Help with Logarithm Manipulation for an Athletics question

[eldrick]

I am interested in track & have found through some hard work that for 800 that 1'42.00 is same value as for 3'44.00 for 1600
( note : for 1600m not 1 mile )

It is a Log relationship but I can't quite settle on what the value for equivalence should be for 1500m.

I have a provisional figure of 3'28.25, but I might be a tenth or so out.

I'd appreciate on any thoughts on "best" log manipulation of "known" 1'42.00 for 800 & 3'44.00 for 1600 to give an exact as possible figure for 1500.

 

[mathman]

Your question is unclear. log(800)=1'42.00 or log(1'42.00)=800? Does 1'42.00 mean 1 minute and 42 seconds?

 

[Mentallic]

I am interested in track & have found through some hard work that for 800 that 1'42.00 is same value as for 3'44.00 for 1600

By hard work do you mean you just took the world record values for each of those races?

It is a Log relationship

How have you decided from two figures alone that it's a log relationship? Why not a linear relationship? A hyperbolic relationship? Any of the other infinite possibilities?

but I can't quite settle on what the value for equivalence should be for 1500m.

I have a provisional figure of 3'28.25, but I might be a tenth or so out.

I'd appreciate on any thoughts on "best" log manipulation of "known" 1'42.00 for 800 & 3'44.00 for 1600 to give an exact as possible figure for 1500.

If you believe it must follow a log curve, then you simply need to solve these equations simultaneously to find the values of a,b:

a\log(800)+b = 102
a\log(1600)+b = 224

So you can then calculate the value of a\log(1500)+b.

 

[eldrick]

Your question is unclear. log(800)=1'42.00 or log(1'42.00)=800? Does 1'42.00 mean 1 minute and 42 seconds?

1'42 = 102 secs

 

[eldrick]

By hard work do you mean you just took the world record values for each of those races?

No

There is no world record for 1600m

How have you decided from two figures alone that it's a log relationship?

No

I know it is a Log relationship from years of research/calculation

Why not a linear relationship? A hyperbolic relationship? Any of the other infinite possibilities?

Then try some to link 1'42.00 for 800 with 3'44.00 for 1600

If you believe it must follow a log curve, then you simply need to solve these equations simultaneously to find the values of a,b:

a\log(800)+b = 102
a\log(1600)+b = 224

So you can then calculate the value of a\log(1500)+b.

Thank you

I normally end up with 3'28.25 with my alternative attempts at solution, but I believe that may be 1 or 2 tenths out

What did you get with your method ?

 

[Mentallic]

No

There is no world record for 1600m

No

I know it is a Log relationship from years of research/calculation

Alright, I'll take your word for it.

Then try some to link 1'42.00 for 800 with 3'44.00 for 1600

There are infinitely many ways to link two points on a graph. Most of them will not even be close to representing an accurate or even an approximate model of the real life situation though.

Thank you

I normally end up with 3'28.25 with my alternative attempts at solution, but I believe that may be 1 or 2 tenths out

What did you get with your method ?

Well I was hoping that you'd check it yourself to see that my model can't be correct.

For 1500m it produces a result of 3'52.60, but extending it out to the extremes is where it goes very wrong. It gives the marathon (42.195km) a time of 13'20 and the 100m sprint has a negative time.

 

[fresh_42]

The method suggested above yields 3'32.64 which I think is too high. (12sec on the last 100 of 1600 is usually not achievable.) My calculation gives me an estimation of 3'28.14 (ln 102/800 ≈ -2.06 and ln 224/1600 ≈ -1.966; so I estimated ln x/1500 ≈ -1.975 which yields x ≈ 3'28.14).

 

[eldrick]

The method suggested above yields 3'32.64 which I think is too high. (12sec on the last 100 of 1600 is usually not achievable.) My calculation gives me an estimation of 3'28.14 (ln 102/800 ≈ -2.06 and ln 224/1600 ≈ -1.966; so I estimated ln x/1500 ≈ -1.975 which yields x ≈ 3'28.14).

I'm interested only in 1500.

Don't bother with extreme distances above/below.

The answer is somewhere in 3'28-low region +/- tenths.

It is refining precision...

 

[fresh_42]

I'm interested only in 1500.

Don't bother with extreme distances above/below.

The answer is somewhere in 3'28-low region +/- tenths.

It is refining precision...

I find 3'28.14 fits pretty well on your target. (I have only been calculating 1500. What I didn't know was how the "energy rate" declines. You started with an exponent -2.06 (on 100m) and ended with an exponent -1.966 on 1600. So to estimate -1.975 on 1500 seemed a good guess to me.)

 

[eldrick]

I find 3'28.14 fits pretty well on your target. (I have only been calculating 1500. What I didn't know was how the "energy rate" declines. You started with an exponent -2.06 (on 100m) and ended with an exponent -1.966 on 1600. So to estimate -1.975 on 1500 seemed a good guess to me.)

I don't have much confidence/interest in your answer.

I want as much precision possible.

I have no interest in "energy rate".

Just 2 parameters with what 1500 is...

 

[Mark44]

I don't have much confidence/interest in your answer.

@eldrick, ow about a little more civility from you? The question you have asked is ambiguous, and the members posting here are just trying to help. Show them more courtesy, and a lot less attitude...