Do you know how?
Depends on what level. NCAA converts it by subtracting .46. I have seen some Assoc. subtract .49.
Another way may be to figure out fast you are running in Meters per second and figure out how long it would take to run 5 extra meters based on that.
55/6.63(55 time) = 8.3 meters per second
Acutally now that I look that would be too slow. It seems to be about a tenth off.
Id like to hear what Mike and others think.
The reason it’s a tenth off is because you are still accelerating until about 65m.
Everyone needs to remember that each plate (10 meters), of your race will almost always have a different time. The less time you’re dealing with (40yards/55m/60m/), it’s espescially much more difficult to make a guessimate of (based on other distances), because the 0-60m distance is much more calculitably unstable mostly due to acceleration phase.
You’re calculating your average m/p/s. You have to view it in chunks.
Run next to your buddy’s car or bike/wspeedometer to find out your top mph, convert that down to m/p/s and you have your top speed.
Based on things I’ve read and heard, a solid sprinter reaches top speed at about 45-60 and can maintain it for a plate of time of about 10-15m.
Your top speed is much faster than 8.anything, and at least over ten m/p/s quik. (which I’m sure you know)
Take your top speed and divide into those last or extra 5 meters, tack this onto your 55 time, and you got it.
use these calculators to make things easier – https://www.onlineconversion.com/length_common.htm – to convert lengths
https://www.onlineconversion.com/speed_common.htm – to convert your speed
Personally, I wouldn’t listen to any ” subtract .46 ” to determine your 55 meter time, because everyone will run a different race. IE some people are great outta the gate while others close like frieght-trains. Some sprinters may be at top speed by 50m, while others get there at 60m, so this particular system of estimation are better off not used at all.
You reach top speed around 30 meters and this can be maintained for 2-3 seconds. After that deceleration happens and each 10 meter split will be slower unless your Flo-Jo.
You had the right idea you probably just got a little mixed up about when max veleocity is reached. And as you probably know the faster you are the longer it takes for you to reach max veleocity. So for a 11.5 runner it may take 20 meters where for a 10.5 runner it may take 30 meters. Think of it as 2 cars. If car A’s top speed is 100 miles per hour and Car B’s top speed is 150 miles per hour if there accel was the same and they hit 100 miles per hour at the same time car B would still be accelerating.
What you said about “Take your top speed and divide into those last or extra 5 meters, tack this onto your 55 time, and you got it” is about correct. I makes sense. My flying 30 is 3.0. That = 10mps and divided by 5 meters is a half second. So 6.63=7.13 which is about right. Still not too far off the .46 or .49 conversion.
Big10 is right. The reason Quik’s original equation didn’t work was because it estimated the final 5m based on the average speed of the original 55m rather than estimating it off of an instantaneous speed closer to the end of the 55m. When you take a split later in the race (say the last 5-10m) and estimate 5-10m after that point, you can get a pretty reasonable estimate for 2 reasons:
1. You are making an estimate based on an instantaneous speed (5m or maybe even 10m) rather than average speed.
2. You are extrapolating the estimation to a very limited distance (5-10m beyond the original distance). The further you attempt to extrapolate the split estimation, the less accurate it is likely to be. This is why it’s extremely difficult to make estimations for the 100m based off of 55m times.
As for the 0.46-0.49 add-on times, they are actually decently accurate. This is because they are based on the 2 points I mentioned above. That is, the given governing body lays out a qualifying time for the standard distance (in this case the 60m). They know that to run the qualifying time for the standard distance, an athlete will almost certainly have to attain a given instantaneous speed. Knowing this, they can make a reasonable estimatation for a short distance. The logical reason that different governing bodies or associations would have different conversion factors is because they are likely basing the split time off of different qualifying times for the standard distance. In the examples given, I’d be willing to bet that the 60m qualifying time associated with the 0.49 second conversion is slower than the 60m qualifying time associated with the 0.46 second qualifying time.
