Created : 20 july 2001
Last updated : 14 november 2001
Classification : Mathematical analysis
Copyright : Restricted shareware
Status : Finished.
Comments : This is an educated but still rough comparison. The tendencies discovered here must be compared to real life experiences too. To date these have confirmed the results of this analysis
2,5 mtr.
14,33 sq. mtr
8,50 mtr.
13,00 sq. mtr.
3,50 sq. mtr.
5,00 mtr.
3,33 sq.mtr.
100 kg
150 kg
17,5 sq. mtr.
7,50 mtr.
250 kg
2,6 mtr.
17 sq. mtr
8,50 mtr.
14,83 sq. mtr.
4,15 sq. mtr.
5,50 mtr.
3,96 sq. mtr.
180 kg
150 kg
21 sq. mtr.
8,27 mtr.
330 kg
Width
Mainsail area
Mainsail luff
Rated main area
Jib area
Jib luff
Rated jib area
Weight craft
Weight crew
Genaker size
Hoist height -
from deck
Total weight
Width
Mainsail area
Mainsail luff
Rated main area
Jib area
Jib luff
Rated jib area
Weight craft
Weight crew
Genaker size
Hoist height - from deck
Total weight
The values given for the F18 class are all fixed and found in the F18 rules; the values given for the F16HP however represent just one possible implementation of the F16 HP 2-up rule. Nethertheless, all important values, apart from the actual sailarea's and sail luff lengths, will remain unaffected when change actual sailarea; this includes the Texel handicap number.
The values given for the actual sail areas and luff lengths represent a setup that is expected to be a general optimum for the Formula 16 HP class. It gives the values encountered on a craft with a small, high aspect, jib that can easily be made to be selftacking. Fitted with mainsail that has a higher aspect ratio than a F18 mainsail but fitted on a equally high 9 mtr. mast. Thus allowing the boom to be positioned high above the trampoline. Both improving handlibility while retaining good light air performance. And it will make sure that a normal sized crew of 150 kg's is double trapped in the stronger winds. The F16 HP is not intended to be a "youth boat" as some might be temped to call it solely based on it's length. (without disrespect to youth classes or other 16 ft classes in general)
The F16 HP setup was first fitted with a max overall sailarea of 16,10 sq .mtr. which leaves 3,10 sq. mtr. rated sailarea for the jib when the mainsail is maxed out udner the rule. Now that the Texel committee has change the Texel calculation formula the class uses 16,33 sq. mtr. (3,33 jib) as the limit.
Anyways, the crew will have to work hard at the higher wind ranges but than we don't call this class HP (High or Hyper Performance) for nothing, do we ?
* in light air it is the amount of wetted surface that completely determines the total amount of hull drag.
* in heavy air it is mainly the prismatic ratio (wave drag) that determines the total amount of hull drag
* both boats have equally sophisticated and optimized sail rigs with the same level of control.
* Both boats have equally high rigs making that both rigs experience the same skewing of airflow along the mast in light air. Ergo both rigs produce sailforces equal to the ratio of sailarea's
* both boats can transform the full sail power potential into thrust without capsizing or pitchpoling
* both boats have similar hull shapes to such a degree that the small differences that exist can be
neglected.
* the dimensions of the hulls of both boats are such that it can be roughly assumed that all
dimensions of a crosssection on a particular position are fixed in ratio to the same relative crossection
(lenghtwise) in the hull of the other boat.
Example : if a F16HP bulkhead at 60 % of the boat length is found to be as wide as 80 % of the width of the same relative bulkhead of the F18 than all other F16HP crossection dimensions, like waterline height, are also at 80 % of the F18 Dimensions .
* Each crosssection in one boats is diretcly comparable to a crosssection in the other boat at the same relative position lengthwise.
