Are you sure it's not the bike?? (Page 2)
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2014-10-10 3:34 PM in reply to: mike761 |
1502 Katy, Texas | Subject: RE: Are you sure it's not the bike?? Originally posted by mike761 Originally posted by gsmacleod Originally posted by 3mar I can feel a huge difference in impacts so it would go to reason that the rubber on the tire is experiencing the same. I'm rolling with it. If you don't think I'll run tests to make sure and differences are truly statistically significant, then you haven't met an engineer. Now, that being said, there are other external factors such as rider ability changing etc. But frankly, since I've had the new bike I have been doing a lot less bobbing and weaving than before. I'm actually plowing through a lot of debris just to see what happens. I'm curious as to how exactly you are going to test this to ensure significance. Shane Is one data point statistically significant?? Obviously not...that's why I said I'd report back when I had more data. |
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2014-10-10 4:14 PM in reply to: 3mar |
Champion 9407 Montague Gold Mines, Nova Scotia | Subject: RE: Are you sure it's not the bike?? Originally posted by 3mar So I don't want to belabor the point on the flats, but follow me on a little mental experiment. Picture two small (about 2" diameter) balls in your head. One is made of a very hard material, such as steel, or even marble. The second is made of rubber. Now, these two balls are both the same size and mass (remember it's a mental experiment...but if you're having trouble with assuming they are the same mass, then assume the rubber ball has a lead core or something). So each ball will have the same mass, size and shape (and therefore density and acceleration when dropped). We take each one and wrap it in the rubber of a bike tire then drop it from a distance of a foot unto a sharpened point. The force of each impact would be the same (f=ma, and since the mass is the same and the size/shape is identical, the acceleration would also be the same). So at the moment of impact, which rubber coating would you assume would bear more damage? I would say the one wrapped in the harder material. Why? Because the force (f) will have to go somewhere...all of it. For the rubber ball, a lot of the force will dissipate in the ball itself and where the rubber and the sharp object meet will have less force to cause the damage than the ball where almost all the force is in that interaction as little to none will be dissipated on the hardened ball material. So which is the carbon bike in this example? Also, how does this apply to a pneumatic tire? Regarding what I will use for statistical significance....probably a Student's T test. I'll look at the average number of flats per tire, or mile, or both with and without the new bike and see if there is a statistically significant difference. I realize by having different tires there isn't a true control group, however, I'm playing Devil's advocate against my theory by using the less puncture resistant tires on the side I feel there will be less flats...i.e. I'm stacking the cards against myself. What type of flats were you getting on the other bike? This will make a huge difference in your analysis. Remember, energy, water, heat, whatever, always take the path of least resistance. Also there is a conservation of mass and energy that a closed system will have to deal with the same amount of energy released. If some of that energy is taken away from the impact point between the debris and the tire, then that will make a tear less likely....just a theory. Good luck with your testing - I doubt you will find evidence to support your theory however. Shane |
2014-10-10 4:29 PM in reply to: gsmacleod |
1502 Katy, Texas | Subject: RE: Are you sure it's not the bike?? Originally posted by gsmacleod Originally posted by 3mar So I don't want to belabor the point on the flats, but follow me on a little mental experiment. Picture two small (about 2" diameter) balls in your head. One is made of a very hard material, such as steel, or even marble. The second is made of rubber. Now, these two balls are both the same size and mass (remember it's a mental experiment...but if you're having trouble with assuming they are the same mass, then assume the rubber ball has a lead core or something). So each ball will have the same mass, size and shape (and therefore density and acceleration when dropped). We take each one and wrap it in the rubber of a bike tire then drop it from a distance of a foot unto a sharpened point. The force of each impact would be the same (f=ma, and since the mass is the same and the size/shape is identical, the acceleration would also be the same). So at the moment of impact, which rubber coating would you assume would bear more damage? I would say the one wrapped in the harder material. Why? Because the force (f) will have to go somewhere...all of it. For the rubber ball, a lot of the force will dissipate in the ball itself and where the rubber and the sharp object meet will have less force to cause the damage than the ball where almost all the force is in that interaction as little to none will be dissipated on the hardened