Calculating the Area of a station

Hi all,
Wondering can you help me with the John Teale book; How to design a boat.
In chapter 1 there's a section on Displacement Calculation. I can see that the Simpson Multipliers are used to multiply the area of each station, and then Levers are used to multiply the product of the Simpson Multiplier x Area. However, what he doesn't explain is how to calculate the area of each station.
Can someone explain how this is done please?
What we have is, say, each station is 2.3 units wide. The half beam is, say, 2 units wide. There are 6 stations. The transom is, say, 1.5 units wide (I'm just using arbitrary dimensions here).

I can see that this calculation is used widely elsewhere so it's important to understand how it's done.

Thanks,
Dermot

 

If you are doing it by hand then you need a planimeter. But they are very expensive. So for a home designer you need to put your film/tracing paper over graph paper and count the squares for each station

Richard Woods

 

A planimeter not necessary. Simpson's method serves well to calculate the area of each cross section. Just working with the breadths of the frames.

 

You can find some pretty cheap used planimeters on ebay (like < $25).

 

There are lots of ways to determine that area of a given station. The graphical box method like Richard said (this is the way AutoCAD does it BTW, it counts pixels), Simpson's or Tchebycheff's method, section curve definition and direct integration (most surface modeling programs), or just get a planimeter (or mechanical integrator, which is far easier especially for professional D&O's and cross curves when you are not using a surface modeling program).

Note that this is a very different question than "what should be the sectional area of a given station?"

 

Hi all,
Many thanks to all of who have replied thus far. I do appreciate your time and expertise.
Let me give some more depth to my question.
The following is a quote from John Teale's book: "How to design a boat".
"The volume of displacement is found by working out the under water area (below lwl) of each half-section and then putting these figures into a table. Volume can then be translated into a weight, which should be equal to the weight of the boat.
On a hard-chine boat areas can be calculated simply by dividing each half-section into rectangles and triangles as appropriate. Remember to measure areas on one side of the boat only for each section." he then goes on to say:
"The working of the calculation is straight forward and the example in table 2 refers to a 20-footer. The Simpson Multipliers (SM) start at 1 on station 0 and continue as 4 and then 2 for as long as necessary. They end up as 4 and finally 1 again."

Now, on the afore-mentioned table 2 he seems to multiply the already worked out area by the SM.

My confusion is around the fact that I thought that the Simpson rule was what was used to work out the area of the half-section, but he seems to be saying that you use the simpson multiplier as a multiplier to the already worked out area. Therefore my question is , how did he find the area of each half section.
Is working out the area as simple as treating like a rectangle and multiplyng the width of the section by the average of the lengths of both sides of the station? (i.e. the trapezoidal rule)
Any ideas?
Many thanks,
Dermot

 

You can do a double column (station) x line (water line) Simpson's tablature; when I first started there were pre-printed spreadsheets.

How you calculate the areas of the sections is independent of how you integrate the sectional area curve. So you can use any area method to calculate the area of the section (i.e. the summation of the halfbredth in L x the waterline in L gives the area in L^2) and then plot the sectional areas (L^2) as a ordinates along the length of the waterline (L), making a sectional area curve.

Next, use any area rule to calculate the "area" of the sectional area curve. The displacement is just the summation of the ordinate (in L^2) at some distance along the hull (station # x station spacing, h, in L). The area under the curve would then be L^2 x L = L^3, or volume.

Read this link, paying attention to the worked examples for sectional area and displacement.

http://www.usna.edu/NAOE/_files/documents/en400/TEXT - 2011/7-Chapter 2 Text 2011.pdf

 

Actually, I found a great video here where this woman explains it all very very well:
https://www.youtube.com/watch?v=0SomDYA4pfI

Thanks all for your input.

 

Thanks jehardiman for that link. I'll take a look at that now.

 

You can always cut out cardboard shapes of your sections and weigh them on an accurate scale. Jeweller's scale ?

 

Dermot; Imagine that you are looking directly at the end of a box. You can see that it is 5 inches high and ten inches wide. That is an area of 5" x 10" or 50 square inches. That does not tell us what the box will hold. We need the third dimension. Look directly at the top of the box, it is 20 inches long. Multiply the area times the length to get the volume of the box which will be 50 x 20 = 1000 cubic inches. That box occupies 1000 cubic inches of space. If submerged it would push up 36 pounds when in fresh water.

If the box was a boat the arithmetic is not so easy because the boat has curvy lines. Divide the boat into an even number of spaces front to back. That will cause you to have an odd number of sections (stations), necessary for the Simpson rule to work correctly. Measure the immersed area of each of those stations.

Make a chart with station NUMBERS 1,2,3,...15 or some convenient number. (the arithmetic is more accurate with a larger number of stations). Multiply the area of each of the stations by the Simpson multiplier. (1,4,2...4.1) to get the product of the area times the Simpson number. Now add all the products to get the sum. Multiply the sum by the distance between stations, divide by three to get a close approximation of the total immersed volume. The result will be in terms of cubic inches, the immersed volume of the boat. Multiply that number by 0.03611 to get pounds of displacement.

A small skiff might have a displaced volume of 9000 cubic inches. so....9000 x 0.03611 = 325 pounds which is in the ball park for a one person fishing boat. Maybe you want a bigger boat that can support 1200 pounds of boat, motor, gear, fuel, occupants, and fishing tackle. ...and beer, don't forget the beer and the cooler. 1200 pounds divided by 0.03611 = 33,231 cubic inches, means that the boat has to be a lot bigger than the small fishing skiff. If you calculated the areas and volume in feet then take the number of cubic feet of volume and multiply by 62.4 for fresh water. In that case the 1200 pound boat would have needed 1200/62.4 = 19.23 cubic feet of volume.

The odd figure of 0.03611 is the result of dividing the weight of a cubic foot of water by the number of cubic inches in a cubic foot......62.4/1728 = 0.03611. That's how much a cubic inch of water weighs and how much a cubic inch of displaced volume will support.


























If you are a weird person, like so many of us, you will enjoy all that brain exercise.

 

There is a misconception, that the displacement of the boat is the volume under the waterline as drawn. The displacement is equal to the total weight of the boat plus supplies and crew. The boat will not necessarily float on her drawn lines either, unless there is a careful distribution of weights.

 

Now I understand everything. No need for a planimeter nor Intergration methods, enough to know the weight of the boat, including supplies and crew. Thank you.
(Then, in this definition, what it is the weight of the boat ?)

 

Tansl, please read Gonzo's comments again. He is merely saying that the design displacement of the boat may not be the actual size of the hole in the water because of errors in weight estimations of all the miscellaneous loads that amateurs, sometimes even professionals, fail to account for.

It appears to be a fact of life that a boat will gain weight over time on account of all those amenities that the user is so obliged to add. Thus the design displacement goes to hell in a hand basket. Not to say that the designer should overestimate....well maybe he should fudge the numbers upward a little bit.

 

messabout, thanks for your work to appease the spirits. I read the comment of Gonzo and I think it has nothing to do with what is in this thread. And also it has several statements that make what he says does not make much sense. For example, it seems to mean that the boat weight (displacement) is equal to the weight of the ship, plus supplies and crew. As you know, the defined subject can not enter into the definition. Is there anything else that demonstrates a certain ignorance of the subject in general, but rather not continue.
Thanks, again, for your peacemaking spirit.
And you are quite right, boats, like humans, gain weight with age and humans, like good wine, gain in quality with age.