The last two assumptions describe a situation where the F16 HP hull can be made to perfectly match the F18 hull when all length parameters are multiplied by a factor (18 foot / 16,5 foot) = (5,49 / 5) = 1,098 and all the other dimensions are multiplied by (any) one single factor. (in the example a ratio 80 % was taken and the factor would then be 1/80 % = 1,25)
Total Weight F16HP / F18 = (150+100) / (150+180) = 0,75757 = 76 %
Length hulls F16HP / F18 = 5 / 5,49 = 0,91075 = 91 %
Mainsail area F16HP / F18 = 14,33 / 17 = 0,84294 = 84 %
Rated mainsail area F16HP / F18 = 13,00 / 14,83 = 0,87660 = 88 %
Luff length (mast height) F16HP / F18 = 8,5 / 8,5 = 1,0000 = 100 %
Mainsail aspect ratio F16HP / F18 = ((8,5)^2 / 14,33) / ((8,5)^2 / 17) = 1,1863 = 119 %
Jib area F16HP / F18 = 3,50 / 4,15 = 0,84337 = 84 %
Rated jib area F16HP / F18 = 3,33 / 3,96 = 0,84091 = 84 %
Mainsail aspect ratio F16HP / F18 = ((5,0)^2 / 3,5) / ((5,5)^2 / 4,15) = 0,9799 = 98 %
Genaker area F16HP / F18 = 17,5 / 21 = 0,83333 = 83 %
Genaker luff (hoist height) F16HP / F18 = about (7,5 / 8,27) = 0,9069 = 91 %
Genaker aspect ratio F16HP / F18 = ((7,50)^2)/ 17,5) / ((8,27)^2/ 21) = 0,9869 = 99 %
Heeling / pitching moment F16HP / F18 = ratio sailarea * ratio mastheight = 84 % * 100 % = 84 %
Righting moment F16HP/F18=ratio (crew weight*(width+1 mtr.)+1/2*craft weight*width)
= 650 kgm / 774 kgm = 0,8398 = 84 %
Texel handicap ratio F16HP / F18 = 102,52 / 102,74 = 0,99785 = 100 %
We will start with a simple mind experiment where we just size down a F18 hull by multiplying all dimensions with the same factor.
Lets assume that a hull shape of a certain length is just as optimal for a 5,49 mtr. long catamaran as it is voor a smaller 5 mtr. long catamaran when all of it's dimensions are multiplied by the length factor 5 / 5,49 = 0,91075 or rounded of to 0,91 = 91%. The hull has been resized to an exact 91 % copy of its bigger brother.
Now lets calculate the volume that this resized hull encloses. Because the smaller hull is an exact (but smaller) copy of its bigger brother we can calculate the exact ratio between the two displacements by taking the third order of the multiplication factor. (for proof of this read the appendix at the end of this page). Ofcourse the F18 hull volume is 100 %.
Volume 91,07 % / Volume 100 % = (0,91075) ^3 = 0,7554 = 76 %
We can do the same for its total surface area by taking the second order of the multiplication factor. This factor is normally known as the square of a value. (for proof of this read the appendix at the end of this page) Again F18 hull surface area is 100 %.
Surface Area 91,07 % / Area 100 % = (0,91075) ^2 = 0,8295 = 83 %
Obviously the same reasoning will apply to the part of the hull that is submerged as long as the submerged volume of the 91 % hull is an exact 91 % copy of the submerged part of the original 100 % F18 hull. The only variable that is left to complete this situation, as described above, is the overall weight of the craft. Only at one specific weight will the submerged volume also be a 91 % an exact copy of the F18 submerged volume.
Now lets determine this weight ratio. This is very simple for it is exactly the same as the submerged hull volume which is in turn equal to the overal hull volume ratio which was the first calculated ratio in this document. The displacement (c.q. weight) ratio therefor is also 76%
And now, look at the ratio "Total weight F16 HP to F18" given at the paragraph "Important ratios" which is (100 + 150) / (180 + 150) = 250/330 = 0,75756 = 76 %. Both ratios are 76 %. Actually, they only differ by a neglectable (0,7576-0,7554)/0,7576 = 0,29 % = say 0,3 % = neglectable.
Or in other words ; when the F16 HP has a weight of 76 % of an F18 : 0,7553 * 330 kg = 249,25 kg both submerged volumes are exact copies of eachother albeit at different sizes. Of course 249,25 kg is easily rounded of to 250 kg (= 100 kg boat weight and 150 crew weight) this without creating any noticable offset. We immediately conclude that the F16 HP hull is an exact scalled down (91 %) copy of the F18 hulls in every respect.The fact that the grandfathered boats were within a few kg's of this calculated weight is just shear luck. Hence the name :"The lucky mind experiment"
What is the result of this coincidence
Light air
Using the same reasoning we can take the ratio calculated for the total hull surface area as the exact value for the wetted surface area. The wetted surface area is therefor equal to 83 %
Now, using the assumption that wetted surface area is responsible for nearly the total hull drag at low speeds (c.q. light air) we can assume that a F16 HP design can be expected to experience around 83 % of the drag of an F18 boat when both are travelling at the same speed. The last implies that the F16HP must have a sail rig that creates around 83 % of the power of an F18 rig under the same conditions. When this condition is met than the F16HP will be perfectly equal to the F18 class in light air.
