tag:blogger.com,1999:blog-13887692.post8462776733548216021..comments2024-03-21T03:15:06.288-04:00Comments on Cozy Beehive: The Rate Of Climbing Uphill ExplainedRon Georgehttp://www.blogger.com/profile/18394865788996482667noreply@blogger.comBlogger27125tag:blogger.com,1999:blog-13887692.post-47201356887623032362012-01-22T14:31:18.658-05:002012-01-22T14:31:18.658-05:00have a look at http://www.cyclingpassions.eu/searc...have a look at http://www.cyclingpassions.eu/search/label/Analytic%20Cycling for an analitic approach.Odyseus3https://www.blogger.com/profile/01301675526164373480noreply@blogger.comtag:blogger.com,1999:blog-13887692.post-84926492932617677162010-08-20T17:37:55.544-04:002010-08-20T17:37:55.544-04:00Hi! I'm not a physicist, but I sort of unders...Hi! I'm not a physicist, but I sort of understand the formula you provided. I built that into a spreadsheet and it gives believable wattage numbers EXCEPT at low or no gradient. On a completely flat road, the formula implies that zero wattage is needed to propel a rider of any weight at any speed!! Maybe I've done something wrong?! I simply wanted to see a power curve for two riders of different weight. For example, one at 100lbs and the other at 200 lbs. Both riders finish several TT courses with identical times. One course is flat. Another at 2%. Another at 4% and so on. What is the needed average power for each rider at each TT course? Would love to see a chart on that! <br /><br />Jeff<br />jshein(at)yahoo.comAnonymousnoreply@blogger.comtag:blogger.com,1999:blog-13887692.post-82731896784133673562010-06-11T18:21:30.184-04:002010-06-11T18:21:30.184-04:00Ok wrong link
Correct online vam calculatorOk wrong link<br /><a href="http://www.cyclingfitness.net/calculate-vam-watts-kg.php" rel="nofollow">Correct online vam calculator</a>Anonymousnoreply@blogger.comtag:blogger.com,1999:blog-13887692.post-37594997779392450632010-06-11T18:18:25.537-04:002010-06-11T18:18:25.537-04:00If anyone else is like me and no good at maths you...If anyone else is like me and no good at maths you can work it out online pretty easily at - <a href="http://www.cyclingfitness.net/what-is-vam-and-how-to-calculate-it/" rel="nofollow">Online VAM Calculator</a><br /><br />Its a bit simple but it gives me the right results ;-)Anonymousnoreply@blogger.comtag:blogger.com,1999:blog-13887692.post-26882758288058929012009-08-03T07:59:06.269-04:002009-08-03T07:59:06.269-04:00@Ron Sorry for the topic highjack. Naked Lunch.@Ron Sorry for the topic highjack. <a href="http://www.imdb.com/title/tt0102511/" rel="nofollow">Naked Lunch</a>.spokejunkyhttps://www.blogger.com/profile/03843522840848008750noreply@blogger.comtag:blogger.com,1999:blog-13887692.post-38559197334565554032009-08-02T19:23:47.455-04:002009-08-02T19:23:47.455-04:00Roads & bike tyres are not smooth & polish...Roads & bike tyres are not smooth & polished steel. Cycle tyres deform into the irregularities of the road surface, forming a cogwheel type effect.<br />Somebody could experiment building an actual cogwheel road & bicycle, but with normal roads, bikes seem to work ok.Jamesnoreply@blogger.comtag:blogger.com,1999:blog-13887692.post-86737082138227887432009-08-02T11:40:05.527-04:002009-08-02T11:40:05.527-04:00Ron,
