BS 7910 assessment curve

[awibroe]

Hello to all,

I am looking to see if anyone has any experience in using the BS 7910 standard or can simply pull me up on my error in the attached.

Within the standard there is a mathematical definition for a pass fail curve (where the variable Lr is plotted on the x axis).

Toward the bottom of the attached I have inserted an exert from the standard, my attempt to define this, my result and what I am expecting to see. 

Obviously something is not quite right. 

Any thoughts or hints welcomed as always.

Thanks,

Andy

PS using MCP3

[Fred_Kohlhepp]

Your equation is wrong.

Redone using Prime 4.0 Express

[awibroe]

Hi Fred,

Thanks for this. I am interested to know how you got the refined formula you have presented as this is obviously quite different to the standard given formula. I appreciate some things are wrong but am surprised if a published industry standard is not correct?

Thanks,

Andy

[Fred_Kohlhepp]

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@awibroe wrote:

Hi Fred,

Thanks for this. I am interested to know how you got the refined formula you have presented as this is obviously quite different to the standard given formula. I appreciate some things are wrong but am surprised if a published industry standard is not correct?

Thanks,

Andy

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The "refined" formula is the standard.  You had incorrectly entered it.

[awibroe]

Fred,

Additionally, any ideas why my plotting fails?

Thanks,

Andy

[Fred_Kohlhepp]

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@awibroe wrote:

Fred,

Additionally, any ideas why my plotting fails?

Thanks,

Andy

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No, I don't know.

[awibroe]

Fred,

Thanks for that I have since worked this out why my plotting failed. But I am interested to know how you have adjusted the formula for when x greater than 1 but less than the maximum range. In this case the standard calls for x^N-1/2N but you have gone for part of the first function?

I do however agree that plotting both, yours seems visually correct?

Andy

[Fred_Kohlhepp]

Andy,

I'm using Express (no longer have access to full Mathcad.)  So I cannot write programs, one of the "higher level" things--I can see what you've written but It won't compute.

How to get around that?  

• function f1 is the basic expression (when L is less than 1, but actually there for both cases.)  
• function ff creates two cases

              1) when L<1  (can you see the boolean expression?  1 when true, 0 when false)

               2) when 1<L<Lmax  (the second boolean.)

So you can see when the second condition kicks in

[awibroe]

Fred,

Yes I understand what you have done. My question is why have you deviated from the standard (what rationale). In the attached image, I have plotted both methods. On the left is the literal interpretation of the standard (I believe) and on the right is what you have programmed.

Thanks,

Andy.

[Fred_Kohlhepp · Accepted solution]

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@awibroe wrote:

Fred,

Yes I understand what you have done. My question is why have you deviated from the standard (what rationale). In the attached image, I have plotted both methods. On the left is the literal interpretation of the standard (I believe) and on the right is what you have programmed.

Thanks,

Andy.

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I deviated from the standard because I typed the wrong thing.  Your equationsf2 and f1p are correct.  When I do it correctly I get the same thing: