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Vent placement on stepped hull.
- Thread starter Bob Rosa
- Start date
Interesting thought? The boat I had seen at the Travers City boat show owner said there was enough vacuum with his step that the hull would not get up and plain properly. So I also had searched all records On file for this design step and this seemed to be the right thing to do.
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Looking at the image showing the framing in the step position, and other pictures, I see no trace of any longitudinal spray rails ending at the outer boundary of the planing bottom or the upper "soft chine".
If that observation is correct, it explains why the step has to get additional venting area. With the double "soft chines" as designed, the flow will not separate cleanly at the lower chine, but follow the increased deadrise upwards, thereby effectively blocking air access from the atmospheric side. Even if there are rails fw of the step, but too low or too close to the centerline, there may be a fluid film collecting and nicely following the bottom radius. This film has to be broken.
Before adding internal vents, I'd suggest adding a longitudinal spray rail forward of the step, its bottom surface to be a horisontal extension of the bottom in the transverse position, where the radius starts. My guess for width, say 50 to 60 mm, outer and aft edges sharp, vertical side height to meet "soft chine radius".
Edit: An easier (and more effective?) way is to add an interceptor at the step. Copy the shape of the frame at the fw edge of the step, up to the outer boundary. Add 5 mm to the profile outwards all along. Fw edge to be square and sharp, aft edge chamfer 45 degrees, leaving 1 mm behind the fw edge. Material SS316 or bronze for the legacy impression, thickness 3 to 5 mm. Make alternate holes for attaching screws to allow about 5 mm further height. Test!
If that observation is correct, it explains why the step has to get additional venting area. With the double "soft chines" as designed, the flow will not separate cleanly at the lower chine, but follow the increased deadrise upwards, thereby effectively blocking air access from the atmospheric side. Even if there are rails fw of the step, but too low or too close to the centerline, there may be a fluid film collecting and nicely following the bottom radius. This film has to be broken.
Before adding internal vents, I'd suggest adding a longitudinal spray rail forward of the step, its bottom surface to be a horisontal extension of the bottom in the transverse position, where the radius starts. My guess for width, say 50 to 60 mm, outer and aft edges sharp, vertical side height to meet "soft chine radius".
Edit: An easier (and more effective?) way is to add an interceptor at the step. Copy the shape of the frame at the fw edge of the step, up to the outer boundary. Add 5 mm to the profile outwards all along. Fw edge to be square and sharp, aft edge chamfer 45 degrees, leaving 1 mm behind the fw edge. Material SS316 or bronze for the legacy impression, thickness 3 to 5 mm. Make alternate holes for attaching screws to allow about 5 mm further height. Test!
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fallguy
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Very glad to see the reply.
I have to admit I am trying to understand and failing; perhaps Bob will understand better.
Bob ~ I told you privately and I offer again publicly to follow the guidance. I do not know all the credentials of the man, but his wisdom on water matters is exceptional and he has helped many people on the forum. If he tells you to break the flow off the hull side; do it. If you are uncertain about details; ask for confirmation.
Good luck.
I have to admit I am trying to understand and failing; perhaps Bob will understand better.
Bob ~ I told you privately and I offer again publicly to follow the guidance. I do not know all the credentials of the man, but his wisdom on water matters is exceptional and he has helped many people on the forum. If he tells you to break the flow off the hull side; do it. If you are uncertain about details; ask for confirmation.
Good luck.
philSweet
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This isn't my area of expertise at all, but I don't think that is normal, and I don't think that is how Hacker did it. I think the hull aft of the step should trim less than the hull forward of the step. This provides more stability as you patter across a light bay chop, where the step should ventilate just fine most of the time. If there is a ventilation problem, it would probably be in a glass-calm sea.
The good news, sort-of, is that the interceptor would probably have pretty much the same effect (making the hull forward of the step seem to have more trim), although I think more than 5mm would be needed to adjust the pressure distribution sufficiently (as opposed to just creating a clean separation to improve ventilation). But it is certainly the easiest and cheapest thing to play with. With an added interceptor, the cg needs to be moved forward from where it should be without one - that is the whole point of the exercise - to be able to move the cg forward and improve dynamic stability (thrust line also plays a role). The cg moves forward more than the center of drag moves forward. The inception of instability is reasonably straightforward to predict for a prismatic hull, but the step and the interceptor will make this harder. I couldn't find any papers on stepped hull critical pitch prediction.
Thank you Gents! Yes I am a bit confused yet but some of the ideas and concepts make sense to me. Unfortunately I'm a way from putting this hull in the water for the first time. I had no idea it would take me so long to build, power, wire and plumb this project. Each week and each day I try to conquer the next step of getting it in the water. Did some of the water plumbing and heat ex-changer plumbing this past week. Still need to do oil cooling, fuel system and wiring yet. I'm using an LS3 6.2 engine with fuel injection so a lot of new technology there to be figured out and learned. This 80 year old brain is challenged but I also have three other old guys trying to help me. Non of us are experts just trying to have some fun and get'er done together.
