Gears & Ratios 101 (sorta, hehe)

G

gone_fishin

Guest
#1
srdavo was looking for some help "visualizing" gearing and ratios as pertains to different drive-types & applications...we did our best, srdavo suggested i post it "as is" for educational & entertainment purposes. if ya can dig thru the rambling, there does seem to be a bit of good info in here.

also, the following is a great example of using "quote" instead of "reply" when using the private messaging system 8)

srdavo said:
augidog said:
srdavo said:
augidog said:
srdavo said:
augidog said:
srdavo said:
augidog said:
srdavo said:
morning bro......
gonna ramble at ya for a bit......okay

on a chain drive motor bike...the smaller the rear sprocket, the faster ya go. I get this part. on my friction drive, the tire IS the rear sprocket....so... the smaller the tire, the faster ya go. am I getting warm?

since I put the friction drive on a 20 inch bike(for now), It's gonna be way faster than a 26 inch. (the dude I bought it from said it would run 22mph on his 26inch bike)
this would explain why I have to pedal so much to get it up to speed....
am I getting warmer? :?

your humble student,
Dave
first, use "quote" instead of "reply" so we can view this whole topic.

there are two kinds of "mechanical advantage"...m/a of speed & m/a of power.

a lower total ratio (1st gear) means m/a of power, a higher total ratio (5th gear) means m/a of speed.

regardless of chain and sprocket, let's look at only tire size: turn a 20" wheel one time fully around...how far did you move along the ground? turn a 26" tire one rotation, you've moved about 20% more distance, yes? turn both sizes at the same speed, the larger outside circumference will cover more ground in the same time.

now, the sprocket diameter versus the wheel diameter means a new transfer point of power and another ratio factor. one rotation of the sprocket means one rotation of the wheel.

same motor, same rear sprocket: larger wheel means m/a of speed, smaller wheel means m/a of power.

we ok so far?

remember to quote instead of reply, then type your response under the quote.
yup...I'm with ya.....
m/a of speed = makes ya go faster
m/a of power = gives ya better take off/hill climbing

what's next?
actually, not much to cover from this point...let's work our way backwards up the drive-train. large driven (rear) sprocket versus very small drive sprocket...the high rpm's of the clutch shaft are drastically changed into a huge m/a of power by that "negative" ratio (10 th to 44 th). raise that ratio a bit (10th to 36th) to get more speed at a cost to power.

yes?
I'm still with ya....
now, on friction drive, to gain m/a of speed, use a larger drive roller. (make ya go faster) :lol: would a smaller diameter wheel do the same? how does wheel size figure in? am I on the right track here?
with as roller drive, the roller has the same effect as if it was on the pavement, bigger roller more speed, no matter tire size, but bigger wheel means less revs on your hub, longer life.

no bicycle engine in the world would drive a bicycle without TOTAL m/a of power, that's why our drive-train starts out at 10000 rpm's for the crankshaft and is gradually reduced thru gearing to 2500rpms for the roller>250rpm's for the wheel itself.

approx m/a-p of engine to pavement= 40:1

that's why that dinky lil piston can even start to move 250lbs

*those are only rough numbers taken out of my skull, prolly not even close, but u get the idea.

am i rambling too much, am i staying on track enuff for ya? what i'm trying to do is get you to where you can formulate and figure the TOTAL m/a because that's where it's really at. each roller&wheel combo/equation will have a different result/total.
hey "teach"....here's where I am: a 26" wheel travels 6.75 feet per revolution. a 20" wheel travels 5.16 feet per revo. doesn't sound like much, but in a mile the 20 inch turns 241 more times than the 26. (btw, you should see my notes. I'm 49 years old and finally found a use for pi. hahaha)
okay, you were talking reduction....going back to the drive roller, a one inch drive roller on a 26 inch wheel would be a reduction of 26:1 right?
same drive roller on a 20" ----> 20:1 so.... more reduction means more power....giving the 26" wheel more m/a of power. (& less wear & tear on the axle)
am I getting it? (please tell me I'm getting it....lol)
on another note....top speed between the 2 wheels....almost the same. how'd I get that...you ask? I started with 8000 rpms....reduced that by 26:1 and 20:1 getting 307 and 400. multiplying 6.75 by 307 (for the 26") and 5.16 by 400 (20") I believe the answer is feet per minute....and there was only 8 feet difference between the two. with the 26 inch coming out on top.
well thats all I got right now, teach.
Dave

:unsure: one more thing....with the larger wheel having the advantage....is that the same as a longer wrench allowing you to have more torque than a shorter wrench?
YES! because, buddy, it's really all about LEVERS! the distance from the centerline to outer diameter. bolt one end of a 2' rod to an "axle"...you hold the lever at the end, i hold the lever further down, we pull against each other: you'd have m/a-p, i'd have m/a-s.

