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A CYCLOCOMPUTER PUZZLER I received an application from a firsttimer. He used a Protégé 9.0 and said he followed Neville’s instructions for setting it up. He says the calibration was set at a wheel diameter of 9999 mm as directed by Neville. He measured a calibration course at 354.98 meters. His average reported calibration constant was 114.80 revolutions per ride. This calculates out to 0.32369 revolutions per meter, or 3.09 meters per revolution He obtained 2587 revolutions on an 8 km ride. This calculates to a length of 7988.8 meters. I checked the length using America’s Running Routes and the distance is very close to correct. However – and here is where I am stuck – his wheel circumference comes out to 1/.32369 = 3.09 meters. This would put the wheel diameter at 98.3 cm or 3.22 feet or 38.7 inches, which is huge. He states that his tire is 700x25C, which has an approximate circumference of 2124 mm. This is a lot smaller than 3.09 meters The wheel size is obviously wrong, but the measurement came out close to correct. What am I missing here? Pete Riegel  

Curious. I cant quite understand the first past of the calcs, cal length 354.98 m cal revs 114.8 I calculate cal const with no scpf 0.323398501 revs/m cal const with scpf 0.3237219 revs/m then I calculate course measure 2587 revs = 7991.427214 m (inc scpf) or 7999.418641 no scpf Neither of these exactly matches up with 7988.8 m which you give. Although the 7999.4`8 is close to 8km , so did he forget to include the scpf? However on your major point about the apparent wheel diameter I note that 3.09/2.124 (ie m/rev divided by estimated wheel circumference) is approximately 1.455. Could the ratio be exactly 1.5 and the error be due to wrong setting up of the electronic counter? At least with Jones counter I have never seen such problems. I will continue to supply and teach only Jones counters to the class of new measurers whom I hope to get qualified this summer. But even so I would be interested to know what you think went wrong. Mike Sandford  Measurement Secretary South of England UK Asssociation of Course Measurers coursemeasurement.org.uk email contact m.sandford at lineone dot net  

Mike, Your calculations are correct. I rounded where I should not have. Also, the measurer calculated meters per revolution, and then multiplied by 1.001, which applied the SCPF in the opposite direction. All this I caught. It was the wheel size that threw me, and I still have no satisfactory explanation. I will post something once I receive the answer to a question from the measurer. Pete Riegel  

Pete: For Bill to get a very low 114.80 counts or “apparent” rev on a 354.98–m calibration course, he could have operated the computer in mile rather than km mode. Even so, it would be perfectly possible for him to get a reasonably accurate measurement on the race course. From the facts I know, I calculate he should have got about 106 “apparent” rev on the calibration course. On reflection, I think Pete's suggestion that Bill set the circumference in the computer to 6999 instead of 9999 probably more accurately explains the 114.80 count. My quickstart instructions require very little reading, but I find that measurers still do not read them completely. Obviously Bill did not read the following, because if he had followed the instruction he would have realized that he had not set up the computer to read exact revolutions: “ Test the installation as follows. Roll the wheel to a ground reading of about 0.9 rev on the rim and zero the trip display by pushing the meter forward in its bracket for about three seconds. Roll the wheel forward for over a revolution until the trip distance increments by one. This should occur precisely at a rim reading of 0.00 rev. “ This message has been edited. Last edited by: Neville,  

This is rather interesting and may points out possible problems with using the electronic counter. I have recently had a few measurers ask about using the Protege and explained what should be done. I am asking them to forward their numbers to me. I hope this situation is not going to be a problem, but we must be give it some time. I have done all the calculations and agree with Mike on his numbers and wonder if the measurer used the SCPF. However, Pete's findings about the bike wheel is difficult to explain. I see Neville's explanation and need to evaluate this further. Let's see what others have to say.  

