stroking Posted June 1, 2012 Report Share Posted June 1, 2012 trying to figure out the difference between et's say one bike runs 6.0 and the other 6.1 how many lengths head would the 6.0 bike be,with both pulling a 1.3 60ft ?? 1/8mile racing !! some say 2 bikes...others say 3 Quote Link to comment Share on other sites More sharing options...
shanYE west Posted June 1, 2012 Report Share Posted June 1, 2012 it would depend on trap speeds of the bikes. at 85mph it would be .0024 miles traveld in .1 which is about 12.5ft.. if both bikes trapped at the same speed. That being said.. itd be safe to say that you could give or take a foot and that would give you a +/- 10mph Quote Link to comment Share on other sites More sharing options...
SlowerThanYou Posted June 1, 2012 Report Share Posted June 1, 2012 There would also be a length difference, depending on reaction time advantage/disadvantage (track) or holeshot advantange/disadvantage (street). Quote Link to comment Share on other sites More sharing options...
shanYE west Posted June 1, 2012 Report Share Posted June 1, 2012 There would also be a length difference, depending on reaction time advantage/disadvantage (track) or holeshot advantange/disadvantage (street). if a guy got the jump on the tree by .3.. that equals about a 2ft head start. If you got 2 good racers.. that wont happen. .03 difference would be more typical scenario with a .03 jump.. assuming he left the line at 10mph.. that would equal about a .5 foot head start or less. Unless rate of acceleration is known.. its impossible to get a exact distance.. The end trap speed is the most important variable if your looking for a quick estimate in distance difference.. I think i edited this like 20 times.. lol Quote Link to comment Share on other sites More sharing options...
SlowerThanYou Posted June 1, 2012 Report Share Posted June 1, 2012 if a guy got the jump on the tree by .3.. that equals about a 2ft head start. If you got 2 good racers.. that wont happen. .03 difference would be more typical scenario with a .03 jump.. assuming he left the line at 10mph.. that would equal about a .5 foot head start or less. Unless rate of acceleration is known.. its impossible to get a exact distance.. The end trap speed is the most important variable if your looking for a quick estimate in distance difference.. I think i edited this like 20 times.. lol I agree speed is the biggest factor. I've ran with these guys back East & a bike or two reaction advantage/holeshot is pretty common with decent rider. Quote Link to comment Share on other sites More sharing options...
RadarRacing Posted June 1, 2012 Report Share Posted June 1, 2012 At 60 mph you are traveling 89.5 feet per second. Each 0.1 would be 8.9 feet lead. At 120 mph the 0.1 would be double that ....17.8 feet. Assuming a bike running a 6.1 is going around 100mph give or take a few mph either way the lead would be roughly 12 feet.... give or take a few feet. Quote Link to comment Share on other sites More sharing options...
shanYE west Posted June 1, 2012 Report Share Posted June 1, 2012 I agree speed is the biggest factor. I've ran with these guys back East & a bike or two reaction advantage/holeshot is pretty common with decent rider. lol yah.. people who race me usually get a bike or 2 lead also... Quote Link to comment Share on other sites More sharing options...
SlowerThanYou Posted June 1, 2012 Report Share Posted June 1, 2012 (edited) Here's what a .291 light vs. .033 light with .000 being perfect. I remember having 3 bikes on him early in the run. The quad in the left lane was a bigger bike & a lot faster. He caught me around 250'. Scroll down a couple rows. http://gallery.competitionplus.com/main.php?g2_itemId=961651&g2_page=5 Edited June 1, 2012 by SlowerThanYou Quote Link to comment Share on other sites More sharing options...
shanYE west Posted June 1, 2012 Report Share Posted June 1, 2012 Here's what a .291 light vs. .033 light with .000 being perfect. I remember having 3 bikes on him early in the run. The quad in the left lane was a bigger bike & a lot faster. He caught me around 250'. Scroll down a couple rows. http://gallery.compe...61651&g2_page=5 a lot of variables. Rate of acceleration would be the biggest. if you left the line at 45mph .3 seconds would get you about 20ft in front before he left. but.. leaving at 45 mph is not even possible.. the rate of acceleration for that .3 would have to be known and a long ass formula that i can't even comprehend would be needed. you could take the 60' time and dissect it to get close. with a 1.5 60' would yeild a avg speed of 27.mph at 27mph you would travel 11ft in .3. but even then.. common physics tells us we're not leaving the line at 27mph. you leaving at 0 and accelerating. so we really need to no know the rate of acceleration to get a more precise number. Kind of a fun topic though.. Quote Link to comment Share on other sites More sharing options...
SlowerThanYou Posted June 1, 2012 Report Share Posted June 1, 2012 I just wanted to provide a visual reference. I happened to have the matching time slip for that series of photos. Quote Link to comment Share on other sites More sharing options...
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