Spiral Clock Configuration

[LeaRebanks]

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

[AndyWesterman]

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

[LeaRebanks]

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...

[AndyWesterman]

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

[LeaRebanks]

Hi Andy,

Many thanks for your help & suggestions to this problem, really appreciated.

I like the way you have plotted all possible results from solve block & the clear comments besides the conditions. All very helpful & certainly going in the right direction for my GOAL.

I really need to clarify the GOAL of this configuration.

1 - The GOAL of the configuration. Is to find the value miss = 0.25

when miss = 0.25 then the 3 equalities are satisfied. ( I will explain 3 equalities later.)

when miss does not = 0.25 then the 3 equalities are not satisfied.

It's all about configuring to find the value of miss = 0.25.

2 - We have 2 slopes

V1 = 1/6 which is a known Fixed slope

AND V2 = 1/(2.75+miss)

In this example for the purpose of setting up the CORRECT configuration let's say we have the value of miss should be = 0.25.

3 - The purpose of the spiral configuration is :-

(i ) - Create & use a common time balance value between V1 & V2 = k2

(ii) - Within this spiral configuration there should be a trigonometric setup, ( not done yet), that defines the k4 angle in such a way that the required angle of 8.9...deg is isolated.

( You have gone some way to doing this with your last attachment No.3, but no Trigonometry yet. It needs the 3 sides trig approach. We know RadiusV1 & RadiusV2 & symbolically can work out the 3rd side, I think. Therefore we should be able to find the angle 8.9...deg.... I think, but I don't know how to do this.)

(iii) - It has already been noted that there will be multiple answers for k4 degree for values of miss value. You have shown this with your plot. Therefore we use the spiral AND repetitions of time / degree and KNOWN volumes from the V1 slope for further equalities to complete the overall configuration.

4 - To complete the configuration & achieve the goal of "extracting" the value of miss = 0.25 we need to introduce a 2nd pair of spirals with similar configuration with the only difference being the slopes.

Still using the known V1 = 1/6 but changing V2 to V2odd = 1/(2.75+miss)*squareRoot of 2.

By using the new V2odd slope another time balance value will created instead of the value for k2 ( we will obviously have to rename k2 to avoid confusion.) But now we have another spiral configuration that produces similar V1 volumes on the V1 slope line using the SAME miss value = 0.25.

If we have found a way to use the trigonometry 3 side approach to "lock" the k4 angle then the miss value = 0.25 can be isolated by setting up 3 equalities as follows:-

Equality No.1 - spiral configuration of V1 & V2 = known VolumeV1 at Time 1.

Equality No.2 - spiral configuration of V1 & V2odd = known VolumeV1 at Time 2.

Equality No.3 - spiral configuration of V1 & V2 known VolumeV1 = spiral configuration of V1 & V2odd known VolumeV1 at Time 3.

The 3 values produced from this system of 3 equalities are :-

miss = 0.25 (ii) k2 from spirals V1 & V2 and (iii) k2(renamed) from spirals V1 & V2odd.

Obviously the miss value = 0.25 is a critical factor in determining the slopes of V2 & V2odd, therefore it is imperative to incorporate a trigonometric "locking" mechanism into the overall configuration.

Therefore, please could you review your worksheet to include the above.

Many thanks for your help & assistance here.

Best regards,

Lea...

[LeaRebanks]

Hi Andy,

I hadn't heard back from you since I posted my most recent reply.

I was just wondering if you were still reviewing this?

Is it possible to configure this by using one variable "miss" as I have discribed?

I would be grateful to hear your thoughts.

Best regards,

Lea...

[AndyWesterman]

Hi Lea,

Sorry for delay replying, difficult both from the work angle and also the fact that what you need is not wholly defined.

I did think about it a little (while in the garden) & probably the next step is to change the M2 equation to a parameterised form { M2(x):1/(.75+x) }

this will need all the following equations to change to match.

I would strongly suggest that You plot a graph of the BalanceV1andV2(x) against x (miss variable) as it appears to give some complex results and perhaps a singularity. Because we don't know the physical reality of your system its impossible to know if this is expected or not.

Instead of using k4p which I had changed earlier to allow the given ... find at the end, I think that a value needs to be calculated here (& this can revert back to the k4 name that it originally had) presumably based on calculation at x (miss) = 0.25.

As this will cause any unmodified equation to turn red it ensures that all affectefd equations are re-written & checked for validity (I find that it helps).

how this affects the rest of the calculation, I'm not sure at the moment, Hopefully you can try some of this out & let the forum know your progress.

Regards

Andy

[LeaRebanks]

Hi Andy,

Many thanks for your recent reply.

I shall try what you suggest & review the results.

Here is the plot you suggested.

It's interesting but I am not sure what I can learn from this.

I really appreciate the help & attention to this.

Best regards,

Lea...

[AndyWesterman]

Hello Lea,

I've modified your plot to show Re(balance...) & Im(Balance...)

Between -12 & -3 the result is complex & at ~3.25 it becomes infinite.

Both conditions will cause a lot of the following equations in your main sheet to fail.

the commoent was simply to check if this is important , or if these are conditions that can never apply.

Regards

Andy

[LeaRebanks]

Hi Andy,

Many thanks for that.

It seems that maybe I should change my formulae, configuration & plot into polar co-ordinates to avoid the malfunction as shown in the existing setup.

Many thanks for your help.

Lea...

[AndyWesterman]

Video Link : 3488

Hi Lea,

Had another look at your spiral question

video taken as an animation of the graph at the end

don't know if it is what you needed or not but either way, hope it helps

Regards

Andy

[LeaRebanks]

Hi Andy,

REALLY IMPRESSIVE... WOW wonderful animation... really cool.

That's exactly my concept.

I will need a while to go through your worksheet.

I still need to "extract" the 'miss x' value to = 0.25

Please review my notes on the goal of this exercise requiring 3 equalities and needs a 2nd similar spiral configuration that you have done, but using a different slope for V2 say V2*root2 but same 'miss x' = 0.25

It's all about "extracting" the 'miss x' value to = 0.25

I am really so impressed with your animation. Many thanks.

Best regards,

Lea...