You have achieved a whole new level of geekdo...Ron,<br />You have achieved a whole new level of geekdom. *he he he*<br />I'm just picking on you. Reading that whole thing just made my brian hurt... I was just saying, Thursday before my after work ride in fact, that I SO long to be an engineer, but math is just painful to me. It's one of those things that I CAN do if I have to, but I hate doing it.Cycling Phunhttps://www.blogger.com/profile/07407316752929288793noreply@blogger.comtag:blogger.com,1999:blog-13887692.post-47846349018333187592009-08-01T23:36:17.553-04:002009-08-01T23:36:17.553-04:00@Spoke - What is Naked Lunch? :)
@ Bikeboy - Tha...@Spoke - What is Naked Lunch? :) <br /><br />@ Bikeboy - Thank you !! I didn't bother looking for too many computers that have this feature. I did a search myself on Amazon and found a barometer from Oregon Scientific (http://www.amazon.com/gp/product/B000EGX30Y) that has the same feature. It costs 220dollars. I think any watch with this feature is going to be really expensive. Looks like buying the cycling specific computer is better than a vario, but a vario is what first came to my mind. I'd still like to see if it would work for a cyclist, because varios are specially made to calculate and record your vertical speed. Hmmm....wish I had some spare cash lying around!!!Ron Georgehttps://www.blogger.com/profile/18394865788996482667noreply@blogger.comtag:blogger.com,1999:blog-13887692.post-89530168829434006112009-08-01T23:19:53.843-04:002009-08-01T23:19:53.843-04:00@Ron - regarding the method number 5 of using inst...@Ron - regarding the method number 5 of using instruments, I did a small Google search and found there's Mavic's 200 dollar cyclocomputer Wintech Ultimate can show you your vertical speed . See - http://www.mavic.com/road/products/wintech-ultimate.996646.4.aspxRonhttps://www.blogger.com/profile/16268869622833968439noreply@blogger.comtag:blogger.com,1999:blog-13887692.post-26251221591405269692009-08-01T22:41:29.317-04:002009-08-01T22:41:29.317-04:00I feel like one of the guys with a straw sticking ...I feel like one of the guys with a straw sticking out of my head in 'Naked Lunch'.spokejunkyhttps://www.blogger.com/profile/03843522840848008750noreply@blogger.comtag:blogger.com,1999:blog-13887692.post-84668085695770676152009-08-01T22:30:10.200-04:002009-08-01T22:30:10.200-04:00Anon 6:45pm : On the debate of walking vs cycling ...Anon 6:45pm : On the debate of walking vs cycling on steep roads, do remember that humans can get to all fours and crawl up when the going gets tough. If its gets ever more steeper, they can be a mountain climber and climb up. You can't do this with a bicycle. A bicycle does not have legs, fingers and an opposing thumb to hold onto things. We all know round stones don't stay put on steep slopes, they roll down! Same for wheels. <br /><br />Your experiment would be theoretically interesting. Not sure if any public development authority will build a special track for you to ride your specially made bicycle on. But remember, this is a specially made bicycle with modified wheels suitable to "hold on" to the track. Its not a bicycle as we know it, more like a terrain buggy. It would be still interesting to find out whether human power alone can ride this HPV up a steep slope. My opinion is because there are no wheels for easy rolling, it becomes very hard to move the vehicle up with human power alone. :)Ron Georgehttps://www.blogger.com/profile/18394865788996482667noreply@blogger.comtag:blogger.com,1999:blog-13887692.post-52109115310395948272009-08-01T18:45:41.283-04:002009-08-01T18:45:41.283-04:00Ah yes, you are correct about the cog railway.
I ...Ah yes, you are correct about the cog railway.<br /><br />I did not mean a cogged wheel bicycle on a traditional road surface, but one a surface that matched the cogs on the wheel, which would unlikely be made of asphalt. Although I suppose it would be possible to create such a surface using asphalt in much the same way that rumble strips are used at the edge of lanes on highways to alter drowsy drivers that they are in danger of driving of the edge of the road.<br /><br />Perhaps I should try and convince the local road authority to build such a road just to test out my hypothesis. Maybe I can find a cogged wheel for a bicycle on ebay! Heh.<br /><br /><br />Cheers!Anonymousnoreply@blogger.comtag:blogger.com,1999:blog-13887692.post-55113424463424341692009-08-01T18:28:21.992-04:002009-08-01T18:28:21.992-04:00For example, if the bicycle wheels had cogs that m...<b> For example, if the bicycle wheels had cogs that meshed into the slope surface (sort of like the cog railway on the west slope of Mount Washington) then the 40-45% limit could be beaten (I believe). Of course, using a human body to propel such a bicycle would result in very low speed that would likely be beaten by a walker/hiker/climber/snail. </b><br /><br /><br />If you have enough "horsepower" and "thrust" you can send anything up. For example, the Top Thrill Dragster at Cedar Point climbs the 420 feet at an angle of close to 90 degrees! It can achieve this only through the hydraulic launching mechanism that generates 120 mph of speed in under 4 seconds. That takes about 15,000 Horse Power.<br /><br />I and most of the world have yet to see a cyclist climb up a grade of exceeding 40 or 45%. Technically, from the ambiguous way I worded it, you're right...its not so much a handicap of the human body but just a handicap of the human-bike machine to cope with such grades on puny human power alone. Hence for our purposes we are only practically concerned with a portion of the curve that goes upto a grade of 40%. <br /><br />Cog wheels?! How would they look? I thought the Cog Railway was named only so because the train used cog gears for transmission, very old school, but pretty innovative at the time they came out. Does it ever head uphill? I really doubt a bicycle with 'cog wheels' can go up an asphalt road of 45%. Let me see one in action :)Unknownhttps://www.blogger.com/profile/12256394060474969622noreply@blogger.comtag:blogger.com,1999:blog-13887692.post-71010147174401757352009-08-01T18:11:10.912-04:002009-08-01T18:11:10.912-04:00"Our human bodies can only use a tiny portion..."Our human bodies can only use a tiny portion of this curve, ranging from 0% grade to 40-45% grade."<br /><br /><br />I am wondering about the above statement. I has been my understanding that this is not a limit based on the abilities of the human body but based on the amount of friction available between a rubber tire and a road surface (whether asphalt, concrete or packed gravel). As the road grade increases, the amount of friction possible between the tire and the road surface decreases. Even if the bicycle was powered by a motor or engine, the bicycle would still come up to this limit. It is not so much that a human could no longer provide enough force to turn the cranks but that the tires would begin to skid on the road. The only way to exceed the limit to to have a greater transfer of the driving force between the wheel and the surface of the slope. <br /><br />For example, if the bicycle wheels had cogs that meshed into the slope surface (sort of like the cog railway on the west slope of Mount Washington) then the 40-45% limit could be beaten (I believe). Of course, using a human body to propel such a bicycle would result in very low speed that would likely be beaten by a walker/hiker/climber/snail.<br /><br />Have I misunderstood this?<br /><br /><br />Cheers!Anonymousnoreply@blogger.comtag:blogger.com,1999:blog-13887692.post-22222393622964234662009-08-01T16:39:13.637-04:002009-08-01T16:39:13.637-04:00Anon @ 5:25 : That's flattering. But see, I...Anon @ 5:25 : That's flattering. But see, I'm not guaranteeing that my analysis is right. This is why I submitted all this information to the public for a 'peer review' you know. If someone catches any of my mistakes and if it makes sense, I'd like to correct things.Ron Georgehttps://www.blogger.com/profile/18394865788996482667noreply@blogger.comtag:blogger.com,1999:blog-13887692.post-9094725168503367082009-08-01T16:25:49.336-04:002009-08-01T16:25:49.336-04:00Woot. Seriously Ron, why don't you make this a...Woot. Seriously Ron, why don't you make this a scientific journal? You really give new meaning to the word "BLOG"! :)Anonymousnoreply@blogger.comtag:blogger.com,1999:blog-13887692.post-26016532957554927882009-08-01T13:55:07.500-04:002009-08-01T13:55:07.500-04:00Here's a rule of thumb I go by. A watt of outp...Here's a rule of thumb I go by. A watt of output will raise one kg a thousand feet in an hour. So, if<br />Tom Danielson and his bike weighed 100kg total, he would have had to<br />put out a minimum of 540 watts continuous.Jenny Bnoreply@blogger.comtag:blogger.com,1999:blog-13887692.post-85308711928247518102009-08-01T13:49:29.184-04:002009-08-01T13:49:29.184-04:00Dan,