By looking at the sail ratios we may conclude that equality to the F18 in light air is indeed achieveable. It's nearly spot on when looking at the jibs, the mainsail ratio however shows that the F16HP might well have a small advantage over the F18's in the lighter conditions. With the emphasis on small because the actual mainsail ratio is 84 % just as the jib and it is only the rated mainsail area that comes out at an 5 % power advantage. This advantage does NOT indicate that the F16 HP will be 5 % faster. Not at all, actually, this advantage will be hard pressed to produce more than just 1 % speed increase with respect to a F18 in light air. But this analysis is for a later time.
Heavy air
The F16 HP at high speeds will have exactly the same hull shape as the F18. And therefor have the exact same important hull shape ratios like bow wedges and stern wedges that the F18's hulls have. Now there are three lines of thought on how to estimate achievable high speeds or hull drag at high speeds.
The first is that hulls of equal shape but of different sizes are capable of reaching the same topspeeds under the same conditions.The reasoning is that both hulls experience the same pressure distribution on their hulls. Now is the F16HP projected surface perpendicular to the onrushing volume of water exactly (0,9107) ^2 = 0,8293 = 83 % (again) of the same projected surface of the F18's. This would mean that "wave making drag" which is the result of this pressure distribution multiplied by its wetted area is equal to 83 % again in the case of the F16 HP with respect to the F18. So there are equal pressures on the same relative places along the hull but the effective area is sufficiently smaller in the F16HP.
So this line of thought predicts that at high speeds the F16HP is likely to experience only 83 % of the total drag a F18 would experience over the total span of (wind)speeds. This Because the F16 HP has both 83 % of wetted surface area as well as 83 % of the projected area with respect to an F18. So both frictional drag as wave making drag are at a perfect 83 % of those of the F18. Effectively giving the F16HP hulls a drag curve that is 83 % of the same curve of the F18 through the hole speed spectrum.
The second line of thought is that the wave making drag can be estimated by looking at its theoretical maximum hullspeed. This theoretical maximum hullspeed is given by the following formula which is derived of Froude's well known formula for heavy monohulls.
Topspeed = parameter (finess ratio dependend) * sq. rt. (length on waterline)
or
Vmax = Pa * sq.rt. (waterline length)
The parameter Pa is the same for both boats for they have the same finess ratio.
The ratio in topspeeds between F16 HP and F18 will therefor be :
Vmax F16HP / Vmax F18 = Pa * sq.rt (5) / PA * sq.rt. (5,49)
or
Vmax F16HP / Vmax F18 = sq.rt (5 / 5,49) = sq.rt. (length ratio)
resulting in :
Vmax F16HP / Vmax F18 = 0,95433 = 95 %
This would imply that the F16HP needs more sailpower than the F18 to be able to travel at the same high speed.
However I have big problems accepting this line of thought for the huge difference in overall weight is not taken into account at all by this formula. Lets do the following mind experiment. Make another hull by resizing the F18 hull by a factor of 10. The finess ratio is still the same so the value of the parameter is also the same for these two hulls. Now re-calculate the topspeed difference and get 0,316 = 32 % apparently this new hull needs even more power to push it up the same speed of a F18. Even though this hull has only a volume of 1/1000 of the F18 and the wetted surface of 1/100 of the F18. Continueing this line of thought produces a miniture hull of say 1/100 of the length of an F18 which would even need alot more power in order to travel throught he water at F18 speed.This cannot be correct. This formula must be heavily based on the assumption that shorter boats always need (alot) more power to achieve the same speeds as a longer boat does and was made to reflect this. Or it has incorporated a mechanisme that takes the linked reduction of sailarea into account. Froude's law has also been proven to wrongle predict multihull performance in the past which makes it less trustworthy now. This is not what we want. We want a formula that gives the hull drag forces at equal speed of two different hull lengths (or maybe even shapes). We could do with a formula that relates a (high) speed to a given thrust but this formula is apparently not doing that.
