Thunk... Crack Test

[Billiard Architect]

Well here are the results of the first test. I downloaded a program called sigview (www.sigview.com). It is a nice little app that analyzes sound. The only thing I was interested in was the time coding but it does it to 10,000 of a second which was sufficient for what I wanted to do. So I set up a rack and broke the balls with the cueball exactly 49 inches away each time. I broke the rack 5 times and recorded the sound. The images attached are what was recorded with a time stamp on each one. Here are the calculations.

17.58 inches per second @ 1 mph
49 inches from cueball to rack
time to travel 49 inches @ 1 mph -> 49/17.58= 2.7873 seconds

break1 2.7873/.1855 = 15.025 mph
break2 2.7873/.1673 = 16.660 mph
break3 2.7873/.1585 = 17.585 mph
break4 2.7873/.1761 = 15.828 mph
break5 2.7873/.1667 = 16.720 mph

How much did this cost me? Zip, Nada, nothing (at least for the next 21 days). And it seems pretty accurate.

 

[mjantti]

Interesting results. I've been thinking about building an inexpensive device for measuring break speed. The idea of having a microphone is a good and cheap one.

 

[Billy_Bob]

Very interesting!

Something else of interest would be to intentionally leave a gap between certain balls and see if that shows up.

i.e. Break with perfect racks - Then break with racks with gaps between balls, then compair...

 

[Egg McDogit]

cool test man. Where are you putting the mic tho? equidistant from your break spot and the 1 ball? Otherwise, I wonder if the sound delay could be skewing your results - b/c of the tiny tiny increments of time you're working with...it might have an effect...maybe you break harder than you think

peace
-egg

Well here are the results of the first test. I downloaded a program called sigview (www.sigview.com). It is a nice little app that analyzes sound. The only thing I was interested in was the time coding but it does it to 10,000 of a second which was sufficient for what I wanted to do. So I set up a rack and broke the balls with the cueball exactly 49 inches away each time. I broke the rack 5 times and recorded the sound. The images attached are what was recorded with a time stamp on each one. Here are the calculations.

17.58 inches per second @ 1 mph
49 inches from cueball to rack
time to travel 49 inches @ 1 mph -> 49/17.58= 2.7873 seconds

break1 2.7873/.1855 = 15.025 mph
break2 2.7873/.1673 = 16.660 mph
break3 2.7873/.1585 = 17.585 mph
break4 2.7873/.1761 = 15.828 mph
break5 2.7873/.1667 = 16.720 mph

How much did this cost me? Zip, Nada, nothing (at least for the next 21 days). And it seems pretty accurate.

 

[ceebee]

17.58 inches per second @ 1 mph
49 inches from cueball to rack
time to travel 49 inches @ 1 mph -> 49/17.58= 2.7873 seconds.

Are you sure your time wasn't .27873 seconds or ????

20 mph = 5280 ft x 20 = 105600 ft an hour

105600 ft / 60 minutes = 1760 ft per minute

1760 ft per minute / 60 = 29.3333 ft per second

29.3333 ft per second x 12 (inches) = 351.999 inches per second

351.999 inches per second / 49 inches = 7 or 1/7 of a second

1 / 7 = .14 of a second

GREAT IDEA, I might have to have one of those software programs, so that I can use it in my BreakRAK demostration shows. If I can get that info to pop up on a computer screen, that will be SUPERB. It is also a great training aid. Thanks for submitting your find.

 

[Billiard Architect]

no you have 63,291 inches in a mile. If you were going one mile per hour you would be traveling 63,291 inches an hour. devide that by 60 to find out how many inches you travel in a minute 1054.333. Devide by 60 again and that tells you how many inches you travel in a second (17.58) so if you take your 49 inches and devide that by the 17.58 will tell you how many seconds it would take to travel 49 inches if you were going 1 mph 2.7873. now if you were going 2 mph that would be 2.7873/2 or 1.3936 seconds etc. So you take the 2.7873 and devide it by the time that you hit the cueball and the time that it hit the rack. Very simple calculation.

Yes the mic was equidistant from the rack and the cueball

JV

 

[ceebee]

no you have 63,291 inches in a mile. If you were going one mile per hour you would be traveling 63,291 inches an hour. devide that by 60 to find out how many inches you travel in a minute 1054.333. Devide by 60 again and that tells you how many inches you travel in a second (17.58) so if you take your 49 inches and devide that by the 17.58 will tell you how many seconds it would take to travel 49 inches if you were going 1 mph 2.7873. now if you were going 2 mph that would be 2.7873/2 or 1.3936 seconds etc. So you take the 2.7873 and devide it by the time that you hit the cueball and the time that it hit the rack. Very simple calculation.

Yes the mic was equidistant from the rack and the cueball

JV

Your equation still comes out to .139635 of a second or app'x 1/7 of a second @ 20 miles per hour.

 

[powderburner]

ceebee, I see nothing wrong with either your's or Johnny "V"'s reasoning. I get 63360 inches in one mile which slightly reduces the mph Johnny "V" reports.

Equation can be described as distance in inches divded by (time in seconds multiplyed by 17.6 inch per sec per mph) where distance is cb to head ball and seconds is cb travel time.

I suspect the cb has some velocity loss and probably isn't linear anyway so the speed obtained is not quite the average speed anyway. My guess is that rounding the mph result to 1 decimal point is quite realistic, assuming all distances measured are accurate to around .005 inch and the event times are good to .1 millisec. Rounding to nearest mph is probably more reasonable, unless your are trying to compare different break speeds with the same exact set up.

 

[AuntyDan]

Never mind nit-picking the equations, if you want to analyze the accuracy of this technique the obvious thing to do is to run it alongside a recognized technique like a Radar gun setup and compare the results.

Johnny "V" - Can this work with a pre-recorded sound file like a .WAV, or do you have to feed the microphone input directly into the software? I ask because I have a voice recorder in my cellphone which I could easily use at a Pool Hall if it can work on the files after the fact. You could also theoretically calculate Pro break speeds from recordings of matches if you could extract the audio during the break accurately enough.

 

[Billiard Architect]

Never mind nit-picking the equations, if you want to analyze the accuracy of this technique the obvious thing to do is to run it alongside a recognized technique like a Radar gun setup and compare the results.

Johnny "V" - Can this work with a pre-recorded sound file like a .WAV, or do you have to feed the microphone input directly into the software? I ask because I have a voice recorder in my cellphone which I could easily use at a Pool Hall if it can work on the files after the fact. You could also theoretically calculate Pro break speeds from recordings of matches if you could extract the audio during the break accurately enough.

You can load sound files into sigview whether it time codes them or not I am not sure. (I would have to believe that it does).

I think a radar gun would show a slightly higher speed because it shows the max speed that the cueball travels where as with sound it would take the average speed. What I mean to say is lets say when you hit the cueball it is traveling 19 mph but because of friction it slows down to 15 mph by the time it hits the rack. The average time would come out to 17 mph with the sound file where as the radar gun would show 19 mph. Just a thought.

JV

 

[Skeezicks]

What I mean to say is lets say when you hit the cueball it is traveling 19 mph but because of friction it slows down to 15 mph by the time it hits the rack. The average time would come out to 17 mph with the sound file where as the radar gun would show 19 mph.

I think a break shot loses only about 1% of its speed. 19 ---> 18.81, an 18.9 avg., but I understand what you are saying.