Jump to content

psyclones

Members
  • Content Count

    28
  • Joined

  • Last visited

Community Reputation

1 Neutral

About psyclones

  • Rank
    Quark

Profile Information

  • Favorite Area of Science
    Mechanical Engineering.
  1. I realise I have not been as involved in the discussion as every one else. I may have created the thread, but in my opinion I'm not the "Arbitor" of the thread, people can say what they want. All insights and comments are welcome! I realised I read you post it out of context - So again, sorry.
  2. This statement was NOT aimed at you studiot. It was talking about at the guy who told me the story - which I realize now maybe I shouldn't have mentioned. Knowing the details of the problem I think is sort of beside the point. What I was trying to find is simple conceptual flaw in the screw design – if there was one. You don't have to insult me.
  3. My apologies I haven't chimed in yet. (I need to switch on notifications from this forum if that possible). To answer your question studiot, I don't have any more information - this was just a story told to me. What made me question it was the credentials of the guy who designed it, who I'm guessing know very little if anything about fluid mechanics. I just thought it was a very interesting problem. But as the discussion has unfolded - modelling the drain pipe as an open channel flow can get quite tricky. So details an assumptions about the problem are critical to solving it, which is what I thought. But thank-you all.
  4. Hi, Someone told me after building a multi storey building. In one section of the building existing downpipe(s) weren't effective at transferring water from roof to ground level - water I believe was pooling on roof. To solve this, a screw design (see attached) downpipe was created. The screw design supposedly helped suck the water down the pipe. There are two problems that appear to me straight away: 1. The velocity ignoring frictional losses down the screw design would be less because gravity forces are not pulling the liquid straight down, the smaller the incline (from horizontal) the slower the inlet velocity of liquid. 2. The internal screw sheeting and core structure subtracts volume the flow could occupy in the straight downpipe, effecting flow rate. Or course this is highly simplified and I'm unsure of how centrifugal effects on water flow swirling around would affect flow rate and head losses. There would be higher head losses per length of pipe, but I'm unsure if those losses would lower inlet pressure to below that of free falling water flow. Trying to fabricate and even clean would be just short of a nightmare. Although I think its novel an interesting idea, but I can’t see the design would be more effective than a standard downpipe? Like I said this was just a story, I didn’t actually get to see this. I’ve attached cad model of how I think it would work. What are your thoughts?
  5. Ok that's interesting. I can only find as copy of Principia Mathematica at the University library, of which I'm not a current member.
  6. There’s a reference in Bertrand Russell's book: Analysis of matter, pg 3, third paragraph - continuing onto Pg 4. He talks about R as relating a term to its successive term in a sequence - using the sum of first "n" odd numbers equaling n^2 as an example. I'm not sure how he defines "R" to start with in this example. Then introduces Rxn, which relates a sequence of numbers together?? I'm not sure if I'm reading his notation correctly. Any input on this passage would be helpful. There is link to the book below. http://strangebeautiful.com/other-texts/russell-anal-matter.pdf
  7. Hi, I've been pulling my hair out trying to solve this. If we treat it as a series similar to proof by induction it can be simplified. Am I on the right track. But the LHS doesn't appear to factorize and reduce in any easy way! 2017-10-24.pdf
  8. Hi, Can you give me any tips of how to begin this proof using the Fibonacci sequence? Fn-1Fn+1 - Fn2 = (-1)n , where Fibonacci numbers can be calculated by Fn = Fn-1 + Fn-2 If I let n= k+1 then, but Fk Fk+2 - Fk+12 = (-1)k+1 If k = 0 , 0x1 - 12 = -1 (true) , but how do I take it further.
  9. I've paraphrased the question: Yes it's 8 filling ingredients. I thought too that it was 1/8, but I don't believe that's the answer. The topic is combination(s) (nCr) within the text. so the sandwich can have ham only, or ham and lettuce, or ham and tomato, or ham tomato and lettuce and so on.. up to 7 other ingredient fillings combined with ham. Does that help to clarify the problem? I can't remember how to use Latex code on your site or I'd show my working using nCr notation. [ nCr(7,0) + nCr(7,1)+nCr(7,2) + ... +nCr(7,6) +nCr(7,7) ] / [ nCr(8,0) + nCr(8,1)+nCr(8,2) + ... +nCr(8,7) +nCr(8,8) ] = 2^7 / 2^8 given: nCr(n,0) + nCr(n,1)+nCr(n,2) + ... +nCr(n, n-1) +nCr(n ,n ) = 2^n
  10. Did I complete the above post incorrectly?
  11. I found a problem, I just like to check it. Question: From 8 total ingredients, (ham being one)- what's the probability of choosing a sandwich with ham? My solution: 27/28 = 128/256 = 1/2
  12. Thanks for your post. I can assure you I've used 'three moment equation(s)' (solving three equations simultaneously) to arrive at the given moments at b and c, and reaction at b (reaction b is under the support b). My question is, given the manner in which I've formulated the M(x) equation. Is it correct- given I have fixed unequal (in this case) BM's at points b and c, with a free moment due to dist load? [edit] Using the integration method to solve for angle = 0, to solve for deflection. [edit]
  13. Please excuse [deflection] error. Could you help me with the following. Using continuous beam theory, constructing BM diagram from points b to c, to calculate the max deflection. I only found a have a single solution, though the BM digram show two points of zero bending. I can provide the solution. your thoughts. [edit: Rb = 685 N] Please find revised attachment. attach1(revised).pdf
  14. Hi, The problem uses a continuous beam, but I'm unsure if I'm correct in solving for the max deflection between pts A and B (I've included the BM and SF diagrams btw pts A snd B, only). Deflection should be a maximum when SF = 0. I've also formulated an equation to solve for max deflection using the second area-moment, where 'x' is measured as the distance from A. The BM at B ([latex] M_{B} [/latex]) is straight forward, but the BM due to load is non-linear, I believe this needs revision, but your thoughts. [latex] \Sigma A\bar{x} = \int_{0}^{X}wx\frac{x}{2} \,dx - M_{B}\frac{x}{L}\frac{x}{2} [/latex] BM cont beam.pdf
×
×
  • Create New...

Important Information

We have placed cookies on your device to help make this website better. You can adjust your cookie settings, otherwise we'll assume you're okay to continue.