Planing Hull at Disp Speeds

Tad, I agree with what you've wrote. I believe that the said paper does present a result of a more comprehensive research done at Wolfson Unit, but at the same time it shows just one small bit of those results. That's why it is, imho, usable as an example which shows the importance of the transom area ratio, but surely cannot be used for a more precise quantitative analysis of a generic hull. It just lacks too many details regarding the geometry of hulls included in the research.

Let me explain my previous words in a more precise way. I'm just trying to read that partial data from Wolfson Unit in a correct way.

The graph in the paper by Van Oossanen shows the dependance of resistance on the transom area ratio, for two Froude numbers: 0.35 and 0.6 . The first Fn rappresents a speed regime corresponding to S/L ratio of 1.18 (in usual knots and feet units), which pretty much corresponds to a cruising speed regime. The second Fn corresponds to S/L ratio of 2.0, which is a semi-displacement speed regime.
The resistance is given through a Telfer Coefficient (Ct), which is defined as Resistance/Displacement ratio divided by S/L ratio (all values must be in consistent units).

What I said in the previous post is that, according to that graph, the Ct of a vessel with a transom area ratio close to 1 (rappresentative of a planing vessel) appears to be 2-2.5 times higher than a Ct of a vessel having a zero transom area ratio (grossly rappresentative of a displacement vessel). That is valid for Fn=0.35 (displacement cruising speed).

Now, a constant Fn means that the S/L will also be constant (they differ by a square root of g, which is a constant). If you also consider just vessels having a same displacement, then Ct (by definition) will be proportional to vessel's resistance only (at that S/L ratio). And since the Ct of a wet transom vessel is claimed to be 2.0-2.5 times the Ct of a vessel with dry transom, then their resistances will differ by the same factor. Again, it is only valid for Fn=0.35 (or S/L=1.18) .

This is just the reading of that infamous graph... We don't have sufficient info about their method of research to discuss the domain of it's validity, as I said before in reply to Tad.
What we can do is to compare these results with a significant number of real-life displacement and planing boats, and see if the numbers correspond... But, as we know - in theory there is no difference between the theory and the practice, but in practice there is. So real-life data will be much more scattered than that simple linear graph. Just consider the unknowns related to propellers efficiencies, amongst many...

Cheers

P.S.
UUUUPSSS!!! I've had to edit the above text, because I had inverted the two vessel cases - dry transom with fully-wet transom. Sorry!

 

Mat-C,
My Willard is a thirty footer. As in 30' w a 27.5' WLL.
That should help the numbers a lot!!
I usually claim 18 hp to make 6.2 knots. At WOT my Willard gets about 7 knots but probably not quite 7.
There is a 40' Willard with the same type of hull that requires 23 hp to make 7 knots. The W 40 is 33000 lbs.

Easy

 

Sorry guys,
I had to correct my previous post because of a very stupid typo - see the P.S. part.
Cheers!

 

.... I don't have figures for the consumption or power required to run at 6 knots, but it is likely to be rather less than at 7. So on that basis, probably a little less than the Willard requires, which is no great revelation as the Willard weighs twice as much.
With a little luck I'll be out this weekend, so will see what I come up with....

But what is obvious to me from all of this is that the most significant factor in all of this is WL length, followed by weight. No surprises there....
Yes, the immersed transom contributes to the drag, but I don't for a moment believe that it results in requiring 2 1/2 times the power to run at displacement speed.

 

Talking about how much power is requited or fuel consumption is irrelevant at a S/LR of 1.3 for a full displacement craft. It's like measuring gas milage of a Honda Civic at top speed. Who cares. My Willard is known as a 6 knot boat and there is no other speed worth considering as a speed to cover distance. Comparing a 40' boat at well below its cruising speed w a 30' boat at its top speed is an exercise in stupidity.
A 40' Willard FD boat requires 23 hp to make 7 knots. How much power does a 36' GB need to make 7 knots? A 42' GB? Split the difference between the 36 and the 42 and one should have apples and apples w size and WLL with the 40' Willard. Or to put it another way put my 37 hp engine in a Willard 40 and the same 37 hp Mitsubishi in a 36' GB. I don't think the GB would even make 7 knots at WOT.

Easy

 

Sorry, I'm a bit confused... who's talking about ddisplacement boats at SL's of 3?

 

From page 1 of this thread...

So no, a 37 bhp engine would be unlikely to be able to push the (much heavier) GB at 7 knots by the time you take into account driveline losses etc... it'd go pretty close though...

There's no doubting that hullform has an effect on the efficiency of displacement craft. Round bilge vs hard chine, immersed transoms, slenderness, etc, all have their contribution. It's the scale of that contribution that's at question here...

 

As to submerged transom area vs. speed - see graph. This graph is actually from my paper on catamarans, but derives from study of monohull shapes.

It can be clearly seen that submerged transom starts becoming justified from FnL=0.35.

 

Yes, it does mean exactly that: to overcome 2.2 -2.5 times resistance, it is necessary to have 2.2 -2.5 times thrust, and to have 2.2 -2.5 times thrust, it is necessary to have 2.2 -2.5 times engine power (for same % of propeller efficiency)

real life example:
this summer I happened to take part in sailboat regatta in fat *** centreboard boat, LOA 6m. Her WL beam at transom is ~80% of WL beam amidships, and transom immersion ~10-20% of canoe body max immersion. In very light air, at Fn < ~0.2, a single person, sent to the bows, so that transom is lifted out of the water, made a difference in speed ~50%. If we assume that resistance is proportional to speed squared, this mean ~2.25 times less resistance, when transom is lifted out of the water.

 

That's not really what I meant PS - I realise that overcoming twice the resistance will take twice the power to overcome. What I meant was that there's more to it than that... the weight (and length) come into it too....
By way of example... say it takes 2 hp to account for the resistance of a given hull without an immersed transom. For arguments sake, we accept the 2.5 times the resistance to push the same boat but with an immersed transom - so 5hp. But it might require 20hp to push the mass of either, regardless of their shape.
So the non-immersed transom hull requires 22hp, the immersed transom 25hp.

Right / wrong?

 

Well, Mat, it's clear that the previous reasonings were done for equal displacement and length (see the post #55). That's how I have intended the initial quest by Easy.
If you change the displacement, it's obvious that resistance will be affected, even for the same boat. Same thing for the length.

 

So what's this a graph of? Range of ideal/suggested transom vs. speed?

 

Yes, it shows recommended submersed transom area as function of Froude number.

 

to move each pound a helicopter requires more power than a fixed wing aircraft and a fixed wing aircraft will require more power than a wheeled vehicle

It's how you move the weight

 

Excuse me sir, are you lost?