I talked with the measurer last evening. To his chagrin he saw that he had programmed the Protege in miles/hr instead of km/hr. When this is done the unit does not increment by 0.01 each time the wheel revolves. The calibration and measurement is close, but not exact. He is going to reprogram more carefully and remeasure today. I asked him to stop at existing marks, and we would work out the adjustments when we have the data. I expect the adjustments to be small.This message has been edited. Last edited by: Pete Riegel, Pete Riegel  

Well done, Pete. But then why am I able to calculate in my post above
Should I not be getting something like 1.6 if you have estimated the wheel circumference correctly? Mike Sandford  Measurement Secretary South of England UK Asssociation of Course Measurers coursemeasurement.org.uk email contact m.sandford at lineone dot net  

Mike, If you look at my post in "how it works" you will see a partial chart for a Protege that is calibrated at 9999 but set to record in mph. The program predicts that 114 indicated revolutions using mph would register as either 184 or 185 if km/hr was used. Similarly, 2587 (mph) = 4164 or 4165 (km/hr) and 2582 (mph) = 4156 or 4157 (km/hr) A bit of calculation reveals that the correct measured length will be about 7989 to 8004 meters. I think. Pete Riegel  

Pete, I understand that, and 4164/2587=1.6095863 which makes sense to me when miles have been used instead of km. But I still dont see why I calculated 1.455 for the factor the wheel circumference is out by. Should it not also be 1.609.. or have I made a mistake? Mike Sandford  Measurement Secretary South of England UK Asssociation of Course Measurers coursemeasurement.org.uk email contact m.sandford at lineone dot net  

Mike, The tire size by reported data = 355/115 = 3.09 m/rev. The tire size by corrected reading is 355/184.5 = 1.92 Not 2.12, but not too far off. 3.09/1.92 = 1.61 close. Does this answer the question? I will be receiving the result of a measured roll over one revolution. I'll post it. I am sending you the file for correcting from km/hr to mi/hr. Pete Riegel  

I received more information from the measurer. He discovered a miscount in the number of tape lengths used in laying out the calibration course. The correct length is now 100 feet longer. He recognized that if he remeasured the course as it stood, the adjustments would be large. He elected to start over. On the second complete measurement he got it all correct. He is a bright guy, comfortable with the numbers, and I have every expectation that he will be a competent measurer with a few more courses under his belt. He does not make the same mistake twice. I asked him to obtain a Jones/Oerth counter and use it to record measurement data, but to continue to use the electronic counter as an aid in course layout.This message has been edited. Last edited by: Pete Riegel, Pete Riegel  

Pete, This now makes sense to me. I calculate a cal course length of 354.98+100/3/1760*1609= 385.453m And the tyre circumference comes out as 385.453/184.5 m which is 2.089m which is much more realsitic for a slightly underinflated 700*25 C tyre  I measured my 700*25 C inflated to 100psi but not loaded with any weight to be 2.14m. 51mm smaller than this seems possibly accounted for differences between nominal and the actual rolling radius. So I now understand the data. But there are three different mistakes by the measurer so no wonder it was a puzzle to solve. How could he have made a mistake in the tape lengths? Even if he did not write down the seperate readings for each tape length (ie he just counted tape lengths putting the intermediate marks at an exact reading of the tape, rather than using the method I teach which is to put down the intermediate mark, normally a PK nail, then measure and write down the actual length from this to the previous nail) surely the procedure is to measure one tape length with the bike then check the whole cal course with that cal reading. I have seen mistakes on checking cal courses through wrong use of the tape but never a whole tape length missing. I thought the US procedures explicitly required cal course layout data which enables the certifier to catch missing tape lengths, unless of course the data is just made up. Mike Sandford  Measurement Secretary South of England UK Asssociation of Course Measurers coursemeasurement.org.uk email contact m.sandford at lineone dot net  

He did not initially do the bike check. I asked him to do it later. He is going to do it today, but based on the rollout reading I think it will check out. For one complete revolution of his unloaded wheel he obtained 2.102 meters. The cal course checks out using this, but he will still do the 100 feet vs full length check. The bike check was initiated in the US because we discovered early on that miscounted tape laydowns are not uncommon. Pete Riegel  

Pete: Before you made the announcement that Bill had measured his calibration course 100 ft too long, I already knew that there was more to the story than just his operating the Protégé in M/H rather than KM/H. I had calculated that the that the “apparent” rev he obtained should have been 105.64 not the actual 114.80. To make my calculation, I first looked up the Cat Eye value for the circumference for my 700x23C tire, 2.096 m. This is in pretty good agreement with the actual value of 2.089 m especially considering I use a relatively low pressure of 84 psi and the tire is worn. Bill’s tire was supposed to be 700x25C so I assumed the Cat Eye value for the circumference, 2.105 m, was accurate: Actual number of rev on crse of 354.98 m = 354.98/2.105 = 168.64 This indicates a rim reading of 0.64 at the end of calibration. The computer senses 168 + 1 = 169 rev since we zero with the magnet just behind the sensor. Since we deceive the computer by giving a circumference of 9.999 m, it calculates distance traveled as follows: Dist in ml = 169 x 9.999/ (0.3048 x 5280) = 1.050 In ml mode the computer only shows down to 0.01 ml so it would read 105 “apparent” rev. Therefore, with the rim reading of 0.64, total “apparent” rev = 105.64.This message has been edited. Last edited by: Neville,  

