mx_599
02-25-06, 07:39 PM
i guess micro gearing is the right term. so for two near equal ratios, you still have to "crank" harder with less teeth in the front/rear?
BMX - Micro gearing requires more torque?
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aches
02-26-06, 07:17 AM
no
KinetikBiker
02-26-06, 07:42 AM
I honestly think beerman was wrong on that one...
Just a gut feeling after riding my friends bikes that have micro gearing ranging from 36-13's to 25-9's...
Just a gut feeling after riding my friends bikes that have micro gearing ranging from 36-13's to 25-9's...
MadMan2k
02-26-06, 11:43 AM
He may have meant there was more torque being applied to the chain with smaller gearing, because of the smaller number of teeth, rather than more torque being applied to the pedals?
FMFBMX
02-26-06, 02:45 PM
yeah he ment to the chain. because theres less teeth to be grabbed. thus more strain on the chain.
mx_599
02-26-06, 04:18 PM
....but he said, "you'd" which would imply "you would" be torquing much harder. so i am still confused
KinetikBiker
02-26-06, 06:04 PM
The mysterious beerman, his words are a mystery (i spelled that wrong)
FMFBMX
02-26-06, 06:53 PM
where is beerman!?
ettsn
02-26-06, 07:49 PM
I'll pre-apologize for the nerdy answer. :)
Well, if you were to look at an FEA of the chainring, you'd see the stress riser at the first fully engaged tooth. The torque dissipates as it moves further from the engagement point until it is fully released on the bottom. This is because the chain in tension (at the top between the cog and the ring) and is stetched to full capacity. On the bottom, it isn't in tension.
Does a smaller ring (but same ratio) require more total torque applied for the same effect at the rear wheel? Yes, but only very slightly. The larger front ring has a mechanical advantage due to being a longer lever (discs - chainrings, for instance - are essentially a continuous succession of levers -- think of each tooth on the ring extending down to the crank arm individually, if that helps). Torque is expressed in lbs/ft, where each unit is equal to one pound at the end of a foot-long lever. If you increase that to a two-foot lever, you'll only need eight ounces to provide the same torque. A six-inch lever would need two pounds to equal the same. The same holds true for crank length, but I don't want to confuse the issue. ;) (The crank arm, chainring radius and cog radius each contribute the leverage effect in the transfer of power from your legs to the ground.)
-Paul
Well, if you were to look at an FEA of the chainring, you'd see the stress riser at the first fully engaged tooth. The torque dissipates as it moves further from the engagement point until it is fully released on the bottom. This is because the chain in tension (at the top between the cog and the ring) and is stetched to full capacity. On the bottom, it isn't in tension.
Does a smaller ring (but same ratio) require more total torque applied for the same effect at the rear wheel? Yes, but only very slightly. The larger front ring has a mechanical advantage due to being a longer lever (discs - chainrings, for instance - are essentially a continuous succession of levers -- think of each tooth on the ring extending down to the crank arm individually, if that helps). Torque is expressed in lbs/ft, where each unit is equal to one pound at the end of a foot-long lever. If you increase that to a two-foot lever, you'll only need eight ounces to provide the same torque. A six-inch lever would need two pounds to equal the same. The same holds true for crank length, but I don't want to confuse the issue. ;) (The crank arm, chainring radius and cog radius each contribute the leverage effect in the transfer of power from your legs to the ground.)
-Paul
FMFBMX
02-26-06, 08:00 PM
...Oh, I guess.
MadMan2k
02-26-06, 08:17 PM
ettsn, thank you for the detailed, intelligent answer. That probably explains why racers don't usually use compact chainrings and appropriate cogs.
FMFBMX
02-26-06, 08:25 PM
So how much different is 16-38 compared to a 9-31 its the same thing kind of right...because 16-38 (Subtract) is 22... add 22 to 9 and its 31 so 9-31 is the same as 16-38...Speed wise, correct?
mx_599
02-26-06, 08:46 PM
I'll pre-apologize for the nerdy answer. :)
Well, if you were to look at an FEA of the chainring, you'd see the stress riser at the first fully engaged tooth. The torque dissipates as it moves further from the engagement point until it is fully released on the bottom. This is because the chain in tension (at the top between the cog and the ring) and is stetched to full capacity. On the bottom, it isn't in tension.
Does a smaller ring (but same ratio) require more total torque applied for the same effect at the rear wheel? Yes, but only very slightly. The larger front ring has a mechanical advantage due to being a longer lever (discs - chainrings, for instance - are essentially a continuous succession of levers -- think of each tooth on the ring extending down to the crank arm individually, if that helps). Torque is expressed in lbs/ft, where each unit is equal to one pound at the end of a foot-long lever. If you increase that to a two-foot lever, you'll only need eight ounces to provide the same torque. A six-inch lever would need two pounds to equal the same. The same holds true for crank length, but I don't want to confuse the issue. ;) (The crank arm, chainring radius and cog radius each contribute the leverage effect in the transfer of power from your legs to the ground.)