chain drive...the point where the chain meets the sprocket is your leverage on the axle. roller drive...the point where the roller contacts the tire is your leverage on the axle. each consequent point of transfer can be viewed as two levers working against each other. i'd bet ya can now look at any drive-train from engine-crank to back wheel-on-pavement and see things in a new light.

back to the friction drive. you're correct, speed won't be affected by the tire size, since we're rolling along the outer circumference. but, m/a-p will be affected...using a larger tire, you've made the lever longer, better low-end torque.

haha, i was trying to use pi in my mind when i threw those figures at ya, it was an amusing process, to say the least :lol:

EDIT: i never really tried to teach this stuff to someone before, i missed a great lesson in roller-drive. now i've had a few beers & i remembered what i forgot...roller drive: there ARE two points of transfer on the rear tire, roller-to-tire & tire-to-pavement. right? BUT, since both points are at the same "lever-distance" away from the axle, these two points make for a ratio of 1:1, thereby negating the tire as a factor...ergo, roller-circumference inches = pavement inches.
Augie.... thanks for helping me understand all of this. at first I just wanted a quick answer, but I'm really glad you helped me think this through & arrive at my own conclusion. (could this be chapter 1 of "gear ratios for dummies" ? hahaha)
I'm off to the garage to snap a couple pics of my friction driven 20" bike to post for giggles! then I'm gonna take it apart & mount it on a grown up bike. Thanks again for all the info & your time . you've been a great teacher!
Dave aka seniordavo = srdavo

wanna post this on the forum? maybe in the DIY section? the gearheads who already understand this oughta get a kick out of it....& those who don't ....need to know. later, bro
:lol: please refrain from making fun of our math skills, or lack of...if you just can't help yourself, try to go easy on a coupla old-timers with limited available RAM :lol:
 


P

psuggmog

Guest
#2
I just had to try out the quote function Great flow of information here. There are a few more factors at work here. I'm still lookin' for that free lunch but it ain't happening. The thing about levers is that there is a trade off. The longer the lever, the less force required at the end of the lever to move said load, but the tradeoff is more motion is required. Also the engines have a torque curve where the engine develops maximun horsepower and torque at certain revolutions per minute. It takes lots of power to get something at rest moving and the less to kept it moving. The gearing makes this possible if it falls within the operating range of the engine. In the case of friction drive vs chain drive the coefficient of friction between the driveroller and the tire plays a big part(ever tried to use one in rain ar sleet?)
augidog said:
srdavo was looking for some help "visualizing" gearing and ratios as pertains to different drive-types & applications...we did our best, srdavo suggested i post it "as is" for educational & entertainment purposes. if ya can dig thru the rambling, there does seem to be a bit of good info in here.

also, the following is a great example of using "quote" instead of "reply" when using the private messaging system 8)

srdavo said:
augidog said:
srdavo said:
augidog said:
srdavo said:
augidog said:
srdavo said:
augidog said:
srdavo said:
morning bro......
gonna ramble at ya for a bit......okay

on a chain drive motor bike...the smaller the rear sprocket, the faster ya go. I get this part. on my friction drive, the tire IS the rear sprocket....so... the smaller the tire, the faster ya go. am I getting warm?

since I put the friction drive on a 20 inch bike(for now), It's gonna be way faster than a 26 inch. (the dude I bought it from said it would run 22mph on his 26inch bike)
this would explain why I have to pedal so much to get it up to speed....
am I getting warmer? :?

your humble student,
Dave
first, use "quote" instead of "reply" so we can view this whole topic.

there are two kinds of "mechanical advantage"...m/a of speed & m/a of power.

a lower total ratio (1st gear) means m/a of power, a higher total ratio (5th gear) means m/a of speed.

regardless of chain and sprocket, let's look at only tire size: turn a 20" wheel one time fully around...how far did you move along the ground? turn a 26" tire one rotation, you've moved about 20% more distance, yes? turn both sizes at the same speed, the larger outside circumference will cover more ground in the same time.

now, the sprocket diameter versus the wheel diameter means a new transfer point of power and another ratio factor. one rotation of the sprocket means one rotation of the wheel.

same motor, same rear sprocket: larger wheel means m/a of speed, smaller wheel means m/a of power.

we ok so far?

