I'm 6 feet 2 and weight 225 pounds. how much weight on the tramp is needed to have the same affect of me being out on the wire?
wire or tramp
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Posted: Jul 06, 2010 - 03:50 PM
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Posted: Jul 06, 2010 - 04:17 PM
That depends on far too many things to get a legit answer, other than, "More than 225lbs."
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Posted: Jul 06, 2010 - 06:05 PM
If you think of a lever and your leeward hull is one end, and your body's center of gravity is the other. 8 foot wide boat plus your center of gravity about 4 feet = 12 * 225lb = 2700 foot pounds. Divide 2700 by 8 to put you back on the windward hull = 337lb. I think this is right but I failed math!
edited by: skarr1, Jul 06, 2010 - 05:06 PM -
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Posted: Jul 06, 2010 - 06:49 PM
if a 245 lbs man is on a train going southbound.. wait.. wrong question...
well the wind is variable... so the answer will change with every change in windspeed
as you move weight out over the side... you change the resistance to the force of the sail/wind... so this is a dynamic question.
the application should be: when weight can be moved outboard... it should be. It adds resistance to the force of the wind on the sails, thus improving lift. when that weight is to much to skim the windward hull, move it back in -
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Posted: Jul 06, 2010 - 06:50 PM
My above is true if the boat is level and you are hiked out level. At 45deg. your on the wire effective weight "since gravity pulls straight down"would be halved and the lever length would also be halved. So 112.5lb x 6' = 675ftlb. If you are on the hull 112.5lb x 4' = 450ftlb
edited by: skarr1, Jul 06, 2010 - 06:01 PM -
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Posted: Jul 06, 2010 - 07:29 PM
Yes but how does your body profile impact the windage across the sails and wind resistance to forward progress.
There are a ton of variables, but Steve is on the right track for a good approximation. In regards to the angle of the boat, it probably gets fairly complex since the center of gravity for both masses will get closer together as the angle changes.
Another explanation, same answer as Steve.
If you assumed an 8 foot beam, that your center of mass is 4 feet from your feet, and the boat is level..... You could take your weight x current distance from leeward hull / alternate distance from leeward hull. 225*12/8 = 337.5
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Posted: Jul 06, 2010 - 07:53 PM
darnit, just as I was starting to get that warm fuzzy feeling.
I guess hell is back to exothermic! :)
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Posted: Jul 06, 2010 - 09:14 PM
Fight those feelings~!
and...
'as if' you don't say "thus improving lift" at every possible opportunity
I thought so ( ;~{P -
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Posted: Jul 06, 2010 - 11:46 PM
You guys all failed math, where's Dave the engineer to properly explain turning moments & lever arms? Yurdle has the most correct answer.
If we isolate the problem to its simplest, the equations are pretty simple. Lets ignore all the little things, & focus on the two major moments.
You basically have a horizontal force that can be thought of as acting through the center of effort of the sail(much the same as the center of lift or mean aerodynamic chord of an airplane wing)trying to tip the boat to leeward. The taller the mast, the longer the lever arm, & the greater the moment. We all understand that the same moment could be achieved with a shorter arm (mast) coupled to a greater force. It is essentially the teeter totter problems you did in grade 8 math, or that you actually did on a teeter/totter when your pal was the fat kid.
It is not correct to think of the mass being halved when at 45*, it is the length of lever that has changed. Gravity is still pulling you vertically, & you would record the same value if you could step on a scale, HOWEVER, if you dropped a plumb bob, then measured from the leeward hull to the bob, you would find the arm to be shorter, resulting in a smaller moment. Draw yourself a right triangle, with the hypotenuse represented by the tramp at 45* then use SOH CAH TOA, or the carpenters can interpolate using the old "perfect square" of 3-4-5.
So, to solve you merely calculate the two different arms, then multiply the mass applied at the effective arm. This will give the righting moment sitting on the hull, or on the wire.
Now, whether this moment keeps you dry, or sends you scurrying for the righting line is an altogether different problem. As you tip, your righting moment is lessened, but so is the tipping moment as the sail is no longer perpendicular, however as the total mass of the sail & mast rotate towards the horizontal they create their own moment, becoming greatest once the mass reaches a horizontal position.
By that time you've probably jumped, & have a split second to analyze what you did wrong before hitting the water.
