Rambunctious

try again to post a screen grab

Rambunctious

i give up for now attaching a screen grab, here is the websight i am referring to

http://www.kent.k12.wa.us/staff/trob...tml#Resistance

Rambunctious

ok tomcat, I know I've about beat all the life outta this topic.

but i found a source, he's not a rocket scientist, but he IS a boat scientist

Dr. John Batenh. PhD in Marine Engineering. His PhD topic was performance prediction of high performance planing watercraft.

He confirms my statement that the drag is a function of velocity squared thus the power is a function of velocity cubed.

YES

BTY,The reynolds number in the equation is a dimensionless number that ends up being a constant for a given boat lenth, water density, etc..

here is his reply:

Hi Rob,
Planing hulls and non-planing hulls produce a combination of dynamic lift and
static lift. The proportions vary, and the resistance is a function of
whether the vessel is going over the water or through it. The two primary
components of resistance are wavemaking and frictional resistance (mainly
frictional for a planing craft). in all cases the frictional drag is a
function of: (1/[Reynolds number (V.L/viscosity) squared]). The effective
power comprises the various frictional resistance values summed and
multiplied by the speed of the vessel through the water. The power required
to drive the vessel depends on the efficiency of the hull form, and hence the
laminar/turbulent flow around the hull. It also depends on the efficiency of
the propulsor and the efficiency of the combination of the propulsor with the
hull, i.e. hull, rotative and 'open-water' efficiencies multiplied together -
called the Quasi Propulsive Coefficient (QPC). Divide the effective power by
the QPC and multiply by the shaft/gearbox efficiency (losses) and you get the
power to drive the vessel through the water. Power is therefore proportional
to velocity cubed.
Got it?
Regards, John Batehn


Hey....We now have a new term to throw at the gear heads, " hey.... what is your quasi propulsive Coefficient".... (said in a nerdie tone)

So v^3 Rules!! right???????
i can't find any more proof than a rocket ... i mean boat scientist


Rambunctions

cobra marty

Yea but power is TORQUE and not HP. TORQUE is what turns the prop. MORE TORQUE = MORE POWER!! = MORE SPEED.

So, Power(Torque) is a function of V ^ 3!!

TORQUE RULES!

Rambunctious

Cobra Marty

See post 19346 Speed vs Power for more discussions. but....

By definition, Work (energy) = moving an object with a given force a given distance. (ex 1 foot- lbs)

Power, by definition = rate at which energy is produced per unit time
= (ft x lbs) / second

or lbs x (ft/sec)

or force x velocity

1 horse can lift 550 lbs at 1 ft/sec. that same horse can lift 1100 lbs at .5 ft/sec ( through a pully system) = both are one horsepower. the force is doubled in one case, while it's velocity is halved.

550 ft-lbs/ sec is the defininition of one horse power (1 Hp)

Torque x rpm is the rotating equivelent to force x velocity. 600 ft-lbs of torque at 3000 rpm is 1/2 the power as the same 600 ft-lbs turning at 6000 rpm.

think of your outdrive gearing. say 1.5:1 for a bravo. if your engine makes 600 ft-lbs at 6000 rpm, then through the outdrive gearing, the prop is making 900 ft-lbs of torque and is turning 4000 rpm. (neglecting drivetrain losses for now)

(600x6000) = (900x4000)

the power is the same (energy is conserved)

Cobra Marty, Your statement is true in the context when one is talking about a particular rpm. at that rpm, more torque = more power.

dean campbell

Rambunctious,
can you put some numbers in your equation to simulate a real world situation & do a 5 or 6 point data entry that will substantiate your hypotheses?

tomcat

I love a good equation as much as the next guy, BUT...sometimes the facts don't fit the theory. Here are some points on the curve.

A 36 Skater with a pair of HP500s will run about 110 mph under ideal conditions. The same hull with a pair of 900SCs will run about 150 mph under ideal conditions. Compare these numbers and you will see a velocity squared relationship for power. Take the same hull and put in 2 X 1500HP and you have the new kilo record of 190 mph, again squared. When I put these numbers into Mercury's (squared) equation I get very similar hull constants and curves that align almost perfectly. I don't know why and I can't explain it, but these actual test results don't agree with the power = velocity cubed relationship.

cobra marty

Interesting in another thread that for a speedo which is really a pressure guage but MPH = 8.1 times square root of pressure reading. So is mph related to square and not cubed?

Rambunctious

Cobra marty,
pressure of water is related to square of mph (velocity ) . Kinetic energy (energy of motion) = 1/2mV^2. thewater rushing at a certain velocity is "Stopped" thus creating a pressure proportional to the velocity squared (transforming all kinetic energy of the water onto potential energy (pressure))

POWER is Force X velocity. (add another velocity term) getting the velocity cubed function for Power.

Tomcat, Dean Campbell, I looked on Powerquest's website ( they build their boats here in Holland) to get speed vs power. Im afraid their speed data is marketing driven ( i.e. 5 mph for each engine upgrade) but they give a range in speed for each engine package. because of their large range, their data fits both curves. Your data would be a good example. my only rebutal to that is like you mentioned when we first discussed this. the hull in the cat may be lifting more and making it more efficient (through the reduction in wetted area).


I would love to find good data for a vhull to answer Dean Campbell. The Powerquest data isn't accurate enough.

ANYONE HAVE BEFORE AND AFTER POWER UPGRADES. LEAVING OUTDRIVE AND POSSIBLY PROP ALONE

Tomcat, could you also overlay the v^3 function to see the difference from the V^2 line you have just for fun?. What are you using for a graph? I was messing with excell for the Powerquest study I was doing to answer Dean C.