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  #1  
Old 04-26-2011, 11:51 AM
Tim Hall Tim Hall is offline
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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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Old 04-26-2011, 12:12 PM
DCockey DCockey is offline
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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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Old 04-26-2011, 12:25 PM
Tim Hall Tim Hall is offline
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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:
Are the 16 sections equally spaced?
Yes, so it looks likes I use the first formula you gave me(?)

Quote:
Do the ends of the LWL for the draft of interest correspond with station locations?
No, the end of the waterline does not coincide with stations. I have two stations, extreme bow and stern, that are not wetted, and I assume they simply don't factor in. I centered the stations on the LOA. Perhaps centering them on the design waterline would be better?

Quote:
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)
Does this work with the wetted perimeter of the stations also to get the SA?

Thanks again for your help.
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Old 04-26-2011, 12:34 PM
Tim Hall Tim Hall is offline
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Quote:
The LWL will change with draft unless the stem and stern are vertical.
I'm working this out with a fixed 4" draft.
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  #5  
Old 04-26-2011, 02:17 PM
DCockey DCockey is offline
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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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Old 04-26-2011, 02:44 PM
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pavel915 pavel915 is offline
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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.
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Old 04-26-2011, 03:38 PM
DCockey DCockey is offline
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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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Old 04-26-2011, 03:43 PM
DCockey DCockey is offline
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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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Old 04-26-2011, 05:20 PM
Tim Hall Tim Hall is offline
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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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Old 04-26-2011, 06:02 PM
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Doug Lord Doug Lord is offline
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Quote:
Originally Posted by Tim Hall View Post
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
====================
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:
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Math problem, estimating disp., SA-simpsons-table-skenes-001.jpg  
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Old 04-26-2011, 06:30 PM
Tim Hall Tim Hall is offline
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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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Old 04-26-2011, 09:38 PM
DCockey DCockey is offline
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Quote:
Originally Posted by Tim Hall View Post
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?
Go ahead an use increments of "ft" when plotting the cross-sectional areas. The results from your software will say ft^2 but the number will be be the number of ft^3.

Quote:
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?
Correct
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Old 04-27-2011, 07:22 PM
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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  
Old 04-27-2011, 07:32 PM
Tim Hall Tim Hall is offline
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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  
Old 04-27-2011, 07:49 PM
Tim Hall Tim Hall is offline
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Quote:
Originally Posted by rwatson View Post
It might be worth remembering, you can get a demo version of Delftship
I have Freeship, which I've only tinkered with a bit. Plan to use it only for double-check and calculating stability. Otherwise I have everything drafted in 2D.
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