The third line of thought (taken from a maritime book) is that the total hull drag force is given by :
F drag = Function (hullshape, speed) * sq. (speed) * (displacement / length)
Where "function (hullshape, speed)" is a function giving a constant which is valid for a particular hullshape and speed combination. This is a so called "fiddle factor" which takes up all the characteristics that can't be easily described by mathematical means. This fiddle factor is often the problem for you don't know its value (or behaviour) and without it you can't obtain accurate results. However in this case we are trying to obtain the ratio between the two drag forces of two very similar hullshapes when both hulls are travelling at EQUAL speed. This fiddle factor will therefor drop out of the function because it is very likely to be the same for both hull. The resizing may have a small influence on the produced constant given but at a resizing factor of 91 % it can be assumed to be small and maybe even considered neglectable especially when being compared to the overall influence of the other important variable which differs alot between the two hull ; the ratio (displacement / length).
We will make this assumption and thus use the following equation :
F drag = constant * sq. (speed) * (displacement / length), with speed(1) = speed(2) so
F(F16HP) drag / F(F18) = constant * (displacement / length)
Now when we analyse the ratio (displacement / length) further we'll find that in this case the ratio is given by :
Ratio of Displacement / Length (F16HP / F18) = Volume ratio / Length ratio
91 % ^3 / 91 % = 91 %^2 = equal to wetted surface area ratio = 83 %
So drag force at high speeds is F (F16HP) / F (F18) = 91 % ^2 = 83 %
We've looked at three ways in estimating the high speed potential of the F16HP with respect to the F18. Two of these produced a high speed drag ratio of about 83% and one will produce a drag ratio that is considerable larger than 100 %. The last seems really unlikely for a hull is that is only 10 % shorter than a F18 and has very similar bow wegde and stern wedges combined with a wetted surface that is only 83 % of that of the F18. I choose to discard the second line of thought and accept the value given by the other two methodes as an estimation of the ratio of high speed hull drag, which is 91%^2 = 83 %
(Hintsight : real life experiments have shown that the grandfathered boats are indeed able to keep up with the larger 18 footers beyond light air windspeeds. This is an indication that the high speed drag of methode 2 is indeed errornous. Oktober 2001)
Input
Low speed drag (wetted surface area) = exactly 83 % of the F18
High speed drag (wetted surface + wavemaking) = about 83 % of the F18
We conclude that the F16 HP experiences 83 % of the drag of an F18 over the whole span of (winds)speeds.
comments
High speed drag estimate is only a rough estimate. We all know that the total drag of a sailcraft is composed of four factors :
Frictional drag (wetted surface)
Wavemaking drag (hull finess ratio)
Aerodynamic drag of the part of the boat above the water.
Aerodynamic drag of the crew
The first three factors can be assumed to be about 83 % of the same factors of the F18. The fourth is probably the same for both crews.
When we look at the ratio of the two engines, the sails, than we conclude that the sails ratio are approximating the ratios of drag. It can therefor be assumed that both designs have an equal performance potential. However there is an important thing that we haven't discussed yet and that is the "heeling and pitching moment" versus righting moment ratio.
The heeling and pitching moment is easy to calculate or rather to approximate ; Mast height ratio * mainsail ratio and was given in the paragraph "important ratios" to be 84 %.
Let's calculate the righting moments :
self righting moment F16HP / F18 = 100 * 1,25 / 180 * 1,3 = 125 / 234 = 0,534 = 53 %
Where 1,25 is halve the maximum width of the F16HP and 1,3 is halve the 2,60 mtr. width of the F18 and 100 and 180 are the individual platform weights. What does this mean ? Remember the heeling moment ratio of 84 %. Well, the F16HP can do the wildthing better than the F18 in the really light air because the crew starts to move up the trampoline sooner than the F18 crew when the wind picks up. Now when both crews are on the luff hull than ratio between righting moments has become :
ratio righting with crew on luff hull = 150*2,5 +125 / 150 *2,6 +234 = 0,801 = 80 %.
This ratio is gettting rather close to the 84 % of the heeling ratio of the rig so both crews are on the luff hull more or less at the same time. And now comes the fun part. Double trapezing !
ratio righting moment doubled trapezed = 150 *3,5 + 125 / = 0,840 = 84 %
But the F16HP rig only produces 84 % of the heeling moment when both sails are put in the same airflow. So both boats will be double trapped in the exactly the same windstrength and at that time the F16HP thrust is maxed out at 84 % of the equally maximized out thrust of an F18. So both boat will start to depower at the same windspeed and probably with the same amount.
So on the drag and thrust percentages given (all 84%) , an F16HP is perfectly equal to an F18 in all conditions ! That is in allout speed potential. The F16HP will definately outperform any F18 on things like manouvres and acceleration when going out of a tack or gibe due to its lightweight. It's just a more optimal design rule. However even though these advantages will help you win a race on tactical grounds they are by far to small to make you really faster; the rating (equal to F18) assigned to the F16 HP will remain unchanged.
