…i was just wondering if anyone has got a formula for determining comparable fin areas with different fin setups ?
[ For example …
a 7" deep x 5" base back fin , with two 3x3" ‘sidebites’ = a pair of ?“x?” TWIN fins ?]
thanks crew ,
any help and sharing of your fin experiment experiences MUCH appreciated
cheers !
ben
[*a guy in the local shop said he calculates a fin’s area by laying a fin on a squared inch / squared centimetred graph paper page , and adding up the squares …is this the “correct” way ? …nowadays , i guess computer programmes would do it automatically , yes ??]
To be super accurate you would need calculus. For one fin you would graph the leading edge curve, take the integral, then graph the trailing edge, and take the integral, then you would subtract the integral of the trailing edge curve from the integral of the leading edge curve and you would have the fin area for that fin.
Other than that, the graph paper way would work too.
just one question , WHAT is an ‘integral’ , exactly ? [pardon my ignorance , please ]
cheers
ben
…and , how then would we figure out the size of say a set of quad fins , to equal say a “2+1” setup’s total fin area ?
it doesn’t have to be 100% accurate , i’m just seeking rough guidelines , in order to make comparable sized and aread [?word?] setups , as i have a few different ideas floating around in my head [or maybe it’s just the salt water …]
Integral is a crazy math equation that helps you determine the area under a curve.
You do it by calculating the area of evenly spaced columns which fit under the curve, and adding them up. The more columns you calculate and add up the more accurate your calculation.
For a simple calculation you could calculate the area under a triangle that runs from the front bottom corner to the back top corner and get an approximation of the area of the fin. You calculate the area of a rectangle that represents the dims of the fin (8"x5", or 3"x3", etc.) and divide it in half to get the area of the triangle that is the rectangle bisected.
Instead of a bunch of little squares, like your friend, you would just use one big square cut in half like a triangle.
So if your fin is 7"x5" you could figure it as (75)/2=17.5" of area. And for the 3"x3" (33)/2=3" of area each. 17.5" + 3" +" 3" =23.5" of total area.
If you had two 5.5"x7" keels you could figure (5.5*7)/2=19.25" of area each. 19.25" + 19.25" =38.5" of total area.
Okay I’m probably just displaying my ignorance but would total fin volume be of any use here? Take three thruster fins and submerge them in a accurate beaker full of a fluid. Note the displacement. Then do the same with a different set and compare the results. Obviously fin thickness would play a part in this too. Of course if the fins haven’t been created yet or are already attached to a board that would screw my whole idea.
based on the fcs and futures fin charts [which give fin areas]
i guess i can now figure out … if say , two twin fins have an area of say 25 square inches each ,
then with a 50 square inch total fins area , a comparable thruster setup would be …divide the 50 by three , for the fin area needed of each thruster fin , yes ??
easiest way to accurately measure the area under a curve (or of a fin):
weigh a piece of paper (say 8 1/2 " x 11" … 93 square inches )
draw the fin on the paper
cut out the template for the fin and then weigh it as well.
ratio of the weights is the same as the ratio of the areas. Back when lab lab equipment was hooked up to chart recorders instead of computers this was the easy way to integrate data.
You can do a pretty simple integral by measuring the height of the fin every inch, sum the two first values and divide by two (you should multiply by the length between the measurements, but since this is one inch multiplying by one get’s you the same value. To be very accurate along the training edge you would have to multiply though if the length of the fin is say 5 1/4" but the area in the last 1/4" will be next to nothing anyway unless you have a fin with a very steep trailing edge. Most likely if you are only after a good estimate, you could ignore this area if the last piece if less than 1/2", if it’s larger just round it up to full inches. Of course you could measure every 1/2", 1/4",etc. and multiply by this distance. to be more accurate). Repeat for every inch and sum all the results. You would of course have to subtract any cutout area typically under the trailing edge of a fin, calculated the same way. This known as the trapezoid rule, should be fairly accurate and is a method commonly used in numerical integration. More info at f.ex. http://www.ugrad.math.ubc.ca/coursedoc/math101/notes/techniques/numerical.html
I had a brief look at it and if my head is screwed on right, a simple way to compute area of a fin would come down to measuring height every inch from leading edge and basically summing up all those measurement only the last measurement should be rounded to nearest inch lengthwise and divided by two. Then subtract the cutout calculated the same way. This would give you a good ballpark measurement, but I don’t have any fins here to test it on.
One the other hand, my head is not always screwed on right.
Chipfish…must say I have had a few fish meals but not one where the fries(cips) are part of the fish!! you are unique dude…Don’t go to England they just love "fish’n’chips’
Back to your post…Math, oh dude not a strong subject for me ( have trouble countin’ the kids, (27 at last count…and not a square root amongst them).
