Horsepower/Sail Area equivalence

What you really mean D is that qualified engineers use the correct terms as taught for it has a specific meaning, so not to be confused with anything else.

Amateurs are unaware and use colloquialisms and as such saying one thing when in reality they mean another. When discussing with another amateur, they are both blissfully unaware and chat away in harmony. Whereas an engineer will say what do you mean by X or X doesn't mean that. Of course said amateur becomes chagrined ...and so the fun starts

 

Matt
Thanks for the lecture on imperial units. I have no idea about pounds and inches, when calculating..all a mystery to me. Nice summary

I have also noted in some of my old text books "tonf" too!!

 

Oh well this is a good thread

these are one of the many questions that my "yacht club experts" could not answer properly either

this will also help you to understand why i am doing what i am doing with the microcruiser - to have a "working model"

http://www.boatdesign.net/forums/boat-building/manies-microcruiser-27869.html

i will keep you guys up to date, thanks

 

thanks daiquiri,
thats the answer,
only 100% wrong. it's the very very tiny group of people who have taken the word power to mean something ultra specific in their tiny tiny tiny little group, when they should have just invented another word.

 

Yes, scientists, mathematicians, engineers, architects - they are all very strange people. So you better not listen to them and don't ask them to give you the answers.
You already know well that 1 HP is 1 mq of sail - which is 1 square feet, because it's obviously the same sail.
So don't bother asking around, they will only confuse you.

 

Power can mean a lot of different things, but we don't need to resort to the dictionary here; right from the getgo the reference was to horsepower and meaning was clear.

 

Jeez why must you guys always RUIN a good thread with drivel

this is boatdesign.net

go and write your nonsense on your back wall

 

You are right Manie, I'm sorry for the sarcasm.
It is because some honest tries by Marshmat and R.W. to give an answer have been criticized for the use of correct language. Or at least that's how I understood it.

Now, getting back to the topic, there cannot be a unique answer to the original question. Depends on type of sail, on it's main dimensions, on rig setup, on heel angle, etc.
Even if you know exactly what sail you are talking about, it's power will also depend on the apparent wind speed... This one, in turn, depends on the boat speed. The boat speed will depend on the hull type, on displacement, on waves, on currents, etc.

So, what is really the number one can give? Even if you want to give an approximate number, you need to specify first the above quantities...

 

It's more like RM(max) equals hp.. the sail area just defines the wind speed where it is..

 

Well.. You could.

Lets say..

Sail on a beam reach (apparent 90deg) Different sail areas, different wind speeds, giving different boat speeds. Then, motor the boat at those different speeds noting the power the engine needs to match these same speeds. Graph it is see what it looks like.

It would be a lot of work, but at least most of it you'd be out on the water doing "research".

-jim lee

 

What part of my previous statement didn't you understand?

 

daiquiri!
apology for any offense.

"So you better not listen to them and don't ask them to give you the answers."

bit over the top isn't it? to ignore them all just for misusing a word?
when they obviously should have just followed my suggestion, and selected a latin or greek word and built a discipline specific word, so as to remove all doubt about meaning? rather than stupidly add to the confusion by doing what they did with the word power?

perhaps i should also rewrite the above post to say,

thanks daiquiri,
thats the answer,
except 100% opposite?, the exact opposite? etc


also, fyi,
("It is because some honest tries by Marshmat and R.W." etc
no.
you're after !ON!
read this
http://www.boatdesign.net/forums/multihulls/catamaran-beams-28196-3.html

any excuse will do. i've read others posts of yours. all the same. any excuse will do. you're not in the right, and never will be until 10 years(?) after you retire. (but only if you play hard, relax etc)

can i suggest recognizing the condition exists is the first step towards manners?

imagine the trouble you'd be in if your job depended on good maners? (rather than the single ability to lift heavy mental weights). imagine if i was a client?
(remember; don't get upset - this is exactly what you're body wants you to do. and, it's an astoundingly common syndrome amongst all professionals)





PAR
i read that, and it's the reason i asked here, cause i calculated 300% different.



rick!
,

"The weightlifter had to produce power to lift the weight but once the weight is in a steady position he is no longer producing power."

ridiculous. i had no idea engineering was so removed from common agreed on reality.

isn't common agreed on reality the absolute BASIS of all science?

i say thats an obviously absurd and patently artificial situation, that NO sane person can agree with. "once the weight is in a steady position he is no longer producing power."

it's obvious to me that legacy math from the stone age was blunt.





also;

now we're all good friends again, tell me this, i can't relate this picture to ricks explanation.



i can't reconcile this chart with the newtons calculated per m2 @ 5ms.

the chart was Kg force 1 m flat square sail perpendicular to the wind

 

In case you were a client Bob, every professional would send you home, when you´re not willing to accept the laws of nature and rules of professional business.
When you misunderstand and misuse a term, you can not in all seriousness claim that others have to stay calm about your argueing about there correct use.
So, it is not astounding, that we send several idiotic "clients" through the door every year.

And your picture shows a incorrect term!

Regards
Richard

 

Mal
The graph gives force various values of Cd for downwind condition at various windspeed. It provides values of Cd for the second equation I provided:
Drag Force = 0.5 x rho x V^2 x Cd x Area (In line with the apparent wind)

For example at 15m/s and C = 1, the force is 15kgf. Using the equation:
Force = .5 x 1.2 x 225 x 1 = 135N = 13.8kgf

A little different. Air density is a function of temperature so within the accuracy of the chart without temperature being specified.

On the first item:

When you use the word "horsepower" you are using a well defined term for power originally specified by James Watt. He estimated that the rate a horse could do work was 33,000ft-lbf/min. The story goes that he needed some way to compare his engines to what others could appreciate.

When you use the word "horsepower" you have carelessly wandered into the engineering realm where the word "power" is narrowly defined. It is not a commonly agreed term, rather it is precisely defined:
"Definition: Power is the time rate at which work is done or energy is transferred. In calculus terms, power is the derivative of work with respect to time.
The SI unit of power is the watt (W) or joule per second (J/s). Horsepower is a unit of power in the British system of measurement."

 

All the SI formula for wind-force will give N/m^2

When you see kg shown as a force it just shows that someone used a mass conversion table forgetting that lb-force actually converts to Newtons NOT to kilograms.

This confusion of units is a problem. Otherwise well founded material from institutes like Westlawn and Dave Gerr unfortunately perpetuate this because they don't understand the SI system. As soon as you are given kg as a force you must convert it to Newtons for it to be of any use. So it’s a confusing and backwards effort to talk of kg for force .

The SI system of units is actually very easy to work with and very logical.
it is a very good idea to get an understanding of the fundamental unit of force called after Isaac Newton ( Newton's work in the 1600's was on par with Einstein's for sheer deductive brilliance ). The Newton(N) is close to the force experienced from holding a decent sized cooking apple on the Earths surface. It’s easy to relate to.

One pound force is 'four and a half' Newtons (4.5N)
It’s much more informative and helpful in the next step in a calculation to use correct units. Material strength for example is given in Newtons per square millimeter AKA megapascal (MPa)

Just to illustrate this
Consider how many apples (N) can I support with a mild steel wire 1mm in diameter with a yield strength of 300 apples(N) per square millimeter ( AKA 300MPa yield ).
.After about 5 minutes of exposure the Newton becomes a real and useful unit which anyone can grasp and which is useable for the next step in the calculation.