Avanti 22 speed vs.hp
07-18-2014, 02:10 PM
#21
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Thanks, SJ.
07-18-2014, 09:08 PM
#24
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lol I think it's funny how 99% of people say "what's an avanti" when you tell them...
been a hell of a Lil boat...
been a hell of a Lil boat...
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07-22-2014, 01:09 PM
#26
The nerd-ist answer is this; The power required is (1/2*p*v)[SUP]3[/SUP]*A*Cd
Where p is the density of the fluid
v is the relative velocity
A is the cross-section Area
Cd is the drag coefficient
So if we put some arbitrary numbers in for effect, we will see the power differential at given speeds. So if the density of the fluid is 1, the speed is 40, the Area is 2 and the Cd is 0.5, you need 8000 units of thrust to maintain the speed. Changing only the speed, it looks like this;
40=8000
50=15625
60=27000
70=42875
80=64000
90=91125
So to double the speed from 40 to 80, you need 8 times the thrust. From 40 to 50 is almost double, requiring 7,625 over the 8000. But 50 to 60 requires 11,375 for the next 10 mph.
If you plot the points of hp required at a given set of speeds (say, every 5 mph. And the hp scale should be available for a given engine at differing rpm) then you can project APPROXIMATE speed increases up the log scale. Where planing hulls differ from cars and planes is as you accelerate, you CAN reduce the area in contact with the water (to a point).
For some reason, this discussion never impressed the girls...
Where p is the density of the fluid
v is the relative velocity
A is the cross-section Area
Cd is the drag coefficient
So if we put some arbitrary numbers in for effect, we will see the power differential at given speeds. So if the density of the fluid is 1, the speed is 40, the Area is 2 and the Cd is 0.5, you need 8000 units of thrust to maintain the speed. Changing only the speed, it looks like this;
40=8000
50=15625
60=27000
70=42875
80=64000
90=91125
So to double the speed from 40 to 80, you need 8 times the thrust. From 40 to 50 is almost double, requiring 7,625 over the 8000. But 50 to 60 requires 11,375 for the next 10 mph.
If you plot the points of hp required at a given set of speeds (say, every 5 mph. And the hp scale should be available for a given engine at differing rpm) then you can project APPROXIMATE speed increases up the log scale. Where planing hulls differ from cars and planes is as you accelerate, you CAN reduce the area in contact with the water (to a point).
For some reason, this discussion never impressed the girls...
07-22-2014, 02:02 PM
#27
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In answer to your question, all else being equal, yes. But that being said, everything else isn't equal. For cars and planes the basic rule says that when speed is doubled, drag is squared. But for a planing hull, the area of wetted surface is reduced as you go faster. Additionally, water is incompressible, so there are a few other issues that come to play. Now dereknkathy is correct that when you reduce the wetted area, you increase the aerodynamic drag, but that is nothing compared to the hydrodynamic drag. You'll see it when you change the trim on your boat at speed. Push the bow down and the boat slows. The more you lift out the water, the faster you go. And we're only talking about Vees here. Cats bring a whole set of additional calculations and considerations to the dance.
Please also note that I just picked some random numbers for fluid density and Cd. However, the principle remains the same.
So there are numerous variables that come into play here. In addition to the change of the wetted surface area, you have a difference in the prop efficiency as well. most props create thrust on a non-linear curve as well. Just as your hull has an efficiency "sweet spot", so will your prop. So "8x" thrust may take 8x hp, or more, or less.
Last edited by Pilotpete; 07-22-2014 at 03:16 PM.
07-22-2014, 02:11 PM
#28
All kinds of variables in you really think about it. At higher speeds, less boat in the water = more boat weight on a smaller area of water, but then again, decks are pretty flat, the bottom has a big curve under the bow, at 70 is the air not giving the boat a bunch of lift by pushing on the curved underside? ever try and hold a peice of plywood on the roof of the car at 70?
07-22-2014, 03:22 PM
#29
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All kinds of variables in you really think about it. At higher speeds, less boat in the water = more boat weight on a smaller area of water, but then again, decks are pretty flat, the bottom has a big curve under the bow, at 70 is the air not giving the boat a bunch of lift by pushing on the curved underside? ever try and hold a peice of plywood on the roof of the car at 70?
07-23-2014, 05:20 PM
#30
Well, first off, I can't take credit for the formula. I went to one of my books on the shelf and copied it all over. I got through some of my math classes by dating the professor's assistant.
In answer to your question, all else being equal, yes. But that being said, everything else isn't equal. For cars and planes the basic rule says that when speed is doubled, drag is squared. But for a planing hull, the area of wetted surface is reduced as you go faster. Additionally, water is incompressible, so there are a few other issues that come to play. Now dereknkathy is correct that when you reduce the wetted area, you increase the aerodynamic drag, but that is nothing compared to the hydrodynamic drag. You'll see it when you change the trim on your boat at speed. Push the bow down and the boat slows. The more you lift out the water, the faster you go. And we're only talking about Vees here. Cats bring a whole set of additional calculations and considerations to the dance.
Please also note that I just picked some random numbers for fluid density and Cd. However, the principle remains the same.
So there are numerous variables that come into play here. In addition to the change of the wetted surface area, you have a difference in the prop efficiency as well. most props create thrust on a non-linear curve as well. Just as your hull has an efficiency "sweet spot", so will your prop. So "8x" thrust may take 8x hp, or more, or less.
In answer to your question, all else being equal, yes. But that being said, everything else isn't equal. For cars and planes the basic rule says that when speed is doubled, drag is squared. But for a planing hull, the area of wetted surface is reduced as you go faster. Additionally, water is incompressible, so there are a few other issues that come to play. Now dereknkathy is correct that when you reduce the wetted area, you increase the aerodynamic drag, but that is nothing compared to the hydrodynamic drag. You'll see it when you change the trim on your boat at speed. Push the bow down and the boat slows. The more you lift out the water, the faster you go. And we're only talking about Vees here. Cats bring a whole set of additional calculations and considerations to the dance.
Please also note that I just picked some random numbers for fluid density and Cd. However, the principle remains the same.
So there are numerous variables that come into play here. In addition to the change of the wetted surface area, you have a difference in the prop efficiency as well. most props create thrust on a non-linear curve as well. Just as your hull has an efficiency "sweet spot", so will your prop. So "8x" thrust may take 8x hp, or more, or less.
I have found that the formula used for speed calculators to be very accurate. Divide current speed by the sq root of (HP/weight) to get your constant
If a boat does 60 mph with 350 hp and weighs 4000 lbs
350/4000 = .0875
sq root of .0875 = .29580
60/.29580 = 202.84 constant
Once you have your constant, you can use square root of (HP/weight ) x constant = speed
Say you add 150hp
500/4000 = .125
sq root of .125 = .35355
.35355 x 202.84 = 71.71 mph
So, in this case adding 150hp would add 11.7 mph
Using these formulas I predicted two engine transplants within 1 mph