ball material. So which is the carbon bike in this example? Also, how does this apply to a pneumatic tire? Regarding what I will use for statistical significance....probably a Student's T test. I'll look at the average number of flats per tire, or mile, or both with and without the new bike and see if there is a statistically significant difference. I realize by having different tires there isn't a true control group, however, I'm playing Devil's advocate against my theory by using the less puncture resistant tires on the side I feel there will be less flats...i.e. I'm stacking the cards against myself. What type of flats were you getting on the other bike? This will make a huge difference in your analysis. Remember, energy, water, heat, whatever, always take the path of least resistance. Also there is a conservation of mass and energy that a closed system will have to deal with the same amount of energy released. If some of that energy is taken away from the impact point between the debris and the tire, then that will make a tear less likely....just a theory. Good luck with your testing - I doubt you will find evidence to support your theory however. Shane I will only be looking at puncture flats (yes I took that detailed of records) but on the grand scheme of things, it doesn't matter. You've got to break this, and any other system, into it's most simple components then work your way back. That's how engineering works. It's unfortunately the opposite of how math and science are taught in high schools. Let's set up some boundary conditions for this example. We are interested in the interaction of the rubber of the tire with the road and debris. So we will draw two mental circles; one around our bike encompassing us, the bike and ending at the outermost edge of the tire. The second circle will be everything else. We are concerned with what is happening at that boundary. The force that makes it's way to our imaginary boundary is not straight forward. For example, if you were to bunny hop your bike, when you land again, if you were to lock your knees, the forces exerted at that boundary would be much greater than if you were to bend your knees. Now, the overall force is the same, it's your mass x acceleration of gravity as you come back to earth. So if the force created from the bunny hop is the same, how can the force exerted at our boundary be different? Simple, some of that force was dissipated in your legs and therefore not all of it made it to our boundary. Same thing here. Except instead of the force being dissipated differently due to movement of your legs, more of the force is dissipated in the carbon frame than the aluminum frame. With less force at that boundary, when the tire comes into contact with a piece of debris, it will have less force to push through the rubber. I used the example of the bunny hop so that you could imagine a specific quantifiable event, however as we ride, we are creating thousands of these interactions. Each creating a specific amount of force that must go somewhere. If more go into the frame, then less will end up between the rubber and that piece of metal, stone, glass, whatever. |
2014-10-10 5:05 PM in reply to: 3mar |
Champion 9407 Montague Gold Mines, Nova Scotia | Subject: RE: Are you sure it's not the bike?? Originally posted by 3mar I will only be looking at puncture flats (yes I took that detailed of records) but on the grand scheme of things, it doesn't matter. It most certainly does matter - if you were getting snake bites (underinflation) or flats due to bad rim tape (very common with cheap wheels) then reducing these types of flats has nothing to do with your theory. You've got to break this, and any other system, into it's most simple components then work your way back. That's how engineering works. It's unfortunately the opposite of how math and science are taught in high schools. Let's set up some boundary conditions for this example. We are interested in the interaction of the rubber of the tire with the road and debris. So we will draw two mental circles; one around our bike encompassing us, the bike and ending at the outermost edge of the tire. The second circle will be everything else. We are concerned with what is happening at that boundary. The force that makes it's way to our imaginary boundary is not straight forward. I understand isolation of systems, however there suspension system on a road bike is critical in this consideration. For example, if you were to bunny hop your bike, when you land again, if you were to lock your knees, the forces exerted at that boundary would be much greater than if you were to bend your knees. Now, the overall force is the same, it's your mass x acceleration of gravity as you come back to earth. So if the force created from the bunny hop is the same, how can the force exerted at our boundary be different? Simple, some of that force was dissipated in your legs and therefore not all of it made it to our boundary. Close but not quite. The flexing of your knees increases the time of acceleration and since the force applied to a system is the product of mass and the rate of change of velocity, by increasing the time, you decrease acceleration and therefore the peak force experienced