-Paul
thanks
Well, if you were to look at an FEA of the chainring, you'd see the stress riser at the first fully engaged tooth. The torque dissipates as it moves further from the engagement point until it is fully released on the bottom. This is because the chain in tension (at the top between the cog and the ring) and is stetched to full capacity. On the bottom, it isn't in tension.
Does a smaller ring (but same ratio) require more total torque applied for the same effect at the rear wheel? Yes, but only very slightly. The larger front ring has a mechanical advantage due to being a longer lever (discs - chainrings, for instance - are essentially a continuous succession of levers -- think of each tooth on the ring extending down to the crank arm individually, if that helps). Torque is expressed in lbs/ft, where each unit is equal to one pound at the end of a foot-long lever. If you increase that to a two-foot lever, you'll only need eight ounces to provide the same torque. A six-inch lever would need two pounds to equal the same. The same holds true for crank length, but I don't want to confuse the issue. ;) (The crank arm, chainring radius and cog radius each contribute the leverage effect in the transfer of power from your legs to the ground.)
-Paul
thanks
MadMan2k
02-26-06, 08:49 PM
Gearing is reffered to by ratios, either by saying the number of teeth on the chainring and cog, or by just saying the ratio of chainring teeth to cog teeth.
To find the gear ratio, just divide the number of teeth on the chainring by the number of teeth on the cog.
For example, 44-16 (44 divided by 16) is 2.75. If you ran 44-16, you could state those two numbers or you could just say that you run a 2.75:1 gear ratio.
Without taking into consideration the points that ettsn brought up, you could change the size of your chainring and, as long as the size of the cog changed accordingly, your gear ratio would stay the same. A 25-9 would be 2.78:1(rounded, plug it into a calculator and you'll see the number is long).
To find the gear ratio, just divide the number of teeth on the chainring by the number of teeth on the cog.
For example, 44-16 (44 divided by 16) is 2.75. If you ran 44-16, you could state those two numbers or you could just say that you run a 2.75:1 gear ratio.
Without taking into consideration the points that ettsn brought up, you could change the size of your chainring and, as long as the size of the cog changed accordingly, your gear ratio would stay the same. A 25-9 would be 2.78:1(rounded, plug it into a calculator and you'll see the number is long).
ettsn
02-26-06, 09:15 PM
You are correct. The direct mechanical advantage of a 44-16 is the same as a 33-12 or even a 22-8, because they all share the same ratio of 2.75:1. Given that, there is a slight mechanical adantage at the point of chainring to chain with a larger chainring. The crankarm and the chainring work together as a Class 2 lever. Consider a prybar jammed under a big rock (also a T2 lever): the deeper you push the prybar under the rock, the better advantage the lever has versus the load (the rock). A longer prybar (crankarm in our case) will prove a much better overall mechanical advantage.
Summary: assuming the same gearing, is there a measurable difference in chainring sizes? Yes. Is there a noticeable difference? Probably not.
-Paul
Summary: assuming the same gearing, is there a measurable difference in chainring sizes? Yes. Is there a noticeable difference? Probably not.
-Paul
FMFBMX
02-27-06, 06:58 AM
Thank you very much.
CMcMahon
02-27-06, 01:14 PM
where is beerman!?
I was out riding.
I was out riding.
CBolt
03-06-06, 08:42 PM
Just use Sheldon Brown's gear inch calculator (http://sheldonbrown.com/gears/).
Prozakk
03-06-06, 08:53 PM
But yet I got insulted for picking 44-16...oh, my bad, it was the gnats...again.
dooley
03-06-06, 09:28 PM
I'll pre-apologize for the nerdy answer. :)
Well, if you were to look at an FEA of the chainring, you'd see the stress riser at the first fully engaged tooth. The torque dissipates as it moves further from the engagement point until it is fully released on the bottom. This is because the chain in tension (at the top between the cog and the ring) and is stetched to full capacity. On the bottom, it isn't in tension.
Does a smaller ring (but same ratio) require more total torque applied for the same effect at the rear wheel? Yes, but only very slightly. The larger front ring has a mechanical advantage due to being a longer lever (discs - chainrings, for instance - are essentially a continuous succession of levers -- think of each tooth on the ring extending down to the crank arm individually, if that helps). Torque is expressed in lbs/ft, where each unit is equal to one pound at the end of a foot-long lever. If you increase that to a two-foot lever, you'll only need eight ounces to provide the same torque. A six-inch lever would need two pounds to equal the same. The same holds true for crank length, but I don't want to confuse the issue. ;) (The crank arm, chainring radius and cog radius each contribute the leverage effect in the transfer of power from your legs to the ground.)
