# Centrifugal vacuum?

Interesting discussion! Any progress since September 2014, blainiac? :)
Yes, there has been a lot of progress. I will be building a large unit soon, but I have to wait until funds become available. Throughout the thread discussion, the formula below describes roughly how much thrust can be produced.

The thrust is in N, pressure is in Pa (N/m^2), rho is air density, theta is the angle for each blade from the root. I will post more when I can. I have written a Navier-Stokes fluid simulation to refine some aspects, but it's complicated. PM me for more details.

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Yes, there has been a lot of progress. I will be building a large unit soon, but I have to wait until funds become available. Throughout the thread discussion, the formula below describes roughly how much thrust can be produced.

The thrust is in N, pressure is in Pa (N/m^2), rho is air density, theta is the angle for each blade from the root. I will post more when I can. I have written a Navier-Stokes fluid simulation to refine some aspects, but it's complicated. PM me for more details.
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Blainiac,

Very nice! In your equation, is rho the density of air? Theta is the angle the fins make with the base (or inner diameter of fin connection point), yes? Wait, I see... You're using Bernoulli Navier-Stokes -- That is, you're assuming that the pressure will decrease based upon the increase in velocity of the air near the rim due to the fins striking the air, yes? Are you willing to question whether Bernoulli applies here? And are you willing to attempt to simulate this phenomenon using only Newton's laws instead? If so, perhaps maybe we can help each other... :)

Steve Quinn
Inventor - Rotating Thrust Producing Apparatus U.S. Patent 5,328,333 (expired -- the patent, not me... yet!)

Well speed of sound depends on the speed at which air molecules move (rms) but they are not held tightly together thus it takes some time for them to interact.

1.: well with given blade velocity you can really only get to a certain pressure differential. Increasing the blade height will help but only to a certain point. In order to calculate the exact pressure differential and best blade height you would have to approach this numerically.
If anyone has a better approach, please correct me.

2.: if the time that takes air to reach the inner cylinder is lower than the time necessary for 1 blade to cover the distance between 2 blades, you will have considerably lower average total pressure between the blades in comparison to max total pressure at the blade.
To properly describe this, it takes a few pictures and a page of text… so again this is an approximation.

3.: the air will move outward with the blade_velocity*sind(blade_back_sweep). Just calculate how long it will take it to exit and multiply that with angular velocity of the blades.

The stresses come from centripetal forces. Do not underestimate them if you intend to spin the blade tips at M˃1. Here is something to get you going in that direction: http://www.ewp.rpi.edu/hartford/users/papers/engr/ernesto/poworp/Project/4.%20Supporting_Material/Books/32669_04.pdf [Broken]

Here the questions just start.
Strive states it well. Notice that he does NOT use Bernoulli's ... "Higher velocity over a surface equals lower pressure on that same surface". I believe his approach is right.... But what do I know, I've been working this thing for over 25 years and still haven't got it off the ground. Help! :)

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