Yes looks like I made an idiotic error assum...Dan,<br /><br />Yes looks like I made an idiotic error assuming 17 to 17.7% change in grade is 10%. Ha. I'm pretty sure I did not even think twice about that. Thank you very much for the insights. Yours was also very good and pointed me in a different direction.Ron Georgehttps://www.blogger.com/profile/18394865788996482667noreply@blogger.comtag:blogger.com,1999:blog-13887692.post-69595184721553177742009-08-01T10:55:42.431-04:002009-08-01T10:55:42.431-04:00Nice work, and thanks for the link!!! One small i...Nice work, and thanks for the link!!! One small issue I noticed, however:<br /><i>Outside of this linear curve after about 10% gradient, a 10% relative increase in grade (say from 17% to 17.7%) should only result in a 3.8% increase in climbing rate.</i><br /><br />Actually, 17.7 / 17.0 - 1 = 4.1%, not 10%.<br /><br />The sine(arctan Δy/Δx) can be simplified as Δy / sqrt[(Δx)² + (Δy)²] (thanks to the Tin Man of Oz). Then one can directly calculate the relative error of using Δy/Δx directly: ε = ( 1 - 1 / sqrt[1 + (Δy / Δx)²] ). For (Δy / Δx)² reasonably small, linearizing the square root yields ε ≈ ( 1 - 1 / [1 + ½(Δy / Δx)²] ) = ½(Δy / Δx)² / [1 + ½(Δy / Δx)²] ≈ ½(Δy / Δx)². So for Δy / Δx = 0.17, this works out to 1.4%. So you make a +1.4% error in climbing power at a given speed on a 17% grade by assuming grade = rise / run. But since rolling resistance is likely increased by the resulting weight redistribution to the rear tire, I'm not sure which answer is better :).<br /><br />There's two roads in San Francisco which are 31.5% for one block each: Filbert next to Lombard and 22nd west of Church. In these cases, the error is 4.8% relative. That's fairly substantial: more than a full tooth on a 26 tooth cog.<br /><br />Thanks again for the enjoyable article!djconnelhttps://www.blogger.com/profile/01484858820878605035noreply@blogger.comtag:blogger.com,1999:blog-13887692.post-30842861769078002202009-08-01T08:35:11.014-04:002009-08-01T08:35:11.014-04:00Anon : Sorry, I had to rewrite some of this and re...Anon : Sorry, I had to rewrite some of this and republish it to put in the error% of linear and non-linearity of the grade in the climbing rate equation.Ron Georgehttps://www.blogger.com/profile/18394865788996482667noreply@blogger.comtag:blogger.com,1999:blog-13887692.post-77519652200301312082009-08-01T01:50:12.514-04:002009-08-01T01:50:12.514-04:00Wow, I just realized how much I forgot a lot of my...Wow, I just realized how much I forgot a lot of my high school trig. Thanks for this whack on my head Ron!!Philnoreply@blogger.comtag:blogger.com,1999:blog-13887692.post-21836727641990114412009-08-01T01:08:47.749-04:002009-08-01T01:08:47.749-04:00You're welcome. I realize that I foolishly lef...You're welcome. I realize that I foolishly left out explaining at what grade the error between sin[arctan(grade)] and simply grade becomes unacceptable. I overlooked this. I think I should put it in when I get time (hopefully tomorrow morning).Ron Georgehttps://www.blogger.com/profile/18394865788996482667noreply@blogger.comtag:blogger.com,1999:blog-13887692.post-87844742041477184392009-08-01T01:05:38.037-04:002009-08-01T01:05:38.037-04:00Thanks. Was confused there.Thanks. Was confused there.Anonymousnoreply@blogger.comtag:blogger.com,1999:blog-13887692.post-31975999806332958992009-07-31T23:14:26.798-04:002009-07-31T23:14:26.798-04:00Anon : Good catch. See, I made a note in the Power...Anon : Good catch. See, I made a note in the Power Perspective section that if grade is steep, G is to be replaced with sine[arctan(rise/run)] which then agrees with the vector analysis. For really really small degrees, sine (angle) is approximated as the angle.Ron Georgehttps://www.blogger.com/profile/18394865788996482667noreply@blogger.comtag:blogger.com,1999:blog-13887692.post-44961842294930195662009-07-31T22:58:27.555-04:002009-07-31T22:58:27.555-04:00Hmm...I don't get understand this this thing. ...Hmm...I don't get understand this this thing. You say from the power perspective that climbing rate = Speed x [rise/run]. But from the vector analysis perspective, climbing rate is Speed x Sine(hill angle). So that would mean rise/run = sin (hill angle)? Am I right? I'm sorry if you cannot see my problem.Anonymousnoreply@blogger.com