Hmmm.... not sure I can let all this go by without a comment.
About 8 years ago (urk) I developed a formula for determining
equivalent times across a variety of distances. I generated this
based on the top 10 times in the world at each distance - then
assumed that the average of the 10 times for each was a fair
indicator of an equivalent time. For a couple of data points, this
could have been skewed by a phenomenal person, such as Michael
Johnson holding several of the best times for the 400 or 200 or
whatever - so I took the top 10 people - not times.
Anyway... Essentially here is the scoop:
1. Several years ago, I spent a lot of time researching the
relationship of equivalent times to run various distances.
I found that doing a regression on time vs. distance was
futile. BUT... using speed vs. distance yielded a very
nice formula which worked well for WRs, U.S. national
records and collegiate records.
2. I backtested the model of speed vs. distance (using distance
as the given and trying to predict speed) and found it did
very well for post-1945 records at the 800 through the 10000.
From 1964 onward it also worked well for the marathon.
3. The problem was converting it back from determining speed
for a given distance back to time. This is where I started
messing up with all the numbers. I (unfortunately) developed
the model using miles not kilometers or meters. A hazard
of living in the U.S. So the inputs are in miles and
seconds. Sorry bout that.... Ambitious types can let
me know how to improve it!
4. An algebraically simpler formula is:
Let a = 13.49681 - 0.048865*olddist + 2.438936/(olddist**0.7905)
b = 13.49681 - 0.048865*newdist + 2.438936/(newdist**0.7905)
Then equivalent time for the newdist;
newtime = (oldtime/olddist) * (a/b) * newdist
As an example:
Suppose a runner can run 3:30.0 for the 1500 meters.
1. olddist = 1500/1609 = 0.932256 miles
2. oldtime = 3*(60 sec.) + 30.0 = 210.0 seconds
We want to predict an equivalent 5000 time.
3. newdist=5000/1609=3.107520
4. newtime=?
To solve for ?
a = 16.02925 (by plugging in the above)
b = 14.34025 (again, substitute)
newtime = (210.0/0.932256) * (16.02925/14.34025) * 3.107520
= 782.447
= 13:02.4
Hence, the model would say that a 3:30.0 1500 and a 13:02.4 5000 are
equivalent.
I apologize for the unwieldiness - this can be algebraically
simplified. I hope many of you try this out!
Based on this, the appropriate conversion factor for 1500 to the mile
is 1.0815 - very close to Patrick Hoffman's figure. I didn't
actually calculate 1600 vs. the mile, since the 1600 is only
important to U.S. high schoolers - but hey - if it weren't midnight
on the east coast in the U.S. I might do it now.
=====
Dave Cameron
[EMAIL PROTECTED]
__________________________________________________
Do You Yahoo!?
Get personalized email addresses from Yahoo! Mail
http://personal.mail.yahoo.com/
