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**toitu****Member**- Registered: 2010-03-16
- Posts: 32

Hi there

I have a problem that I need some help on. I have two sets of values where the ratio between the two varies in time - one is a measure of larger rock sizes and the other of smaller rock sizes from the same sample. What I am trying to do is come up with some type of index value that reflects the relative difference in sizes e.g. the Index value approaches +1 when the larger rock sizes are significantly higher than the small ones and -1 (or 0) when the larger rock sizes are closer to the smaller values. However, I'm not sure how I can do this. Any help or suggestions would be fantastic!

Thanks.

Toitu

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**mathsyperson****Moderator**- Registered: 2005-06-22
- Posts: 4,900

Forgive me if I've misinterpreted, but I think you want a function like f(x) = 1 - e^(1-x) would work.

When the ratio is 1 (ie the rocks are the same size), then f(x) = 0.

As the ratio rises, so does f(x). For example, f(2) ≈ 0.632 and f(3) ≈ 0.865.

As the ratio gets very large, f(x) tends towards 1.

A more general version of this function would be 1 - e^(a(1-x)).

"a" is some number that you could change according to your needs.

f(1) will always be 0 and f(x) will always tend to 1 as x gets large, but changing a will change the speed at which it does so.

For example, if your big rocks tend to be about 1000 times bigger than the small ones, you might want to make a fairly small so that you don't get a lot of index values in the region of 0.999.

Why did the vector cross the road?

It wanted to be normal.

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**bobbym****bumpkin**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 109,606

Hi all;

To me the question seems to be like calling up your mechanic and asking to fix your car over the phone? It is way too general. How about hauling the car in so he can get his hands on it?

Unless you provide data, I mean ordered pairs, how can this question be answered? Unless you know or suspect the underlying law, the DE, a fit is arbitrary. We first try a polynomial fit, why, because everything is known about them. They are easy to compute and manipulate, in short they are easy.

Same reason we linearize everything, it makes it easy.

How about providing the data or at least the locus. Or do you already suspect the relationship, the differential equation?

It is true, sometimes you can guess the law by looking at the data. The most famous curve fit in history, Kepler and Brahe. But Kepler had volumes of data to look at.

**In mathematics, you don't understand things. You just get used to them.****If it ain't broke, fix it until it is.**** Always satisfy the Prime Directive of getting the right answer above all else.**

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**toitu****Member**- Registered: 2010-03-16
- Posts: 32

Hi all

Thanks for your suggestion mathsyperson and point taken, bobbym.

*Last edited by toitu (2010-12-17 03:52:45)*

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**bobbym****bumpkin**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 109,606

Hi toitu;

It will take many questions before something can be done. We have the time column that is fine. Where do you see the relationship where something is approaching 1? One or two examples of your thinking here will provide a hint as to what you are trying to do and how I can convert that to math.

**In mathematics, you don't understand things. You just get used to them.****If it ain't broke, fix it until it is.**** Always satisfy the Prime Directive of getting the right answer above all else.**

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**toitu****Member**- Registered: 2010-03-16
- Posts: 32

Hi bobbym

My thinking, which was more than likely incorrect, was that it would make it easier to understand the size mix index if it could be scaled so that 1 represented a large difference between the 50th and 90th values and values closer to 0 represented less of a difference. This was just an arbitrary idea on my part, as, obviously, a non-mathematician.

What I am trying to do is plot/find some sort of relationship between the instrument outputs (I have two that respond in slightly different ways) and the rock size size mix using the 50th and 90th to represent the entire sample. The idea is that if this relationship is reasonable, I can then use just the instrument outputs to roughly determine the rock size mix e.g. proportionally more larger rocks or smaller rocks.

The instruments both respond in slightly different ways especially if there is a lot of larger particles so my thinking was to use these differences to 'predict' changes in the mix of finer and coarser material. So I guess I have two ratios/relationships that I am trying to compare: one between the 50th and 90th rock sizes and one between the two instruments outputs.

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**bobbym****bumpkin**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 109,606

No problem. Your idea is logical but difficult. Please bear with me, I will need much information to understand.

So we have two instruments and two rock sizes? Large and small 90%, 50%.

You want to predict a mixture based on the two instruments readings?

Where are the two instruments readings?