The advantages of a stepped hull is a reduction of wetted surface area/drag and the introduction of another stagnation line, ie the highest lift pressure on a hull. Much has been written on this subject but it appears that the consensus is that the lower overall drag is more significant in faster hulls where the skin drags begins to dominate. The gap in the graph of the stepped hull track where the pressure goes to zero (not exactly zero) from -1 to
-.5 is where the step occurs on the hull. I do not know why your are trying to add additional vents. If the step is vented on step, why bother.
The total lift should be the area under the curve.
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In your earlier post (#6?) you mention examples of similar design in need for additional venting area. With the step height 2.5" there is very little "channel area" from the side of the boat into the step for air transport; mind you, there is a lot of air required to avoid a vacuum-trap, so any blockage of the inlet has to be avoided, hence my suggestions.
As to PhilS's comments, I fully agree, but since I have no numbers to refer to (other than "15' from bow"), I'd start with a (probably too) low interceptor height and providing means for easy adjustments. The height is a function of the boundary layer thickness, which in turn is a function of speed and length of the wetted bottom area ahead of the step.
The hull is 25 feet OAL. The WL is almost the same. The original hull design is John Hackers Zipper 225 racer. I used Excell to factor the length tp 25 foot. So from the bow to the step is 15 feet. If I have built it correctly when on plain aprox. 3 feet forward of the step and 4 ft forward of the transom should be in contact with the water. I also added 6 inches to the beam to make the hull wide enough for two persons side by side. There are a couple of other things I've do to the bottom to create an almost sponson effect to the bow 15 foot. only in the water at speed will tell if I did good.
To Baeckmo
A question
The question of drag from a completely ventilated transom has come up on different topics.
I am curious what the force (accumulated negative pressure) would occur on ventilated transom, say 8 foot beam 4 foot height, 12 degree deadrise would be at 30 knots. Or how to determine this?
A question
The question of drag from a completely ventilated transom has come up on different topics.
I am curious what the force (accumulated negative pressure) would occur on ventilated transom, say 8 foot beam 4 foot height, 12 degree deadrise would be at 30 knots. Or how to determine this?
fallguy
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Thanks.
Somehow, I got stuck on tracing the step shape which I knew was wrong and filled the vent space.
I think I read another post and started to think the 5mm was an outboard adjustment and not downward, too, in case 5mm wasn’t enough.
So, this plate is roughly 2” wide by 2” high and say 6mm thick (3/16 to 1/4”)? Or a bit longer than the vertical dimension? Thanks, again.
I deleted my post which I qualified as wrong in the post itself.
Setting a chunk of bronze on the side of the boat at or just below the running waterline intuitively seems like a splash maker…but given the way interceptors work on a transom makes perfect sense. Sadly, ain’t my rorschach.
BoatBuildLog
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Are you trying to face air to a single step or working with multiple steps ? curious how you are deciding the vent location relative to the running surface.
Hi Barry,
First, since fluids cannot transfer tension, you should not use “negative pressure”; it often causes misinterpretations. You have to consider static pressures in the form of ambient (atmospheric) and hydrostatic (geodetic), plus dynamic pressures from/in moving fluids (water and air).
Second, the transom losses are part of the shape-dependent combination of events, including pressure differences and turbulence.
Let’s look at a rectangular barge at rest, floating at equilibrium. We focus on the events close to the hull, neglecting the wave systems in the more distant range. Since it is not moving, the sum of all horizontal forces is zero. At the transom, as well as the other sides, the pressure above wl is atmospheric. From wl and down there is the geodetic pressure added, it goes from zero at the wl to (rho*g*h), where rho is water density, g is earth acceleration and h is the local depth along the transom (and bottom).
If the barge is set in motion, the water (and air) in front of and behind the body has to move; ie it must get some dynamic energy/pressure, corresponding to the velocity: (pdyn=V^2*rho/2), where V is velocity in the horizontal plane. At the transom, this energy is taken from the hydrostatic pressure, resulting in a lowering of the surface level along the transom; there is less hydrostatic pressure pushing the transom forward.
Now, if we travel with the barge, the observed flow is analogue to the internal flow across a sudden area increase, where the transom represents the step change. It has been shown that there is a degree of energy recovery after the flow has left the step. In the case of a transom piercing the surface, close to 50% of the initial dynamic pressure is recovered, manifesting as less surface sinkage and a vortex rolling towards the transom at the surface, and down.
