What is the conversion for 100 yard tome to 100 meter time? 220200; 440400?
Thread: 100 yards 9.8 = 100 meters at ?
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04182005 02:37 PM

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04182005 03:19 PM>What is the conversion for 100 yard tome to 100 meter time? 220200; 440400?
Tafnut'll have to answer that one: he's the expert on tomes.
100yds + 0.9 = 100m
220yds  0.1 = 200m
440yds  0.3 = 400m
All are approximate.

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04182005 04:00 PM:)
100y at 9.8?
I'll give my usual .30 (.24 is for 'good' timers  very few of those guys left) to get to FAT, so now we're at 10.10. He is running more than a second off WR pace (which is close to 10m increments of .84), so we'll go the nominal 10% more and make it .93. 100y is 8.56m short of 100m, so multiply .93 by .856 nd you get .80, so add that to the 10.10 and you get:
10.90 +or .1 seconds I'd think.
signed,
eldy jr.

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04182005 11:21 PMIn the 100, T&FN has always used 0.85 as teh conversion from yards to meters.

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04192005 12:08 AM>In the 100, T&FN has always used 0.85 as teh conversion from yards to meters.
My models suggest that the differential should be more like 0.75 or 0.76s for world class sprinters. That is, a 9.70s 100m (no reaction) produces an 8.94s 100 yd (91.44m) split.
These figures would indicate that a 9.85 FAT performance would be equivalent to about a 9.10 FAT 100yd, or roughly an 8.9h. Seems reasonable considering the top lists for 100 yd.
I would think 0.8s would be a more "realistic" converssion figure. 0.85 seems to underestimate the 100m equivalent (but the 0.05s doesn't make a big difference, since I'm sure there's some variability here and there). 0.9s would be too high.

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04192005 05:26 AMI use a conversion of x1.0856, so 9.8y = 10.04/ auto; x 1.0856 = 10.90m (or 10.97, if =l0.10y)

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04192005 07:45 AM>I use a conversion of x1.0856, so 9.8y = 10.04/ auto; x 1.0856 = 10.90m (or
>10.97, if =l0.10y)
I gave up on yards in about 1960. :)

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06072019 06:17 PMDog,
JRM actually gave you the clue you need to answer your question.
100 yards = 91.44 metres
That's the conversion.
So how do we use it?
So Dog wants to know what the equivalent time is over 100 metres, for a 9.8 second 100 yard time?
All Dog needs to do is simply divide 9.8 / 91.44, then multiply by 100.
What that does is break the total elapsed time down to seconds per metre, then that rate is simply applied to the other distance.
By using 91.44, we automatically convert yards to metres, since both 100 yards and 91.44 metres is the same distance.
...and what did we find out?
The converted time becomes 10.717 seconds over metres.
It would probably be expressed "officially" as 10.72 seconds.
You can do this going in the other direction, too....
Let's say you're curious just how Usain Bolt's WR 9.58 100M time works out over yards?
Ok, same process as above, only in reverse, so....
9.58/100 = .0958
.0958 x 91.44 = 8.759952
Bolt's record is equivalent to a rounded off 8.76 second 100 yard time.
So, in a nutshell....divide by the distance the time is given for, then multiply by the other distance.
Divide, then multiply.
Start with the distance given and remember, 100 yards = 91.44 metres.
...you're welcome. =)
DCmanLast edited by DCman; 06072019 at 06:22 PM.

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06072019 06:50 PMIt can't be that cut and dried, however, because of the speed decline in the last 9 meters of a 100 meter race. Will they be going the exact same average speed over that last 9 meters that they were going the first 91? I'd also say Bolt's 100 yard would be faster than 8.76 for the same reason.
You there, on the motorbike! Sell me one of your melons!

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06072019 07:06 PMYou do have a point, scott, but there are SO many variables to consider, I'm not sure you could ever be totally correct.
I mean, think about it....
No one runs the same race the same way twice.
Some days you're a dynamo, other days, you'd rather be back in bed.
It happens to all of us.
Any conversion you use (like the other ones listed in this thread) will be based on the average rate of speed and you're simply comparing distances.
Mine is no more or less accurate than the others, but I what I noticed, reading through this thread is how much everyone got hung up on "conversion factors", when really, it was just very simple math.
Divide, then multiply.
...that and I noticed that JRM gave us all the golden nugget, but it seemed no one else noticed (maybe because they're all hung up on "conversion factors"?).
Anyway, that's all I was really saying.
We don't know the acceleration / deacceleration rates of the person that Dog asked about, so we can never make a truly accurate conversion. All we can do is base it on the average rate and convert for distance.
A lot of simplified conversions work that way, so to answer your opening statement...yes, its that cut and dried. That was my whole point.
DCmanLast edited by DCman; 06072019 at 07:20 PM.