**In mathematics, you don't understand things. You just get used to them.****If it ain't broke, fix it until it is.**** Always satisfy the Prime Directive of getting the right answer above all else.**

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**bobbym****bumpkin**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 109,606

First line 513 and 363 what does that mean? 513 of what and 363 of what?

I know you are columnizing the data to be readable to humans. You probably have it in this form:

513,363,525,387...

Can you provide it like that cause then it goes directly into the machines?

**In mathematics, you don't understand things. You just get used to them.****If it ain't broke, fix it until it is.**** Always satisfy the Prime Directive of getting the right answer above all else.**

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**bobbym****bumpkin**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 109,606

I assume you are using a light beam and a photodiode or phototransistor of some sort.

Let me graph the data, so I can look at it first.

**In mathematics, you don't understand things. You just get used to them.****If it ain't broke, fix it until it is.**** Always satisfy the Prime Directive of getting the right answer above all else.**

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**toitu****Member**- Registered: 2010-03-16
- Posts: 32

Yes that is correct - I have two turbidimeters, each from a different manufacturer.

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**bobbym****bumpkin**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 109,606

Just to ask, has nothing to do with the problem. What is the smallest size particles they can handle?

**In mathematics, you don't understand things. You just get used to them.****If it ain't broke, fix it until it is.**** Always satisfy the Prime Directive of getting the right answer above all else.**

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**toitu****Member**- Registered: 2010-03-16
- Posts: 32

They are most sensitive, generally speaking, to the 5 um to 62 um range so coarse clay to fine silt. Depending on how the optics are arranged and wavelength, you get varying sensitivity to the really small clays

and to sand.

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**bobbym****bumpkin**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 109,606

When I take your first instruments reading and call that x and the second y we get this. It does not look like there is any relationship between the two instruments.

**In mathematics, you don't understand things. You just get used to them.****If it ain't broke, fix it until it is.**** Always satisfy the Prime Directive of getting the right answer above all else.**

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**toitu****Member**- Registered: 2010-03-16
- Posts: 32

Oh woops - my apologies, I sent an incomplete dataset - if you add these values the relationship between the two instruments should improve towards the end (best seen plotted as two time series). This is really the crux of my issue as both instruments are "seeing" the same thing but the difference in response is likely caused by the ratio of coarse to fine particles.

214.773,209.800,211.853,174.199,189.173,191.371,192.880,185.553,190.166,192.646,193.968,194.669,195.047,191.664,191.809,194.483,199.939,201.024,206.703

210.832,187.377,169.235,159.652,163.636,156.078,147.651,141.822,139.128,136.263,137.929,139.762,135.770,137.610,146.714,181.534,213.340,222.791,235.928

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**toitu****Member**- Registered: 2010-03-16
- Posts: 32

and the remainder of the size data

16.601,18.126,16.717,18.734,19.829,18.677,21.798,16.067,16.643,19.997,20.474,19.652,20.983,18.882,18.712,19.87,18.825,20.268,21.573

108.259,111.85,84.324,98.898,78.927,96.461,98.442,91.044,75.171,92.825,81.87,101.233,89.004,85.324,76.127,85.562,90.995,103.068,89.413

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**bobbym****bumpkin**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 109,606

You want me to add both of those to the original lists?

**In mathematics, you don't understand things. You just get used to them.****If it ain't broke, fix it until it is.**** Always satisfy the Prime Directive of getting the right answer above all else.**

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**toitu****Member**- Registered: 2010-03-16
- Posts: 32

If it will help - sorry, I wasn't paying attention to where the data ended!

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**bobbym****bumpkin**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 109,606

That is okay. I will now graph them with respect to time. Time will be the x axis. Can you please provide the times in the same format.

**In mathematics, you don't understand things. You just get used to them.****If it ain't broke, fix it until it is.**** Always satisfy the Prime Directive of getting the right answer above all else.**

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**bobbym****bumpkin**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 109,606

In post #16 that data goes in the front of the list I have or in the back? Same with post #17? Because now they must line up with the time data correctly. There also appears to be less time data?