I sit …I watch as the surfers glide over head …simple observation the more of those fins things, the slower they go, like they have their brakes on!
If I am slow…I’m mornay dude. So I wondered why use fins… then I noticed that some surfers don’t go in straight line, they go rippin all over (on drugs or drunk, I figured).
So I say that if ya wanna go fast and straight, forget fins and if ya wanna ride all over, like ya drunk, get some fins!
Small area fins= fast big area fins=rippin drunk goin all the place.
So fins ==drag dude…Hang on phones ringin and I think it’s Derek Hynd on the line…Gotta Go!
The following method is similar to the suggested integration methods already posted.
Buy a piece of acetate (or a thin sheet clear plasic) and draw a grid on it that reflects the kind of accuracy you want –i.e. smaller grid more accurate, larger grid less accurate, see diagram below.
Trace the outline of fin on grid, then count up whole squares. Make your best call on the remaining sections, see diagram.
For example, say you decide to mark off a 1 cm sq grid, each square would then have an area of 1 cm sq… For the those regions of the fin (edges) that do not completely fill a square just make a guess as to how much they fill in terms of fractions e.g. 1/4, 1/2, 3/4, etc. Then add up all the fractions, times the sum by 1 cm sq and then add that total to the total of complete squares.
Suggestions and Notes
Use a permanent marker to make the grid, but use a erasable marker to make the outline of the fin so it can be reused.
Using a grid less than 1 cm sq is likely to be far more accurate than meaningful, but that’s your call.
Cracked fin repair questions??!! This method will also lend itself to calculating the centroid or the geometric center The geometric center can be use to estimate how the moments and forces will tend to resolve for a given load. See KCasey post in
Here’s the same basic idea, but with a slight modification to yield not only the fin area but also its aspect ratio:
Trace outline of fin on paper.
Draw a “base line” from the leading edge at the base of the fin (i.e. not including the mounting tab) to the trailing edge of base.
Draw series of equally spaced lines from the base line to the tip of the fin. Each line should be parallel to the base line.
Measure the distance from the leading edge to trailing edge along each line (Easiest to do if a ruler measuring in decimal values is used – can usually find such a ruler in the stationary section of a drug store, etc.) Write length of each line adjacent to it.
Sum up the lengths measured in step 4.
Divide sum by the number of lines to get average line (chord) length.
Measure distance from base line to tip of fin (span).
Multiply fin height from step (7) times the average chord (step 6) → Area of fin
Multiply fin height (step 7) by 2 to get the effective span (Note: This is a good approx for center fin, less so for side fins. For the latter, the multiplier will be somewhat less than 2 and depends on the distance of the fin from the rail of the board).
Divide length obtained from step 9 (effective span) by the average chord (step 6) → Gives aspect ratio.
Area relates to maximum lift generated before stall.
Aspect ratio relates to change in lift as angle-of-attack is changed.
Long chord length gives more acceleration but may yield lower maximum speeds than high aspect ratio foils. The latter being more critical about the angle of attack and hence more sensitive to rider input. For the average joe, this means a long chord fin (keel) will on average yield a better (speed) result imho. Fatter foils quicker to establish laminary flow whereas thin foils are slower to get flow and more critical in keeping flow but can handle a higher flow rate once established. For average joe? Fatter fins are faster.
Long chord length gives more acceleration but may yield lower maximum speeds than high aspect ratio foils. The latter being more critical about the angle of attack and hence more sensitive to rider input. For the average joe, this means a long chord fin (keel) will on average yield a better (speed) result imho. Fatter foils quicker to establish laminary flow whereas thin foils are slower to get flow and more critical in keeping flow but can handle a higher flow rate once established. For average joe? Fatter fins are faster.
Shit that cat of yours is an aggressive feline, might need Da Goat to give it a quick left hoof and a head butt to, well…you know, settle it down.
Back to business…you are, in my opinion, spot on!
It is all a matter of what you want to on a wave, what boad you are riding are how good you are.
Racing motor bikes have big fat rolled tyres, not for speed but for grip in turns. You wanna turn a surfboard (sorry Derek…TURN…not slide or skid), you need some kinda fin dude. You turn, you choose!
Back to your post…Math, oh dude not a strong subject for me ( have trouble countin’ the kids, (27 at last count…and not a square root amongst them).
maybe as they get older ??
[maate you seem an interesting thinker too…are you mates with / hanging out with the alien abductee ['fatbaslardarse] or something …
keep up the good werk , anyways !]
also , thanks everyone else , for your input on the fin areas comparisons ,
i was hoping it would generate some discussion , guidelines , and of course , practical experimenting [something i do a bit of from time to time]
please keep it coming , and feel free to share fin comparisons / ride reports / photos of your different fin setups on this thread too , if the mood / stoke so moves you