by the system. Same thing here. Except instead of the force being dissipated differently due to movement of your legs, more of the force is dissipated in the carbon frame than the aluminum frame. With less force at that boundary, when the tire comes into contact with a piece of debris, it will have less force to push through the rubber. I used the example of the bunny hop so that you could imagine a specific quantifiable event, however as we ride, we are creating thousands of these interactions. Each creating a specific amount of force that must go somewhere. If more go into the frame, then less will end up between the rubber and that piece of metal, stone, glass, whatever. For this to be true, you'll need to consider both the stiffness of each frameset and each wheelset and determine which system has more flex. It may very well be the new bike is not as stiff but it isn't as simple as saying carbon is less stiff than aluminum. You could simplify this, as was suggested earlier, by using your old wheels on the new bike to further control variables. Shane |
2014-10-10 5:13 PM in reply to: gsmacleod |
1502 Katy, Texas | Subject: RE: Are you sure it's not the bike?? Originally posted by gsmacleod Originally posted by 3mar I will only be looking at puncture flats (yes I took that detailed of records) but on the grand scheme of things, it doesn't matter. It most certainly does matter - if you were getting snake bites (underinflation) or flats due to bad rim tape (very common with cheap wheels) then reducing these types of flats has nothing to do with your theory. You've got to break this, and any other system, into it's most simple components then work your way back. That's how engineering works. It's unfortunately the opposite of how math and science are taught in high schools. Let's set up some boundary conditions for this example. We are interested in the interaction of the rubber of the tire with the road and debris. So we will draw two mental circles; one around our bike encompassing us, the bike and ending at the outermost edge of the tire. The second circle will be everything else. We are concerned with what is happening at that boundary. The force that makes it's way to our imaginary boundary is not straight forward. I understand isolation of systems, however there suspension system on a road bike is critical in this consideration. For example, if you were to bunny hop your bike, when you land again, if you were to lock your knees, the forces exerted at that boundary would be much greater than if you were to bend your knees. Now, the overall force is the same, it's your mass x acceleration of gravity as you come back to earth. So if the force created from the bunny hop is the same, how can the force exerted at our boundary be different? Simple, some of that force was dissipated in your legs and therefore not all of it made it to our boundary. Close but not quite. The flexing of your knees increases the time of acceleration and since the force applied to a system is the product of mass and the rate of change of velocity, by increasing the time, you decrease acceleration and therefore the peak force experienced by the system. Same thing here. Except instead of the force being dissipated differently due to movement of your legs, more of the force is dissipated in the carbon frame than the aluminum frame. With less force at that boundary, when the tire comes into contact with a piece of debris, it will have less force to push through the rubber. I used the example of the bunny hop so that you could imagine a specific quantifiable event, however as we ride, we are creating thousands of these interactions. Each creating a specific amount of force that must go somewhere. If more go into the frame, then less will end up between the rubber and that piece of metal, stone, glass, whatever. For this to be true, you'll need to consider both the stiffness of each frameset and each wheelset and determine which system has more flex. It may very well be the new bike is not as stiff but it isn't as simple as saying carbon is less stiff than aluminum. You could simplify this, as was suggested earlier, by using your old wheels on the new bike to further control variables. Shane Here is one of my favorite quotes: "The difference between theory and practice is that in theory they are not different, but in practice they are." i.e. we could go on for ever and ever on this one. How about we just see if the flats reduce? If it does, and I'm on the side of the road less, then I'll be happy with any theory. |
2014-10-10 5:23 PM in reply to: 3mar |
Champion 9407 Montague Gold Mines, Nova Scotia | Subject: RE: Are you sure it's not the bike?? Originally posted by 3mar Here is one of my favorite quotes: "The difference between theory and practice is that in theory they are not different, but in practice they are." i.e. we could go on for ever and ever on this one. How about we just see if the flats reduce? If it does, and I'm on the side of the road less, then I'll be happy with any theory. I have no issue with that - however claiming that you are looking for statistical significance implies much more than being happy with your theory. Shane |
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