-Paul
You are correct. The direct mechanical advantage of a 44-16 is the same as a 33-12 or even a 22-8, because they all share the same ratio of 2.75:1. Given that, there is a slight mechanical adantage at the point of chainring to chain with a larger chainring. The crankarm and the chainring work together as a Class 2 lever. Consider a prybar jammed under a big rock (also a T2 lever): the deeper you push the prybar under the rock, the better advantage the lever has versus the load (the rock). A longer prybar (crankarm in our case) will prove a much better overall mechanical advantage.
Summary: assuming the same gearing, is there a measurable difference in chainring sizes? Yes. Is there a noticeable difference? Probably not.
-Paul
I'm not sure what ettsn is on about...
A 25/9 will put almost twice as much force on a chain as a 44/16, (if you can apply 1000 newtons - 3600newtons for the 25/9 as opposed to 2000 for the 44/16) Most of that is at the rear, being the smaller of the 2, which is why we have aluminium up front, and steel at the back. On a 9t driver 4teeth are sharing the load, if it were shared equally it might be ok, but it isn't as that fella said one of those teeth takes most of the load.
As for the lever, you've got that wrong too, with a larger sprocket you have less of a mechanical advantage, that's why there's more forc on the chain with a micro gearing. Consider a larger gearing the same as moving the load away from the fulcrum, and a micro drive as moving it closer.
Micro drive has been done before in the mtb world - it's a ****ty way to lose weight, if you need it for clearance and are happy with the knowledge that your chain will be more prone to snapping then go for it. Oh yeah, and wear can be an issue too, I know someone who wore out a 25t imperial in 3 weeks, I have had a 44t imperial on one of my bikes for year.
Stolen almost entirely from G at gsportbmx.
Well, if you were to look at an FEA of the chainring, you'd see the stress riser at the first fully engaged tooth. The torque dissipates as it moves further from the engagement point until it is fully released on the bottom. This is because the chain in tension (at the top between the cog and the ring) and is stetched to full capacity. On the bottom, it isn't in tension.
Does a smaller ring (but same ratio) require more total torque applied for the same effect at the rear wheel? Yes, but only very slightly. The larger front ring has a mechanical advantage due to being a longer lever (discs - chainrings, for instance - are essentially a continuous succession of levers -- think of each tooth on the ring extending down to the crank arm individually, if that helps). Torque is expressed in lbs/ft, where each unit is equal to one pound at the end of a foot-long lever. If you increase that to a two-foot lever, you'll only need eight ounces to provide the same torque. A six-inch lever would need two pounds to equal the same. The same holds true for crank length, but I don't want to confuse the issue. ;) (The crank arm, chainring radius and cog radius each contribute the leverage effect in the transfer of power from your legs to the ground.)
-Paul
You are correct. The direct mechanical advantage of a 44-16 is the same as a 33-12 or even a 22-8, because they all share the same ratio of 2.75:1. Given that, there is a slight mechanical adantage at the point of chainring to chain with a larger chainring. The crankarm and the chainring work together as a Class 2 lever. Consider a prybar jammed under a big rock (also a T2 lever): the deeper you push the prybar under the rock, the better advantage the lever has versus the load (the rock). A longer prybar (crankarm in our case) will prove a much better overall mechanical advantage.
Summary: assuming the same gearing, is there a measurable difference in chainring sizes? Yes. Is there a noticeable difference? Probably not.
-Paul
I'm not sure what ettsn is on about...
A 25/9 will put almost twice as much force on a chain as a 44/16, (if you can apply 1000 newtons - 3600newtons for the 25/9 as opposed to 2000 for the 44/16) Most of that is at the rear, being the smaller of the 2, which is why we have aluminium up front, and steel at the back. On a 9t driver 4teeth are sharing the load, if it were shared equally it might be ok, but it isn't as that fella said one of those teeth takes most of the load.
As for the lever, you've got that wrong too, with a larger sprocket you have less of a mechanical advantage, that's why there's more forc on the chain with a micro gearing. Consider a larger gearing the same as moving the load away from the fulcrum, and a micro drive as moving it closer.
Micro drive has been done before in the mtb world - it's a ****ty way to lose weight, if you need it for clearance and are happy with the knowledge that your chain will be more prone to snapping then go for it. Oh yeah, and wear can be an issue too, I know someone who wore out a 25t imperial in 3 weeks, I have had a 44t imperial on one of my bikes for year.
Stolen almost entirely from G at gsportbmx.