The total problem becomes a summation of vectors, you must add up each individual lever arm to get a final resultant, & that is assuming you can quantify each one involved.
In a nutshell, & a simple one at that, you have a mass of 225 lb(I know I know, we should convert that to stone, lbs is actually a force, not mass, why don't you Americans endorse the metric system?)on an arm of either 8', or 8+ "X", with "X" being how much further horizontally his center of mass acts through when on the wire.
Now, are we all clear, can we sleep tonight? That is as simple as I can explain it in writing, diagrams & a beer would be easier. The devil of course is in the details & once we add those we can argue forever.
edited by: Edchris177, Jul 06, 2010 - 11:42 PM
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Posted: Jul 07, 2010 - 12:20 AM
So, Are you a politician, 8 paragraphs and no answer in lb or stone!
I reserve the right to change my answer!!!!!
edited by: skarr1, Jul 06, 2010 - 11:23 PM -
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Posted: Jul 07, 2010 - 12:25 AM
I love Professor Hilliard's answer. Reminds me of the convoluted logic and math that came from the old hobielist/beachcats mailing lists in the 1990's.
Somewhere I've got a copy of the famous "Butt Cheek Differential" white paper by a Mr. Chris VanEpps who could always drop a nuclear bomb made up of equations to settle any argument.
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Posted: Jul 07, 2010 - 01:04 AM
Yeah what he said. I'm a hydraulics engineer so I can only legally take moments of fluids (beer is best) and I'm half cut right now.
Basically you can figure out the overturning/righting moment by taking the wieght (force not mass) of each piece of the the boat (hulls, mast, cross bars, you etc) and multiplying by the horizontal distance from the center of mass to the pivot point horizontally. It's pretty easy to do really uou learn that in grade 12 physics, and learn it REALLY well in the first 3 months of engineer school (that is essentially what structural engineers do all the time).
So assuming that the weight of the boat and mast is essentially negligible (not exactly true) you have to balance the amount of sideways force applied to the sail by the wind (lift). Fortunately you also have a lee hull, rudder and daggerboard in the water providing resistance to against heeling to help you.
You have a friend in that moment is force times distance so it is multiplicative. So if you are level and 10 feet from the lee hull while on the windward hull weigh 200 lbs you are applying a righting moment of 2000 ft-lbs (talking lbsF here). If the wind gusts and you fly the hull the (horizontal) distance between you and that leeward hull gets a little shorter, say 8 feet, now you are providing a righting moment of 8x200=1600 ft lbs, you lost 20% of your lenght and 20% of your righting moment. So to arrest the tendancy to capsize you have to do a few things, usually you initially dump wind with the main or traveller, or you hike to move your center of gravity back away from the lee hull and put things back into balance (you are usually also going faster so you get some extra help from the rudders and daggers).
Now comes the fun bit, whey you trap, horizontally away from your boat, all the sudden you have significanly moved your center of gravity away. Depending on your leg length it is usually and estension of about 3-4 ft (measuring to your bellybutton if you are a guy). So multiplying that out 200 lb x 13 ft = 2600 lbs or righting moment plus whatever the rudder/dagger/etc give you.
The moral of the story of going fast:
#1) You want to fly the windward hull to reduce drag but not too high. WHY!? because the flatter the boat is the longer the moment arm to resist the wind force when you hike or trap. You can't change your the weight of your ballast (you) but you can change its poition which is proportionally just as good.
#2) In heavy winds you can depower, dump wind OR trap out. Trapping out is fastest because you can use more wind power, trapping out with the boat close to horizontal provides the ability to use the MOST wind power and is therefore fastest.
So in the books when they tell you the flying the hull at 45 degrees isn't the fastest way to sail, those are some of the primary reasons. Of course I am ignoring the fact that when you heel, the sail depowers itself and becomes less and less efficient the farther away from vertical it gets but the juste is the same. It also explains why you can be trapping and near a capsize but stretch your arms way out, you can sometimes save it.
Now see if we can get EC to explain all the lift and shape characteristics of cloth sails and we will have it made. I can talk about cavitation until the cows come home but turbulent flow on bluff bodies boggles my mind(been meaning to read Prandtl for a number of years).
D.
