resistance scaling

[previah]

Thanks for the responses guys! Obviously there are a lot of different opinions on this subject. What I have in mind is to use the admiralty formula to estimate the resistance:

R ~ disp^(2/3) x V^2

Thus:

R2 = R1 x (disp2/disp1)^(2/3) x (V2/V1)^2

Assuming V2 = V1 (in reality V2 < V1) then:

R2 = R1 x (disp2/disp1)^(2/3)

Obviously the real answer is far more complicated than this. But can this be used as a first quick and dirty estimate?

[DCockey]

Originally Posted by gonzo

The resistance on a strut or keel is in part from the tip vortex. The would be the same regardless of displacement-draft.

Only if there is leeway/angle of attack, assuming the usual situation of lateral symmetry.

[Leo Lazauskas]

Originally Posted by gonzo

The resistance on a strut or keel is in part from the tip vortex. The would be the same regardless of displacement-draft.

Induced drag is not relevant to this discussion.

Using a surface-piercing parabolic strut allows the wave resistance to be
calculated analytically so that the effects of draft (and hence displacement) can be easily discerned. .

There is a fairly detailed discussion and examples using "Tuck's Strut" in Chapter 4, Section 3 of:
"Hydrodynamics of high-speed marine vehicles"
Faltinsen, O.M.

[Leo Lazauskas]

Originally Posted by DCockey

Only if there is leeway/angle of attack, assuming the usual situation of lateral symmetry.

Just to be obnoxiously pedantic, that's only strictly true if there is no free-surface.
With waves created by the hull (and/or ambient waves) there is a vertical
component of velocity that can create vortices at the bottom of the strut, even at zero AoA.

[DCockey]

Originally Posted by Leo Lazauskas

Just to be obnoxiously pedantic, that's only strictly true if there is no free-surface.
With waves created by the hull (and/or ambient waves) there is a vertical
component of velocity that can create vortices at the bottom of the strut, even at zero AoA.

I assume you are talking about vortices due to separation along the tip. If so then a free surface is not necessary for a span-wise component of velocity. Such vortices can also occur at sharp chines, tight radius bilges, etc.

[DCockey]

Also can be vortices at the root of the strut, etc due the boundary layer along the hull rolling up and separating. Again, no free surface necessary.

[Leo Lazauskas]

Originally Posted by DCockey

I assume you are talking about vortices due to separation along the tip. If so then a free surface is not necessary for a span-wise component of velocity. Such vortices can also occur at sharp chines, tight radius bilges, etc.

No, I was talking about (potential) flow on the flat bottom of a strut.
I take your (equally pedantic) point about separation and other viscous effects.

[DCockey]

Originally Posted by Leo Lazauskas

No, I was talking about (potential) flow on the flat bottom of a strut.
I take your (equally pedantic) point about separation and other viscous effects.

No vortices are generated in purely inviscid (theoretical) flow. There needs to be at least vanishingly small viscosity for separation to occur and for vortices form. There are several types of separation in three dimensions. One is essentially the same as two dimensional separation with the velocity on/near the surface going to zero and a "bubble" forming. Another is when the streamlines on/near the surface lift off and form trailing vortices without a "bubble". Classic case is along the leading edge of a delta wing at higher angles of attack. It can also occur on a tip of a wing with finite thickness even when there isn't an angle of attack or "lift".

[Leo Lazauskas]

Originally Posted by previah

Thanks for the responses guys! Obviously there are a lot of different opinions on this subject. What I have in mind is to use the admiralty formula to estimate the resistance:

R ~ disp^(2/3) x V^2

Thus:

R2 = R1 x (disp2/disp1)^(2/3) x (V2/V1)^2

Assuming V2 = V1 (in reality V2 < V1) then:

R2 = R1 x (disp2/disp1)^(2/3)

Obviously the real answer is far more complicated than this. But can this be used as a first quick and dirty estimate?

To test your idea, see:
Robards, Simon William, "The hydrodynamics of high-speed transom-stern vessels"
http://unsworks.unsw.edu.au/vital/ac.../unsworks:3426
Appendix D contains the resistance graphs.

Specific resistance (Rt/Weight) for a variety of drafts and displacements
is shown as a function of Froude number for many hull series (e.g.
NPL, NOVA-I, II, III and IV, D-Series, Sklad, Series 63 etc.)

Good luck!
Leo.

