BladeRunner 101

[MrRob]

Thats still more than a 25% increase. I think it can be done!

[bert]

Originally posted by Burty
as JBD said tuned up race 2.5's produce 400hp each but 800hp prob aint enought to push the blade to a ton

how fast do you reckon then can`t be to far away don`t know much about bladerunner hull but provee have 950hp inboard and go over a ton

[MrRob]

Cant be far off a ton with 800HP+

["Old beardy"]

Originally posted by glen76
there was a 4ltr or 6ltr batboat b28 for sale(last week but cant find it now), its open but maybe you could get pascoe to wack a canopy on it off a v24, or cut into for a leisure layout... who made the race b28 anyway. depends if u want someting off the shelf or if ur like me and always see potential ideas and projects (that are usually out my league) in everything

If anyone is interested in a 4ltr B28, Mick Archer has his up for sale. The whole rig on the trailer for less than £20K.

[JamesM]

I wasnt really being serious about the highly tuned 2.5's they wouldn't be any good for that application at all. I dont reckon you could do much better than a pair of 300X's, genuine 320hp each, and lots of torque but they arent much better than whats fitted now.

[Jon Fuller]

Originally posted by MrRob
Thats still more than a 25% increase. I think it can be done!

As a rough guide (when the only change is amount of power), to double the speed, you must quadruple the power.

So to increase the speed by 25%, it'd need approx 900hp.

---------------------------------------

Levi speed formula

Speed = C x sq-rt hp/weight

Speed(in mph) is equal to the “constant” times the square route of the “power” divided by the “weight”

So, for my boat the speed in “mph” is equal to: 4.8 (C) X the square root of 650 (“bhp” divided by 3 (tonnes))

4.8 X 14.7 = 70.56 mph

speed = 'C' X square of (power divided weight)

..................................power
Speed = 'C' X SR of -------------
..................................weight

-----------------------------------------------

Anyone know what 'Blade 101' weighs?

Of course, it may all be a little different with 'air entrapment' hulls, but the numbers worked well for my/roger severy's P28's, and our Extreme race boats, with differing power/weights.

mebbe our resident Naval Architect could comment

[BluFin]

BluFin (Fatboy) will take between 3 - 4 MPH of your speed

looks like no Sugar, low fat spread for next year !!!

[Jon Fuller]

Originally posted by BluFin
BluFin (Fatboy) will take between 3 - 4 MPH of your speed

looks like no Sugar, low fat spread for next year !!!

mm, I reckon you're a fake! (phoney! )

You don't look like one the "weigh a metric ton" mob to me!

[FrenchPhil]

34 Bladerunner : 2.5 tons roughly from website. How do you get 14.7 from 650 hp and 3000kgs ?

[Adam]

Not quite as simple as that - but shows why you have to work at all areas of the equation to see meaningful gains - for example decreased weight and increased power plus an overall increase in the value of efficiency - in this case 'C'.


In our office we use a similar formula - but use a factor of 'K'. This is then broken down into a number of other factors which gives us more ability to control the formula for changes between designs to show different drive units or numbers of drives for example.

We also use the formula with knots as our speed factor and the K value is where the name of the comapny comes from.

Originally posted by Jon Fuller
As a rough guide (when the only change is amount of power), to double the speed, you must quadruple the power.

So to increase the speed by 25%, it'd need approx 900hp.

---------------------------------------

Levi speed formula

Speed = C x sq-rt hp/weight

Speed(in mph) is equal to the “constant” times the square route of the “power” divided by the “weight”

So, for my boat the speed in “mph” is equal to: 4.8 (C) X the square root of 650 (“bhp” divided by 3 (tonnes))

4.8 X 14.7 = 70.56 mph

speed = 'C' X square of (power divided weight)

..................................power
Speed = 'C' X SR of -------------
..................................weight

-----------------------------------------------

Anyone know what 'Blade 101' weighs?

Of course, it may all be a little different with 'air entrapment' hulls, but the numbers worked well for my/roger severy's P28's, and our Extreme race boats, with differing power/weights.


mebbe our resident Naval Architect could comment

[Jon Fuller]

Originally posted by Adam
Not quite as simple as that - but shows why you have to work at all areas of the equation to see meaningful gains - for example decreased weight and increased power plus an overall increase in the value of efficiency - in this case 'C'.


In our office we use a similar formula - but use a factor of 'K'. This is then broken down into a number of other factors which gives us more ability to control the formula for changes between designs to show different drive units or numbers of drives for example.

We also use the formula with knots as our speed factor and the K value is where the name of the comapny comes from.

OK, so using a weight of 2.5 tons and currently 600hp, to achieve 80ish, how much power does Marc need to do a ton? (in mph)

[BluFin]

Originally posted by Jon Fuller
mm, I reckon you're a fake! (phoney! )

You don't look like one the "weigh a metric ton" mob to me!

Damm, the secrets out...........

[Jon Fuller]

Originally posted by FrenchPhil
How do you get 14.7 from 650 hp and 3000kgs ?

Sorry, i'm not too good at explaining things.

the figure of 14.7 is:

the square root of "the sum of": ...650 (hp) divided by 3 (tons)


650 divided by 3 = 216.66666...

the square root of 216.6666 is: 14.7 (closest to one decimal place)

14.7 is actually the square root of 216.09

but if you go down to 14.6, that's the square root of 213.16.

And if you go up to 14.8, that's the square root of 219.04

remember, it's only an example

[FrenchPhil]

So if my Argo weighs say 400 kgs and hp is 90 hp so :

4.8 * sqr (90/ 0.4) = sqr 225 => 15
4.8 *15 = 72 mph, sounds good

[Jon Fuller]

No, the 'constant' ( 4.8 in the case of my 28 )will most likely be very different for a cat with high drive height etc.

Our Extreme 26 race boat was I think 5.1 (more efficient) as it was a single drive, flatter deadrise etc.

You need to suss out what the 'C' is for your boat type/setup

to do this, you need a known speed/weight/power for an Argo 18 (with the same hull form as yours)

[John G]

Very Tricky - have you got to the bottom of whether your rev gauge is telling the truth yet?

[verytricky]

Cool!

Thanks. - Now I know I will not affordably top 100 in a Blade, I can move on.



I still want to hit 80+ comfortably every time I take her out before I sell her though!

[Jon Fuller]

Marc, if you're gonna be playing this coming weekend, I'll bring me digi-tacho along.

For what it's worth, my tachometers are about 300rpm adrift (bear in mind my motors only rev 3800 rpm, so it's fair sized % of the rev range), they won't adjust up any closer, and to address what you said to me in smallhampton, mine read the same as each other also, but they are still wrong!

[Johnny Rocket]

Originally posted by Jon Fuller
OK, so using a weight of 2.5 tons and currently 600hp, to achieve 80ish, how much power does Marc need to do a ton? (in mph)

By my reckoning this makes C for the Blade = 5.16

So to do a ton will require approximately 938.9 hp

This could be a nice little programming task for someone (Matt?). How about a web page where you can enter any three of S, C, P and W and get the missing figure? Then everyone could play the "how many horses do I need" game

[Jon Fuller]

Originally posted by Jon Fuller
As a rough guide (when the only change is amount of power), to double the speed, you must quadruple the power.

So to increase the speed by 25%, it'd need approx 900hp.

So to go full circle, my rough stab at 900hp wasn't a bad guestimate