While this system isn’t perfect, the conversion from 55m to 60m isn’t likely to be too far off for a given speed range (in this case, those people who could potentially qualify). Making an estimation for 100m based on 55m would be another story though.
This is an interesting topic to me. We kicked this around last indoor season as well, but I’m still not satisfied. What Mike said about the .49 qualifying time being slower than the one that uses .46 is not always true.
I’ve re-posted something I wrote to the board last year to reiterate my point. Although I was concerned primarily with hurdles, the premise still holds:
I’m still not comfortable with this conversion business. It seems that the NCAA lacks consistency, changing the conversions from year to year and division to division.
Here is a summary of 60mh times and corresponding conversions currently used by the NCAA. Times for D1 women and D1, D2, and D3 men are included.
8.43 second range: .56
8.26 second range: .52
8.21 second range: .53
8.15 second range: .54
8.02 second range: .52
7.95 second range: .56
7.93 second range: .53
7.70 second range: .52
You would expect that as the total time goes down, so should the conversion factor. The faster you run the total 60m distance, the faster you’ll cover the last 5m. But that isn’t the case.
For example, an 8.26 second hurdler is subject to the same .52 conversion as a 7.70 second hurdler. Surely they don’t both cover that 5m in .52. But which one does? And does the 7.95 second hurdler really cover the last 5m .04 slower than an 8.02 second performer?
I don’t know what the answers are, but it appears neither does the NCAA. When national qualifying spots consistently come down to one one-hundredth of a second, we need to be sure that the conversions we’re using are correct. It’d be a shame to see an athlete run a superior 60m race and not get into the nat’l meet because a 55m time got generously converted. Of course, it works both ways.
Can anybody point us toward truth? I’ve also seen 55mh times multiplied by 1.07 to arrive at a 60mh equivalent. Is this a better way? Questions abound, who has the answers?
Informative post Duck. Are all the times you listed for the same gender? If not, that could explain some things because run off times would be siginificantly different for men and women. In any case, the conversion explanation is how it should be done (IMHO) not necessarily how it is done as you seemed to have pointed out. I would say if one were trying to make a conversion based to compare race times the only logical method to use would be the one I described. Any one else have any ideas for conversion?
bust out your physics or calculus books people. you will need to find acceleration. it depends if he has constant acceleration. the 8.3 m/ps is his velocity which is the integral of his acceleration. near impossible to find out what it is exactly without a lot of information. if you really want to find out go to a local high school and ask a physics or calculus teacher.
I’m not sure if your response was to Quik’s first post or to everyone but in reality,we can in fact assume that acceleration is relatively constant by 55m especially given the 2 constraints I listed above. Hopefully, your physics and / or calculus teacher would agree….if not, they have difficulty applying their knowledge.
I think that there may be some merrit to using the total 60m time to calculate 55m time from 60m. Here is my reasoning:
-We know the athlete’s total 60m time.
-We do not know the amount of time the athlete took to cover the last 5m OR the athlete’s instantaneous velocity during those 5m, or any point in the race for that matter.
Therefore… if we use a conversion like the divide by 1.07 mentioned above at least we are converting based on concrete data rather than estimating the athlete’s average velocity over the last 5m (which is what the .46 and .49 are stipulating).
The 1.07 factor is also more accurate IMHO because we are not subracting a set amount of time for every athlete, the amount of time subtracted is based on the individual athlete.
But then again i just posted this for arguments sake because i am a nerd :tumble:.
I’m a nerd at heart too. What you suggested may be true, but it still doesn’t really answer the original question: conversion of 55m to a 60m performance. It may work for coming up with equivalent 55m performances if you know a 60m performance but to do the inverse you run into the same issues that my proposed solution does…..estimating a time for a segment that was never run. Also, IMHO the 1.07 multiplier is just as flawed if not more so than estimating an additional 5m fly time based on 55m of acceleration because it uses a constant rather than a somewhat instantaneous velocity (albiet quasi) as in the method I mentioned.
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