J.C.MacSwell

Reading some of the Wiki suggested by Timo I found this in one of the links: "The visible disk of the Milky Way Galaxy is embedded in a much larger, roughly spherical halo of dark matter. The dark matter density drops off with distance from the galactic center. It is now believed that about 95% of the Galaxy is composed of dark matter, a type of matter that does not seem to interact with the rest of the Galaxy's matter and energy in any way except through gravity. The luminous matter makes up approximately 9 x 1010 solar masses. The dark matter halo is likely to include around 6 x 1011 to 3 x 1012 solar masses of dark matter.[6]" http://en.wikipedia.org/wiki/Dark_matter_halo It mentions the halo being "roughly spherical". How would it manage to maintain a spherical shape?

What evidence is their that there may be dark particles, rather than, say, a necessary adjustment to gravitational theory based on the detectable mass and energy that we know is there? For instance, do we see any spiral arms with significant gaps that we can point to and ask; what is holding that together? Or, how sure are we of the distances we are basing the calculations on, and could at least some of the required dark matter required be explained in this way?

Are you suggesting the forces on a plate in a fluid stream are somewhat equivalent to those from a jet of fluid impacting on the plate?

No. Can you not keep it straight? I described it correctly. There is significantly less turbulence at the lower Cd's. Why would I say "significantly less boundary layer turbulence". THAT would not be correct. THAT is the case at higher Cd's on the left of the graph where the skin friction is significantly lower....but due to the earlier separation and increased turbulence....higher Cd's Let's stay with what I said. There is significantly less turbulence at the lower Cd's. Now can you explain why you believe otherwise? What makes you think the turbulence is greater at the lower Cd's in the graph?

This is generally true (the bold part I bolded) for the boundary layer, and for streamlined objects where shear forces (friction) dominate the drag. It is generally not true for bluff bodies, such as in the example, where pressure drag (form drag) predominates, and for reasons I have explained a number of times,,,the transition to a turbulent boundary layer leads to delayed separation, decreased drag and less turbulence overall.

Good. We agree about what I said. This is wrong. Why do you think this is true? Do you believe that at the lower values of Cd the separation stars earlier and has more turbulence? (like the upper smooth golfball) Why (if that is your belief) would you think that would lead to a lower Cd, and the opposite case of delayed separation, (lower dimpled golfball) lead to a higher Cd?

No. You are confusing boundary layer turbulence with the much greater turbulence at separation. The increased turbulence of the energized boundary layer leads to delayed separation, substantially less turbulence overall and a lower Cd

...and the context was for the the opposite case. So you really are genuinely confused?

Perhaps you can point out where you thought I asserted that. It's getting harder and harder to believe that your confusion is genuine.

Don;t forget 12. All fundamental assertions, including this one, are highly suspect

I think "no matter what" may be a somewhat overstatement based on the simplest model, but it is usually fairly accurate in that range. Keep in mind that static friction is usually greater than that once the motion has started. http://en.wikipedia.org/wiki/Friction#Coefficient_of_friction

It is also of course related to the forces acting parallel to that, resisting the motion. I believe the coefficient is fairly stable within the yield points of the surfaces for most materials. Beyond that, with the materials stressed beyond their elastic and ultimate limits, things would change considerably with a number of factors coming into play.

What do you think happens at separation? Do you believe it is not turbulence? Turbulent, but less so than a turbulent boundary layer? For bluff bodies earlier separation without reattachment is significantly more turbulent than delayed separation and lower drag

"As we can see from figure 2.4, there is a drop in drag at a certain Reynolds number (approximately 4 · 105 Re 6 · 105). This is the critical region, where the boundary layer around the body transitions from laminar to turbulent flow, explained in section 2.3.5. The turbulent mixing that takes place in the boundary layer gives the fluid a higher momentum toward the surface of the body, thus moving the separation point farther back. In the critical range, small variations in the Reynolds number cause considerable changes in the drag coefficient." Can I just bold his words and hope you can understand them?

Read carefully what the Author is describing where you quoted him...more turbulence in the boundary layer (micro turbulence) moves the separation point (the start of much more significant larger turbulence) back further on the body...overall significantly less turbulence and drag

I guarantee you that in that example their is significantly less turbulence at the lower coefficients of drag in this example. The earlier transition to a turbulent boundary layer, which happens at higher Reynolds Numbers, triggers less turbulence overall and reduces form drag in examples like this one, which is probably a bluff body similar to a sphere.

As I said earlier: In both cases, the wing and a flat plate parallel to the flow, form drag is much less significant than skin friction. A change to a turbulent regime simply means more drag.

If you can assume the bullets alignment remains horizontal, and the vertical component of the velocity changes from 0 due to gravity, the pressure below the bullet will become greater than the pressure above it. A lift component will result which will slow a bullets fall relative to that of a dropped bullet, which should have a lesser, also upward, vertical drag component.

Does what? Turbulence does not equate to form drag. Form drag is due to pressure differential on the leading and trailing ends of a body in a flow. In a straight pipe, regardless of whether laminar or turbulent, all drag is affected as shear along the pipe wall. Obviously if a pipe has irregular buildup and needs cleaning as in the article you referenced on flushing procedures...then it's not really a straight pipe. Also note that your reference from the World of Physics - Drag Coefficient...includes a table where the drag coefficient of spheres (golf and tennis balls) increases at lower Reynolds numbers. I don't see where it helps your claim to the opposite.

I think that is the key. The friction is greater. But often with the drag of objects the form drag is significantly more important and makes up a higher portion of the drag. If it does not trigger less form drag (a straight pipe would not have any) you would just have the greater friction drag.

What Janus said. You are just displacing the water and converting the available energy in a more complicated way. Essentially you need even more complications to hide the flaw and that leads to even less efficiencies....so better to just use the turbines that are already there.

High velocities and larger objects lead to higher lead to higher Reynolds numbers- the boundary layer transitions to turbulent relatively earlier on the object. This boundary layer turbulence/mixing with the outer flow generally reduces the overall turbulence of the flow (swept downstream more readily and less build up of larger eddies) Though shape dependant this generally means that higher Reynolds numbers lead to lower coefficients of drag. At lower Reynolds numbers turbulence in the boundary layer are often induced for this reason, the earlier transition from laminar reducing drag (or increasing lift of a wing by delaying stall or assisting reattachment of the flow)

Did A and B both agree to the synchronization according to C? Because neither would agree on that unless starting from rest in C's frame. If they did, they would see a very non symmetric acceleration and speed paths of themselves vs the other, such that the clocks would only be considered synchronized at the start and at the end and never agreeing in between. Only C would see the paths as symmetric and the clocks continuously in synch..

Why do you think it should work?

The entry has the backing of something fairly solid. The exit has the backing of... just air