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#1
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| Math problem, estimating disp., SA Hi all, I have what is probably a simple math issue for the more experienced here. Just drafted a canoe design (my first) and I'm trying to estimate the displacement and wetted surface area for a given draft before I take the plunge into learning how to bring the design into Freeship. I believe what I'm doing is sorta using the trapezoidal rule, but I'm not sure if I'm employing it properly. I can accurately get the area and perimeter of my station sections. So I assume I can average these figures and multiply by the LWL to get decent estimates. Assuming I'm on the right track: 1) should I simply take the average over the number of stations (16 of them)? 2) or factor in the '0' areas at the keel as fractional stations (16.33 or whatever)? The first number gives me a reasonable Cp. The second gives me an absurdly low Cp. Thanks for any assistance here. Tim |
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#2
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| The LWL will change with draft unless the stem and stern are vertical. Are the 16 sections equally spaced? Do the ends of the LWL for the draft of interest correspond with station locations? Special trapazoidal rule for N+1 equally spaced stations with spacing D; value at each station A(i); first station end of length is 1, last section is N+1: D/2*A(1) + D*A(2) + D*A(3) + .... + D*A(N-1) + D*A(N) + D/2*A(N+1) General tapazoidal rule for N+1 stations with spacing D(i) between stations i and i+1, value at each station A(i); first station end of length is 1, last section is N+1: D(1)*[A(1)+A(2)]/2 + .... + D(i)*[A(i)+A(i+1)]/2 + .... + D(N)*[A(N)+A(N+1)]/2 |
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#3
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| DCockey, really appreciate the reply. Hmmm, slightly more complicated formula than I thought. Guess a simple average doesn't work. Quote:
Quote:
Quote:
Thanks again for your help. |
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#4
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| Quote:
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#5
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| Use the stations which are wetted. Add a station at each end of the LWL. Unless one has a submerged transom use zero area for each. Then use the second formula since the spacing between the stations along the LWL is uneven when the first and last spacings are included. |
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#6
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| Tim Hall,,, Use Simpson's rule, it is not that much difficult than trapezoidal rule. What you have to do is just integration, and Simpson's Integral is best for numerical integration.
__________________ Md. Al-Amin Pavel http://www.paveldesign.tk "A question that sometimes drives me hazy, am i or are the others crazy?" Albert Einstein |
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#7
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| Simpson's rule is more accurate than the trapezoidal rule for smooth curves. However it needs an odd number of equally spaced stations over the domain of integration which is the LWL in this case. For Tim's purposes I doubt the difference in accuracy is meaningful, and he'd have create new stations. |
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#8
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| An alternative approach if you have a good method for finding the area under a two-dimensional curve. Create a curve of areas. Use the distance along the LWL for the x-axis and the cross-sectional area (below the LWL) along the y-axis. Plot the points of the area for each station (y-value) vs location along LWL (x-axis). Include the zero area points at either end of the LWL. Create a curve through the points and then determine the area under the curve. That will be the displacement volume. Likewise for the wetted surface but use wetted length along the y-axis, and points representing wetted length of the section vs location along LWL. Find the area under the resulting curve and that will be wetted area. |
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#9
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| DCockey, thank you! Area under the curve method using LWL for 'x' and section area for 'y' makes perfect sense in my mind. Once I go beyond basic analytical geometry the abstract nature of mathematics quickly eludes me. Solving problems graphically I can do...er, well that is what I do. My drafting software allows me to calculate curved ares very accurately, so this won't be a problem. One thing I'm not picturing is the relationship of units between the 'x' and 'y' coordinates. Say my units are all in feet. Are the increments for ft^2 the same as the increments for linear feet? In other words, if I go from 0 ft at the WL with 0 ft^2 to 1 ft by 1ft^2, I would have a 45-degree angle on the plot? |
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#10
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Tim, here is a table direct from Skenes for the Simpsons rule. It will give you, displ., CB, and CP. The table works with any even number of evenly spaced stations starting at zero(intersection of bow with wl). If you make up a table like this so you can use it in the future it will be a quick and easy way to accurately estimate these parameters. Might even be able to do an Excel version. I've used it for 40 years and it is accurate. PS- I didn't understand if you have 16 stations including sta. zero as described above or 16 not including sta. zero. . You have to have 0+ 16 stations......(or any EVEN number of stations-usually 10) At any rate this may be of some help down the line.... Good Luck! click on the image for best detail:
__________________ yes, it is a revolution ---"So (yet) another new world begins." Seahorse 2011 My Gallery: http://www.boatdesign.net/gallery/sh...0&ppuser=31218 |
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#11
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| Doug, thank you. Yes, I can see how an excel table would work very nicely with this. Looks like I'll have to do a little more drafting here. I'm not at all familiar with naval architecture conventions. Used to drawing buildings. |
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#12
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| Quote:
Quote:
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#13
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| It might be worth remembering, you can get a demo version of Delftship software for free, and it will read in the station offsets, and calculate the displacement to the nth degree. I still cant handle manipulating hull shapes in Delftship, but the 're-jigging' of the station sizes and importing them works good. |
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#14
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| Ok, so I got my curve and my displacement is pretty darn close to what I had guestimated when starting the project. So I'm looking at this curve, and it occurs to this is possibly a nice analytical tool (I really appreciate DCockey for pointing this out, BTW). Is it possible to estimate my LCB from this? If I, by trial and error, find the x-coordinate (distance on the WL) that divides the displacement in half? |
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#15
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