When the wavemaking drag ratio is slightly more than the calculated 83 %, we can allow the F16HP to become wider and thus allow the F16HP crew to develop more thrust in the stronger winds without affecting the equality in the light and medium wind strengths. This way we can make the F16HP equal again to the F18 class. However this approach is limited to correcting out 1% or 2 %.
We conclude therefor that the F16 HP designs are comparable to the F18 designs to such an extend that sailing a F16 HP against a F18 is the closest one-design approximation of a race after sailing another F18 or the same F18 yourself. Both class might behave a little bit differently to conditions like waves or gusts but the raw speed potential will be equal nevertheless, both classes will only require different ways of sailing to get there.
Is this as fast as the F16 HP can go or will go ?
At the end of this analysis I really want to stress that the F16 HP may have been made equal to the F18 performance but that the F16 HP as a design can be made faster than the F18. We have limited the genaker area, genaker luff length, platform width and the jib area to luff combination to force the F16 HP design into the F18 performance. We can very easily boost the F16HP performance considerably by deregulating genaker size, genaker luff lengt, the jib area to luff combination and platform width. Just flying the maximum allowed jib and a bigger genaker of 21 sq.mtr would put the F16 HP class performance somewhere between the F18 and iF20 class. This is substantiated by real life experiences. The name HP is therefor well deserved.The costs to go All-out in this way are very small. Just buy a F16 HP with a larger genaker and jib instead of the smaller ones, keeping the purchase costs about the same.
Nevertheless the "equal to F18 class setup" will be used for pure F16 HP races in the startup fase of the class. This because the F18 is THE popular class of the moment and ofcourse we set ourselfs to goal to design a class that would follow the regulations set out by the Equal Performance Classes principle. This forbids increasing performance unless the performance of the next Equal Perfomance Class is achieved. For the F16 HP this would be the iF20 class performance and for now that is a considerable step up that requires alot of analysis first.
There are indications that that it is achievable but ehhh, lets stick to the current F16 HP that also allows 1-up sailing with that other well know class, the A-cats; we would surely loose this aspect when we go for the iF20 performance. And this dual role aspect of the F16 HP is one of the most exiting aspects of the class.
It will be "proven" in this appendix that a surface area ratio between two equally shaped but differently sized 3-dimensional objects is exactly equal to the second order (squared) of the size factor between the two objects.
It will also be "proven" in this appendix that a ratio of enclosed volumes between two equally shaped but differently sized 3-dimensional objects is exactly equal to the third order of the size factor between the two objects.
Lets start by looking at a simple 3 dimensional object, a perfect cube. The following equations can be
given for the cube's surface area and enclosed volume.
Surface area = 6 * height * width
Enclosed volume = height * width * depth
Now when the cube is entlarged by a factor of 3 the new surface area and new enclosed volume will become :
New surface area = 6 * (3 * old height) * (3 * old width)
New enclosed volume = (3 * old height) * (3 * old width) * (3 * old depth)
This can be written as :
New surface area = 6 * 3 * 3 * old height * old width
New enclosed volume = 3 * 3 * 3 * old height * old width * old depth
Which can be expressed as :
New surface area = (3) ^2 * ( 6 * old height * old width) = (factor) ^2 *old surface area
New enclosed volume = (3) ^3 * old height * old width * old depth = (factor) ^3 * old volume
Now if we devide the new areas and volumes by the old than we are left with a perfect second and third order of the entlargement factor.
This was simple. Now to proof that any 3-D shape follows the same principles. This is also remarkable simple. Entvision that the 3-d shape is made up entirely of little cubes which are packed together like lego blocks. Now make the blocks smaller, you find that the 3-D shape is more accurately approximated with degreasing cube size. The approximation of the volume of the 3D shape by adding the volumes of all those little cubes is also becoming more accurate when the cubes are made smaller and smaller. Eventually the volume of the shape and the total volume of all the infinitely small cubes added up will become perfectly equal, this proces is called integration in mathematics.
Now when we enlarge the 3D object by a factor 3 than we can do so by enlarging all the little cubes by a factor of three. The endresult will be the same. The total volume enclosed by the cubes is (3) ^3 = 9 times larger than before. And we may assume this to be the exact value of the enlargement of the enclosed volume by virtue of integration.
The same reasoning will apply to the surface area. Only then we'll approximate the 3d shape using little squares instead of cubes.