bearing on the problem. And even if it did, that would account for X 2, not X 8. $\endgroup$ – Phil Sweet Oct 3 '17 at 1:46
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It's a pretty rough approximation, and only holds where skin friction is the dominant source of drag throughout the speed range considered. Skin friction increases about as speed ^2. So for double speed, the resistance force is quadrupled. Plus the rate is doubled. Power is the rate of doing work, so 4 X force and 2 X rate equals 8 X power. This doesn't work if wave making is significant, and it isn't very good in the planing regime where surface area is changing with speed.

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  • $\begingroup$ Thank you, I have read explanations like that before but what seems unintuitive to me is that because you increase the speed you have to quadruple the resistance force so multiplying times two after that because of the rate change seems like double counting to me. It is like saying when you go faster you will reach your goal faster and because of going faster you will reach your goal twice as fast altogether... you know what I mean? $\endgroup$ – vonjd Oct 3 '17 at 9:24
  • $\begingroup$ Hydrodynamics isn't famous for being intuitive. At the body surface, there is a boundary layer that forms at the body wall. The boundary condition is zero-slip, meaning the atoms of the fluid in contact with the body are moving with the body, and thus the velocity profile moving outward in the boundary layer starts at zero (body frame) and increases to free stream velocity across a very thin layer. The shear rate right at the body defines the local drag. As you go faster, not only does the velocity profile get scaled with velocity, but the thickness of the boundary layer actually gets smaller. $\endgroup$ – Phil Sweet Oct 3 '17 at 9:54
  • $\begingroup$ Thus shear,the slope of the velocity profile moving outward, goes up almost as velocity squared (on average, over the body, discounting flow separation and a bunch of other changes). Many use 1.667 in stead of 2 as the power factor. $\endgroup$ – Phil Sweet Oct 3 '17 at 9:58
  • $\begingroup$ What do you think about the intuition of @ItWasLikeThatWhenIGotHere's answer? $\endgroup$ – vonjd Oct 3 '17 at 10:34
  • $\begingroup$ @vonjd If you have doubts about the answer why did you accept it? Prematurely accepting an answer inhibits others from posting possibly better ones. $\endgroup$ – agentp Oct 3 '17 at 13:22

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