[Leo Lazauskas]

Originally Posted by DCockey

No vortices are generated in purely inviscid (theoretical) flow. There needs to be at least vanishingly small viscosity for separation to occur and for vortices form. There are several types of separation in three dimensions. One is essentially the same as two dimensional separation with the velocity on/near the surface going to zero and a "bubble" forming. Another is when the streamlines on/near the surface lift off and form trailing vortices without a "bubble". Classic case is along the leading edge of a delta wing at higher angles of attack. It can also occur on a tip of a wing with finite thickness even when there isn't an angle of attack or "lift".

I consider the flow with a vertical component on the bottom of the (cut-off) strut to be similar to the flow over a low-aspect ratio wing at AoA. The bottom of a parabolic strut is like a lenticular wing, i.e. with a pointy leading edge at midspan and rounded wingtips. Therefore, there is a small amount of lift, and hence there will be induced drag due to wingtip vortices.
There's no need for viscosity here.

I'm happy to be disabused of this opinion, though!

[DCockey]

To expand a little further about potential flow, vorticity, and viscosity.

For any object in a potential flow field, there is a potential flow solution which has continuous velocity potential everywhere exterior of the object and no trailing or other vorticity. However if there are zero radius corners/edges on the surface of the object then the velocities of that solution will be infinite at the zero radius corners/edges except for the special cases such as the flow aligned with the edge or the edge being a stagnation location. I'll stress that this is a theoretical solution and may or may not have physical significance.

Infinite velocities don't actually occur so the theoretcial solution needs to be reconciled with reality. One way to do this is to introduce a sheet of vorticity which originates at the surface where the velocity is infinite and goes downstream to infinity. Outside of the sheet the flow is still irrotational (no vorticity). By adjusting the strength and distribution of vorticity in the sheet the infinite velocity can be made finite, and in fact there will not be any flow across the edge. This is one way to describe the Kutta condition.

So why don't infinite velocities at sharp corners/edges occur in reality. The answer is viscosity. Any amount of viscosity, even an a tiny, tiny bit, would cause infinite stresses which leads to separation. The vorticity sheet is the idealization of the separation as the visicosity becomes vanishingly small.

[DCockey]

Originally Posted by Leo Lazauskas

I consider the flow with a vertical component on the bottom of the (cut-off) strut to be similar to the flow over a low-aspect ratio wing at AoA. The bottom of a parabolic strut is like a lenticular wing, i.e. with a pointy leading edge at midspan and rounded wingtips. Therefore, there is a small amount of lift, and hence there will be induced drag due to wingtip vortices.
There's no need for viscosity here.

I'm happy to be disabused of this opinion, though!

If you are talking about "leading edge suction" and the like that will have to wait. We're headed out to a concert. But I will say that can also be related to vanishingly small viscosity.

[Leo Lazauskas]

Originally Posted by DCockey

If you are talking about "leading edge suction" and the like that will have to wait. We're headed out to a concert. But I will say that can also be related to vanishingly small viscosity.

No, I wasn't talking about LE suction.
I over-thought myself into a mistake by imagining the (flat) bottom of a strut
as a lifting surface. Of course, there is no pressure difference as with a real
wing.

I understand your points about vanishing viscosity.
Without viscosity there can be no starting vortex so planes could not take off, let alone fly. Verified by experiments in zero-viscosity, super-cooled helium, I believe.

[DCockey]

A with a flat bottom, even a prismatic constant cross-section strut, would have flow across the edges between the sides of the strut and the bottom. If the edges are sufficiently "sharp" then in the real world (as opposed to the ideal mathematical world) separation would occur along the edges and could result in trailing vorticity.

[previah]

Leo,
Thanx for the link. Unfortunately the data are for high speed vessel with Fn > 0.2, while the range I'm interested is for Fn < 0.2 instead. Moreover the admiralty formula, I think, is only applicable for low speed.

Best regards,
-Arman-

Originally Posted by Leo Lazauskas

To test your idea, see:
Robards, Simon William, "The hydrodynamics of high-speed transom-stern vessels"
http://unsworks.unsw.edu.au/vital/ac.../unsworks:3426
Appendix D contains the resistance graphs.

Specific resistance (Rt/Weight) for a variety of drafts and displacements
is shown as a function of Froude number for many hull series (e.g.
NPL, NOVA-I, II, III and IV, D-Series, Sklad, Series 63 etc.)

Good luck!
Leo.