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Posted: Jul 07, 2010 - 01:33 AM
Sorry Skarr1, your original math for the level boat, (teeter totter) is correct, I meant to include the simplified figures at the end...it was a long hot day sailing, I made nearly a mile with a hull in the air
a first for me, & since my guts are getting back in normal mode, the first cerveza tasted like another, then a whiskey front moved through.
SOH CAH TOA is the abbreviation for sine, cosine, tangent formulas involving right angle triangles. If you make the hypotenuse 8 (width of boat) & the angle 45*, you can use a cosine function to find the effective arm, then multiply by the force (225). Your mistake is that the cosine of 45* is .707 giving an arm of appx 5.7' The horizontal arm does not get halved at 45*, you have to go to 60*. It is the VERTICAL distance that gets halved at 45*. Often what at first seems obvious, isn't.
VERY SIMPLIFIED 225 lb sitting on the end of an 8' lever yields 225lb x 8' = 1800
On the wire, lets assume his centre of mass is 4 feet further out, & he keeps the hull just on the water, horizontal for our purpose, we now have:
225lb x 12' = 2700
What force would it take on an 8' teeter totter to balance the 225 lb kid sitting 12' out?
2700/8 = 337.5
And you guys thought math & physics were boring
edited by: Edchris177, Jul 07, 2010 - 12:59 AM
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Posted: Jul 07, 2010 - 01:43 AM
I was just fine until i tried to read this thread... no I have a splitting headache
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Posted: Jul 07, 2010 - 01:44 AM
Ok lets see how many paragraphs I can right and not come up with an answer.
If my boat is level and I am setting on the windward hull mast length has no effect.
If I was to attach a seat 4 feet out from the hull mast length has no effect. " boats with wings "
If I hang on the wire and place a weight scale between my feet and the boat the scale will read less than I weigh.
If I attach a weight scale to the trap line it will read less than I weigh.
I think "not know" if you add the two scales the answer would equal my weight.
The higher up the mast the trap line attachment, the trap line scale will read closer to what I weigh, and the scale between my feet an the boat will read less.
As the boat rotates the scale at my feet will read higher and the scale on the trap wire will read less.
At some point both scales will read the same. The longer the mast the closer to 45deg this will be. I think "not know" that they will both read half my weight.
When the boat reaches 90deg, the scale at my feet will read my weight. and the scale on the trap line will read 0.
To sum up in my original answer the 2700ftlb should be less. The 337lb on the hull would be less.
If I move my trap lines up to my spinnaker hound I will has an advantage. "if the mast does not break!"
edited by: skarr1, Jul 07, 2010 - 12:49 AM -
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Posted: Jul 07, 2010 - 01:57 AM
Sounds like he has quite a beam, 16' or so?
My guess is that the final question should be more like:
What force would it take on a 2*sqrt5' teeter totter to balance the 225lb kid sitting 4*sqrt5' out?
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Posted: Jul 07, 2010 - 02:10 AM
skarr1 please see apology above, only your second math stuck in my head.
Yurdle, don't quite follow 2*sqrt5??? The lever is 8' as the fulcrum is the far hull, well actually the center of the hull, & not really because as Dave alluded to, as the rudders & daggers (skegs can be thought of as very short daggers) rotate it creates resistance that gives a net result different than the actual width of the boat...but we are trying to simplify.
Christ, we are politicians...the questions original author is nowhere to be found!
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Posted: Jul 07, 2010 - 02:19 AM
hmmm...I was thinking from his center of balance to the base of the mast. Disregard.
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Posted: Jul 07, 2010 - 10:00 AM
Pretty simple. Just enough to get the cerveza out of the cooler for your Nacra chick!
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Posted: Jul 07, 2010 - 10:28 AM
Oh, I didn't even see the original post I thought we were talking physics. :)
The answer to the original question. If he is 6'2" and 225 lbs sitting on the windward hull and wants the same righting force as being out of the trap. Assuming the boat has an 8' beam and his center of gravity is 3' of his height.
Righting force sitting on hull = 8x225 = 1800 ft-lbs
Righting force out on trap = 11x225 = 2475 ft-lbs
Righting force (torque) is directly proportional to distance and weight therefore:2475/1800 = 137.5%
So you would have to be 137.5% heavier on the hull than on the trap to have the same effect. In other words you would have to be my sister-in-law.
D.
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