Hull Speed Question

S

seabug

Member
Joined
Aug 27, 2005
RO Number
18861
Messages
147
The thread posted on hull speed calculation needs explaining: Where does the constant (1.34) come from? Water-line length to the 1/2 power???? What's this? For "Educational Purposes" --not very helpful.
 
There is no easy answer, it's likely 4 pages of work that boils down to the equation. To the 1/2 power means square root. I'm an engineer I know what it takes to end up with an easy equation, spent hours and hours in school doing it.
 
Actually, that calculator was quite helpful: it gave you the answer. And should you wish to dig deeper, it also cited the source doculemt so you could look it up yourself!

The "Edu purposes" statement neatly solves two important issues: potential copyright violation, and also potential liability issues should someone risk a life on the "condensed formula". If you need better than a "loose rule of thumb" then you should consult a marine architect/engineer.

I find it useful.
 
I also find it usefull.
Close enough for "government work" / close enough to get you near where you should be, and in addition now i will be watching my wake closer, something i never did before, damn good boatered.
 
It's very helpful. If your not getting it, put a little more effort into trying to listen to what people are saying. Not everything in life can or should be dumbed down to just a simple boolean result. Reade Beebe for an explanation. Gerr has already been mentioned, but I suspect that while accurate, might not be what you are looking for.
 
In acquiring my 3 degrees in SCIENCE, we always used a RADICLE SIGN to designate SQUARE ROOT. I've never in any physics or math class heard of using 1/2 as a square root symbol. I'm also associated with several medical engineers who are in agreement. Effort? You must be kidding!
 
<<sigh>> I learned this one in HS... ( and reinforced in getting my degree ):

The multiplication property of exponents thus shows us that the 1/2
power is the same as a square root. In the same way,

(5^3)^(1/3) = 5^(3*1/3)
125^(1/3) = 5^1
125^(1/3) = 5

SO, yes, raising a number to the 1/2 power is the same as deriving the square root of the number.

I have done a lot of programing, and one of the very first things you learn is that is pays ( big time ) to remember ( and use ) all possible representations of a calculation. Many functions are handled diferently by a computer to achieve the same result, at varying degrees of computational effort. ie: a binary "shift left" operation, where possible, is much faster than a " divide by 2". It pays to be flexible.
 
OHHHHHHH, this is some kind of brain contest. I made the mistake of thinking it was about learning about boating. My mistake. I for one will kindly exit from any help as its quite obviously beside the point. I'm sure in a brain contest I would have nothing to contribute with somebody so smart as to have three degrees and wishes to shout down anyone willing to help them.

Oh wait...on second thought. Can I make a tiny suggestion? How about when you want to come and argue with everyone, you start by make the statement "I'd like to argue the point". That way, everyone can just ignore you instead of getting caught up in actually thinking they were here to help.

KMA
 
Bill - Thanks for your analysis and insight. One of my best friends is the chair person in Computer Science here at the University; every time that I talk to him about his work I can't help wishing I could learn something about pc's; I think it would be great to be able to create a program or software; it's a fascinating field. Thanks for the info, but I'm sorry about the Tempest in a Teapot.
 
Bill - Thanks for your analysis and insight. One of my best friends is the chair person in Computer Science here at the University; every time that I talk to him about his work I can't help wishing I could learn something about pc's; I think it would be great to be able to create a program or software; it's a fascinating field. Thanks for the info, but I'm sorry about the Tempest in a Teapot.
 
I should note that a "shift left" would actually multiply, not divide, by two. Divide is shift right. OOps.
 
Haven't read the parent thread to this one but hull speed calculations are covered in the Power Squadron Seamanship course, even have test questions on it.
 
quote:
Originally posted by JimPend

There is no easy answer, it's likely 4 pages of work that boils down to the equation. To the 1/2 power means square root. I'm an engineer I know what it takes to end up with an easy equation, spent hours and hours in school doing it.

Click to expand...

I thank Bill Gates for .XLS spreadsheets and their calculators they make my life easier Jim.
Bill
 
quote:
Originally posted by seabug

In acquiring my 3 degrees in SCIENCE, we always used a RADICLE SIGN to designate SQUARE ROOT. I've never in any physics or math class heard of using 1/2 as a square root symbol. I'm also associated with several medical engineers who are in agreement. Effort? You must be kidding!

Click to expand...


NO HE IS NOT KIDDING.

I do not know your age or where you were educated. However, back in the 1950's in North Carolina where I graduated high school and received a BS in Nuclear Engineering at NC State and later in the mid 19 60's when I got my Masters at Texas Tech the expenent 1/2 was recognized as the square root radical.

George
 
You guys are showing your age. The radical sign was fine when everything was handwritten and we used slide rules.... but I can't find one ANYWHERE on my keyboard.
 
. . . enough degrees here to make a whole circle; Seems this thread is right back were it started!
Q.E.D.
 
Good observation Gene, I also note many of the "waterlines" on some of the older ones here are getting a little wider too,,,;)
 
<< I learned more math during my 1 week "how to use the slide rule" class than I did in all years previous... >>
 
I can't wrap my head around the concept that one pretty precise hull speed formula applies to all hull configurations , regardless of displacement & draft , length to beam ratio , bow and stern shape & full keel vs various deadrise up to 0*. Is that formula really essentially for "average" sailboat or builtdown D/E type powerboat hulls with approx length to beam of 3/1 or a little less? Can anyone offer any insights here? And how does it apply to multi-hull vessels?

I get curious about this every time someone mentions that hull speed calculation, so if some wiser person can wrap my head better that should help. :D
 
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