**In mathematics, you don't understand things. You just get used to them.****If it ain't broke, fix it until it is.**** Always satisfy the Prime Directive of getting the right answer above all else.**

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**toitu****Member**- Registered: 2010-03-16
- Posts: 32

Hi bobbym, sorry for all this hassle - here is the complete dataset and all should line up (the data goes overnight): time, 50th, 90th, Ins 1, Ins 2

1030,1100,1130,1200,1230,1300,1330,1400,1430,1500,1530,1600,1630,1700,1730,1800,1830,1900,1930,2000,2030,2100,2130,2200,2230,2300,2330,2400,0030,0100,0130,0200,0230,0300,0330,0400,0430,0500,0530,0600,0630,0700,0730,0800,0830,0900,0930

11.351,12.998,12.746,10.233,9.739,13.811,10.571,10.947,10.344,11.019,12.096,13.916,13.919,16.268,16.487,15.389,14.482,14.722,13.56,17.116,10.815,14.5,17.234,18.523,18.121,17.275,19.9,13.05,16.601,18.126,16.717,18.734,19.829,18.677,21.798,16.067,16.643,19.997,20.474,19.652,20.983,18.882,18.712,19.87,18.825,20.268,21.573

51.228,68.811,51.105,34.98,35.117,43.299,46.998,46.477,45.175,54.552,71.006,71.07,93.512,120.178,117.319,102.573,86.405,92.763,73.336,123.759,24.549,77.028,92.082,167.491,158.174,86.375,114.518,108.781,108.259,111.85,84.324,98.898,78.927,96.461,98.442,91.044,75.171,92.825,81.87,101.233,89.004,85.324,76.127,85.562,90.995,103.068,89.413

513,525,556,755,740,856,1139,1218,907,534,536,479,520,249,189,182,168,167,187,265,154,143,152,124,168,127,529,221,215,210,212,174,189,191,193,186,190,193,194,195,195,192,192,194,200,201,207

363,387,409,439,488,519,573,628,737,859,886,733,679,635,576,548,496,495,589,693,632,599,561,529,1557,617,400,225,211,187,169,160,164,156,148,142,139,136,138,140,136,138,147,182,213,223,236

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**bobbym****bumpkin**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 109,606

Hi;

That looks pretty good. What I would like to do is scale down the time.

0 = 10:30

1/2 = 11:00

1 1/2 = 11:30

2 = 12:00 etc.

Is that okay?

**In mathematics, you don't understand things. You just get used to them.****If it ain't broke, fix it until it is.**** Always satisfy the Prime Directive of getting the right answer above all else.**

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**toitu****Member**- Registered: 2010-03-16
- Posts: 32

No problem at all

0,0.5,1,1.5,2,2.5,3,3.5,4,4.5,5,5.5,6,6.5,7,7.5,8,8.5,9,9.5,10,10.5,11,11.5,12,12.5,13,13.5,14,14.5,15,15.5,16,16.5,17,17.5,18,18.5,19,19.5,20,20.5,21,21.5,22,22.5,23

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**bobbym****bumpkin**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 109,606

Okay let you know what it looks like in just a few minutes.

They are on the bottom, each one in relation to time, t is the x axis.

What are seeing here as a possible match?

If you want I will superimpose them.

**In mathematics, you don't understand things. You just get used to them.****If it ain't broke, fix it until it is.**** Always satisfy the Prime Directive of getting the right answer above all else.**

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**toitu****Member**- Registered: 2010-03-16
- Posts: 32

To my eye the match between the two instruments occurs post-time step 15 where there is a relative increase in the 50% particle size and a slight decrease into the 90%. What is happening here is that the overall particle size range is closer to what the instruments both 'like' to see, hence the similarity in response.

What I was hoping to do was to develop some index that describes the relationship between the particle size and some index of the instrument response. For example, I tries, perhaps naively, plotting the geometric mean of the two instruments vs. just the 50% - the resulting relationship, although weak'ish, showed that as the geometric mean increased, the 50% particle size value dropped. I interpret this as meaning an increasing influence of the 90% but have no idea how you would differentiate between the two instruments e.g. if Ins 1 > Ins 2 or other way round. But I was hoping to be able to something more sophisticated involving the 90% as well!

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**bobbym****bumpkin**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 109,606

Hi;

Here is a superimposure of instrument 1 in blue over instrument 2.

What do you make of that?

**In mathematics, you don't understand things. You just get used to them.****If it ain't broke, fix it until it is.**** Always satisfy the Prime Directive of getting the right answer above all else.**

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