remember to quote instead of reply, then type your response under the quote.
yup...I'm with ya.....
m/a of speed = makes ya go faster
m/a of power = gives ya better take off/hill climbing

what's next?
actually, not much to cover from this point...let's work our way backwards up the drive-train. large driven (rear) sprocket versus very small drive sprocket...the high rpm's of the clutch shaft are drastically changed into a huge m/a of power by that "negative" ratio (10 th to 44 th). raise that ratio a bit (10th to 36th) to get more speed at a cost to power.

yes?
I'm still with ya....
now, on friction drive, to gain m/a of speed, use a larger drive roller. (make ya go faster) :lol: would a smaller diameter wheel do the same? how does wheel size figure in? am I on the right track here?
with as roller drive, the roller has the same effect as if it was on the pavement, bigger roller more speed, no matter tire size, but bigger wheel means less revs on your hub, longer life.

no bicycle engine in the world would drive a bicycle without TOTAL m/a of power, that's why our drive-train starts out at 10000 rpm's for the crankshaft and is gradually reduced thru gearing to 2500rpms for the roller>250rpm's for the wheel itself.

approx m/a-p of engine to pavement= 40:1

that's why that dinky lil piston can even start to move 250lbs

*those are only rough numbers taken out of my skull, prolly not even close, but u get the idea.

am i rambling too much, am i staying on track enuff for ya? what i'm trying to do is get you to where you can formulate and figure the TOTAL m/a because that's where it's really at. each roller&wheel combo/equation will have a different result/total.
hey "teach"....here's where I am: a 26" wheel travels 6.75 feet per revolution. a 20" wheel travels 5.16 feet per revo. doesn't sound like much, but in a mile the 20 inch turns 241 more times than the 26. (btw, you should see my notes. I'm 49 years old and finally found a use for pi. hahaha)
okay, you were talking reduction....going back to the drive roller, a one inch drive roller on a 26 inch wheel would be a reduction of 26:1 right?
same drive roller on a 20" ----> 20:1 so.... more reduction means more power....giving the 26" wheel more m/a of power. (& less wear & tear on the axle)
am I getting it? (please tell me I'm getting it....lol)
on another note....top speed between the 2 wheels....almost the same. how'd I get that...you ask? I started with 8000 rpms....reduced that by 26:1 and 20:1 getting 307 and 400. multiplying 6.75 by 307 (for the 26") and 5.16 by 400 (20") I believe the answer is feet per minute....and there was only 8 feet difference between the two. with the 26 inch coming out on top.
well thats all I got right now, teach.
Dave

:unsure: one more thing....with the larger wheel having the advantage....is that the same as a longer wrench allowing you to have more torque than a shorter wrench?
YES! because, buddy, it's really all about LEVERS! the distance from the centerline to outer diameter. bolt one end of a 2' rod to an "axle"...you hold the lever at the end, i hold the lever further down, we pull against each other: you'd have m/a-p, i'd have m/a-s.

chain drive...the point where the chain meets the sprocket is your leverage on the axle. roller drive...the point where the roller contacts the tire is your leverage on the axle. each consequent point of transfer can be viewed as two levers working against each other. i'd bet ya can now look at any drive-train from engine-crank to back wheel-on-pavement and see things in a new light.

back to the friction drive. you're correct, speed won't be affected by the tire size, since we're rolling along the outer circumference. but, m/a-p will be affected...using a larger tire, you've made the lever longer, better low-end torque.

haha, i was trying to use pi in my mind when i threw those figures at ya, it was an amusing process, to say the least :lol:

EDIT: i never really tried to teach this stuff to someone before, i missed a great lesson in roller-drive. now i've had a few beers & i remembered what i forgot...roller drive: there ARE two points of transfer on the rear tire, roller-to-tire & tire-to-pavement. right? BUT, since both points are at the same "lever-distance" away from the axle, these two points make for a ratio of 1:1, thereby negating the tire as a factor...ergo, roller-circumference inches = pavement inches.
Augie.... thanks for helping me understand all of this. at first I just wanted a quick answer, but I'm really glad you helped me think this through & arrive at my own conclusion. (could this be chapter 1 of "gear ratios for dummies" ? hahaha)
I'm off to the garage to snap a couple pics of my friction driven 20" bike to post for giggles! then I'm gonna take it apart & mount it on a grown up bike. Thanks again for all the info & your time . you've been a great teacher!
Dave aka seniordavo = srdavo

wanna post this on the forum? maybe in the DIY section? the gearheads who already understand this oughta get a kick out of it....& those who don't ....need to know. later, bro
:lol: please refrain from making fun of our math skills, or lack of...if you just can't help yourself, try to go easy on a coupla old-timers with limited available RAM :lol:
 
5

5-7HEAVEN

Guest
#6
:cool:Thanks for the math lesson, guys. it will definitely come in handy in choosing the correct gearing.

Myron
 

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