I received the calibration course bike check and have certified the course. Pete Riegel  

An Article from Team San Angelo email Newsletter by Bill Cullins Certifying a Running Course Most of the top running events are held on "certified" courses, meaning that the course distance has been accurately measured as per procedures approved by USA Track and Field. I recently had the pleasure of conducting the rigorous measurement process as we obtained a certification for the 2007 Run in the Sun 8K course, and I now have a much greater appreciation for what "certified" means. Like most runners and cyclists, I had always assumed that an automobile odometer, bike computer, or a hand held GPS unit was fairly accurate. To measure a course as per USATF standards, the first step is to measure a “calibration" course. This has to be at least 300 meters but can be longer. In our case, we used fixed endpoints for the calibration course (two fire hydrants) and so ended up with a 354 meter calibrations distance. This calibration course had to be measured with a steel tape, stretched with exactly a set amount of force, and then adjusted for thermal expansion. After laying out the calibration course, the next step was to install and program a USATFapproved bike computer. The USATF measurement committee prefers that a traditional mechanical device (Jones counter) be used, but the certification process does allow for the use of a bike computer (electronic counter) if specific guidelines are followed and if an approved computer is used (Protege 9.0 recommended). The biggest difference in the application of the bike computer is that it isn't used to directly measure miles or kilometers; instead, it is programmed to measure "counts", which are revolutions of the bike wheel. The bike rim is also marked so as to allow partial wheel revolutions to be measured. In my case, I marked the rim into 40 equidistant increments which allowed a measurement accuracy of 1/40th of a wheel revolution. After setting up the calibration course and bike computer, etc. the next step was to ride the calibration course four consecutive times and then average the readings. Remember that the data collected was full and partial "counts" (wheel revolutions), so at the end of the calibration course rides we could divide the number of counts by the calibration course length to calculate what is called the "working constant", or counts/meter. This preliminary working constant was then multiplied by 1.001 ("short course prevention factor") to get the real working constant. In the case of our calibration course, this resulted in .4798 counts per meter. This working constant number was then used to predict where the 4K (4000 meter) turn around point and the mile split points would be. For example, .4798 counts/meter times 4000 meters equaled 1,919 counts plus 2/10 of a count (8 of the 40 rim divisions). After (finally!) getting the target values for the turn point and mile splits, the next step was to do a measurement ride of the actual and place preliminary marks at all critical locations. Since the major reasons for certifying a course are to (1) make sure the distance is accurate and (2) insure that the course measurement can be validated in case a record is established, the actual course ride had to follow the shortest path for the route. That meant that the measurement ride had to follow the tangents; i.e., straightening out the curves as much as possible. I can assure that riding the exact tangents is a fun exercise in concentration while on River Road with traffic. The target values (previously calculated) were used to place preliminary split and tam around marks on the course. A second (and 3rd, and 4th, ...) measurement ride was then done to insure that the initial preliminary marks were accurately placed and could be replicated. The amount of allowable variance between measurements could not exceed 8/100 of one percent. My repeat measurements resulted in a difference of 6/100 of a percent(notmuch  about 15 feet over the entire 8K distance). The course turn point was then adjusted to reflect the difference between the measurements to insure that the final distance was at least 8K. After doing the initial calibration course and measurement rides, it was back to the calibration course (same day) for four more passes over the 354 meter distance. Again, the golden standard was to document that pre and post measurements along this calibration distance were within a set margin of error. If the numbers had too much variance, it was back to square one and start the process over. Unfortunately, I made a number of rookie measuring person mistakes and ended up repeating the complete process several times with long distance guidance from Pete Riegel, a national USATF measurement guru. The last step was to complete a ream of paperwork that documented the entire process and all of the measurements and calculations. A detailed course map with explicit descriptions for the start, turn, and each split mark location had to be prepared. It all paid off in the end, however, and we now have Certification Measurement Certificate # TX07012PR for the 2007 Run in the Sun course.This message has been edited. Last edited by: Pete Riegel,  

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