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L
LeaRebanks1-Visitor
1-Visitor
October 10, 2012
Question

Spiral Clock Configuration

  • October 10, 2012
  • 1 reply
  • 4204 views

Hi All,

Please review the enclosed worksheet & plot.

# - I have 2 contra rotating spirals with their starting degree position 180 deg to each other.

# - Both spirals share the same 360 rotation radius value (k2)

# - The GOAL of this worksheet & my question to you is :- Are you able to setup the CORRECT configuration here.

Many thanks for your help & attention.

Any help or advice would be greatly appreciated shown with working example please.

Best regards,

Lea...

PS - I am using Mathcad version 15

1 reply

A
AndyWesterman12-Amethyst
12-Amethyst
October 11, 2012

Hello Lea,

I'm not sure if I have understood the problem correctly , but I am assuming that you want to optimise the value of K4 for any given value of miss (0.25 seems to be the critical point).

It is possible to parameterise the equations to allow the use of a solveblock to achieve this.

hope the attached reworking of your sheet will help

Regards

Andy

L
LeaRebanks1-VisitorAuthor
1-Visitor
October 12, 2012

Hi Andy,

Many thanks for your reply, really appreciated.

# - Although I don't completely understand your approach, I should point out that it does not quite achieve what I am trying to do. ( But maybe very close to what I want!)

1 - I have enclosed a reply to your attachment & highlighted the a few points.

2 - Firstly I have marked a fixed red plus sign on the V1 slope & outer circle intersection, this is where I want the blue circle of the dynamic trig configuration to equal. ( I sure you already know this, but only when miss = 0.25. But if miss = any other value other than 0.25 then blue circle is = + / - the red plus sign.)

At first look at your attachment k4 = 33 & configuration did not look right, but I changed k4 to = 0

3 - I have marked in red a couple of definitions that concern me. I see that you have defined k4 twice, please could you change this. And k4p with global definition = 8.972.... This value of 8.972.... must not be used as I am hoping that the trigonometric configuration defines this value 8.972 only when miss = 0.25.

On your attachment when miss = 0.25 to blue circle does not equal the red plus sign on the V1 slope, when k4 = 33.

But I noticed that when I change the value of k4 = 0 then the visible result is achieved, but not the solution I am looking for. Because if miss = any other value other than 0.25 then blue circle should be = + / - the red plus sign.

4 – Lastly, just as a reminder, that I only have one missing variable (miss) which should = 0.25 to produce desired result of trigonometric configuration.

Please could you make the alterations to your worksheet to achieve my desired result if possible.

Many thanks for your help & attention.

I really appreciate your input so far.

Best regards,

Lea...

A
AndyWesterman12-Amethyst
12-Amethyst
October 12, 2012

Hello Lea,

I've added a few comments to the sheet to try to help.

The given ... find solve block needs to have one or more conditions that it has to satisfy with the control variable, in this case k4(a), and a start value for the control variable.

The result of the solve block can be used to overwrite the original guess & mathcad will use the 'new' value from that point downwards in the sheet.

If it is easier to understand then alternate naming conventions can be used.

I only added the one condition 'volumeofV1' to demonstrate the principal.

If you want to add extra conditions here then you will need (for example) need to define the position of the blue circle relative to the red plus.

for the new sheet I have added equations to define that these should be coincident , but I understand for what you have said that this needs to vary dependant on the value of miss.

At the moment they are disabled, & the sheet seems to be able to converge to an answer irrespective of the guess value given to the solve block.

If you enable them, the number of discrete solutions that you will be able to find for any given value of miss will probably decrease (at miss = 1.25: 4 solutions without & 2 with),

AND I have noted (in the case of miss=1.25 and the guess value = 70 ) the solve block fails to find a solution, (see the graph of VolumeofV1) .

Depending on the complexity of the system it becomes increasingly likely that local minimums can lock the solve block into a state that does not meet one or more of the conditions set .

The disadvantage of the solve block is that it will give a single solution for the system, dependant on the initial (guess) condition.

It is then necessary to find all of the <real> solutions and verify that they correspond to the physical model.

Usually I like to plot a graph of any function that I am solving this way & this allows me to se if there are potential solutions that I may not otherwise detect.

In the sheet I posted earlier I had 2 global definitions for k4p which represented the solutions I found when miss was set to 0.25.

I don't know anything about the system that you are modelling so it is difficult for me to know if either of these are sensible answers or not.

but if you enable either one of these definitions it means that you can look back at some of the equations that have been modified to include the parameter k rather than the fixed value k4p (or k4 as your original sheet) to check if the resuts match the values that you would expect.

Hopefully I can point you towards a solution for your problem, if I can help further please let me know. & I would be interested to see what final solution you eventually produce.

I have learnt most of what I know from the members of this forum & I'm still finding lots more that I need to understand.

Problems like yours can be solved in many ways, this is just one.

Regards

Andy

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