This means that the velocity required to achieve a certain “venting depth” is higher than that ideally corresponding to the actual loss of hydrostatic pressure (ie surface sinkage).
The reduction of the hydrostatic pressure on the transom is now a part of the increased pressure difference in the direction of the vessel movement (remember there is a pile-up of water up front in order to accelerate the fluid there). When the transom is fully ventilated, the only pressure acting on it is the atmospheric (but there is still a counteracting pressure from the fw hull zone).
With a transom depth of 0.4 m the “ideal” velocity for complete venting would be: rho/2*V^2=rho*g*h; which boils down to V=(2*g*h)^0.5; or about 2.8 m/s, which corresponds to a Froude number based on depth (=V/(g*h)^0.5) of about 1.41.
BUT: Due to the pressure recovery in reality, the Froude number for venting depth is about 3.2 (with some variation due to 3d-effects); in this case corresponding to 6.3 m/s for complete unwetting.
Now, if we check the conditions along the step as discussed in the thread, the area change is confined under a “roof”, thus there is no free surface and consequently no change of geodetic pressure available to set the fluid in motion. The energy must be supplied from the static pressure (sum of atmospheric and hydrostatic). Again, there is a vortex set up, but in this case the pressure may be reduced down to the vapour pressure of the fluid if the speed is high enough (slightly above 30 knots) and the ventilation is blocked off. This means a much higher drag than would occur on a corresponding transom.
So, what is the ideal transom depth then?
Since the flow losses along the aft buttocks and the transom venting depend on different flow mechanisms, the combined loss from the two must be speed dependant, ie the optimum transom/hull depth ratio will vary with design speed.
First, since fluids cannot transfer tension, you should not use “negative pressure”; it often causes misinterpretations. You have to consider static pressures in the form of ambient (atmospheric) and hydrostatic (geodetic), plus dynamic pressures from/in moving fluids (water and air).
Second, the transom losses are part of the shape-dependent combination of events, including pressure differences and turbulence.
Let’s look at a rectangular barge at rest, floating at equilibrium. We focus on the events close to the hull, neglecting the wave systems in the more distant range. Since it is not moving, the sum of all horizontal forces is zero. At the transom, as well as the other sides, the pressure above wl is atmospheric. From wl and down there is the geodetic pressure added, it goes from zero at the wl to (rho*g*h), where rho is water density, g is earth acceleration and h is the local depth along the transom (and bottom).
If the barge is set in motion, the water (and air) in front of and behind the body has to move; ie it must get some dynamic energy/pressure, corresponding to the velocity: (pdyn=V^2*rho/2), where V is velocity in the horizontal plane. At the transom, this energy is taken from the hydrostatic pressure, resulting in a lowering of the surface level along the transom; there is less hydrostatic pressure pushing the transom forward.
Now, if we travel with the barge, the observed flow is analogue to the internal flow across a sudden area increase, where the transom represents the step change. It has been shown that there is a degree of energy recovery after the flow has left the step. In the case of a transom piercing the surface, close to 50% of the initial dynamic pressure is recovered, manifesting as less surface sinkage and a vortex rolling towards the transom at the surface, and down.
This means that the velocity required to achieve a certain “venting depth” is higher than that ideally corresponding to the actual loss of hydrostatic pressure (ie surface sinkage).
The reduction of the hydrostatic pressure on the transom is now a part of the increased pressure difference in the direction of the vessel movement (remember there is a pile-up of water up front in order to accelerate the fluid there). When the transom is fully ventilated, the only pressure acting on it is the atmospheric (but there is still a counteracting pressure from the fw hull zone).
With a transom depth of 0.4 m the “ideal” velocity for complete venting would be: rho/2*V^2=rho*g*h; which boils down to V=(2*g*h)^0.5; or about 2.8 m/s, which corresponds to a Froude number based on depth (=V/(g*h)^0.5) of about 1.41.
BUT: Due to the pressure recovery in reality, the Froude number for venting depth is about 3.2 (with some variation due to 3d-effects); in this case corresponding to 6.3 m/s for complete unwetting.
Now, if we check the conditions along the step as discussed in the thread, the area change is confined under a “roof”, thus there is no free surface and consequently no change of geodetic pressure available to set the fluid in motion. The energy must be supplied from the static pressure (sum of atmospheric and hydrostatic). Again, there is a vortex set up, but in this case the pressure may be reduced down to the vapour pressure of the fluid if the speed is high enough (slightly above 30 knots) and the ventilation is blocked off. This means a much higher drag than would occur on a corresponding transom.
So, what is the ideal transom depth then?
Since the flow losses along the aft buttocks and the transom venting depend on different flow mechanisms, the combined loss from the two must be speed dependant, ie the optimum transom/hull depth ratio will vary with design speed.
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