How different it is to drive with same gear ratio with 26“ and 28â€
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Do you know how much of a difference it is to drive with specific gear ratio (example 48/16) with 28" wheels and 26" wheels?
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Ari
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Do you know how much of a difference it is to drive with specific gear ratio (example 48/16) with 28" wheels and 26" wheels?
single-speed
asked 3 hours ago
Ari
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Do the math. Circumference = 2 pi r.
– Daniel R Hicks
2 hours ago
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up vote
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down vote
favorite
Do you know how much of a difference it is to drive with specific gear ratio (example 48/16) with 28" wheels and 26" wheels?
single-speed
asked 3 hours ago
Ari
61
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Ari is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Do you know how much of a difference it is to drive with specific gear ratio (example 48/16) with 28" wheels and 26" wheels?
single-speed
single-speed
asked 3 hours ago
Ari
61
New contributor
Ari is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 3 hours ago
Ari
61
New contributor
Ari is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 3 hours ago
Ari
61
New contributor
Ari is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 3 hours ago
Ari
61
asked 3 hours ago
Ari
61
61
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Ari is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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New contributor
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Check out our Code of Conduct.
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Do the math. Circumference = 2 pi r.
– Daniel R Hicks
2 hours ago
add a comment |Â
Do the math. Circumference = 2 pi r.
– Daniel R Hicks
2 hours ago
Do the math. Circumference = 2 pi r.
– Daniel R Hicks
2 hours ago
Do the math. Circumference = 2 pi r.
– Daniel R Hicks
2 hours ago
add a comment |Â
2 Answers
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up vote
2
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Yes. You can do the math on this: (chainring ÷ sprocket) × wheel size. With a wheel that has an outer diameter at the tire of 26", you'd get 78 gear inches. With a 28" wheel, it would be 84 gear inches, or about 7% higher.
On a multi-gear setup, I've found that a 7% step is just about the smallest step that's really noticeable.
answered 2 hours ago
Adam Rice
4,7881330
add a comment |Â
up vote
2
down vote
A larger wheel (including the tire) just gives a higher effective gear ratio.vBicycle gear ratios are often specified in gear inches, which takes the diameter of the driving wheel into account.
Gear inches = diameter drive wheel (in inches) × size front sprocket / size rear sprocket
[Source: https://www.sheldonbrown.com/gain.html]. Use of inches is traditional, you could of course use metric units.
Obviously, you can calculate the sprocket ratios which give comparable gear inch values for each wheel size.
Also, you can directly calculate the difference in effective gearing by simply dividing one wheel diameter by another. Assuming wheel with tires have an actual diameter of 28" and 26", a 28" wheel gives 28/26 = 1.07 i.e. 7% higher gear ratio over a 26" wheel.
Choosing 26 and 28" wheel diameters for comparison is a little strange. A 559mm rim MTB wheel with a 2" tire would be about 26" in diameter. A 622mm MTB wheel with a 2.25" tire would be about 29".
Assuming you would use similar tire sizes on each size wheel, you can simply divide rim diameters. 622mm / 559mm = 1.11, i.e. the larger wheel gives about a 11% higher effective gear ratio.
answered 2 hours ago
Argenti Apparatus
29.6k23377
Not sure if this is region-specific, but in German discussions I’m often see the term „Entfaltung“ (“developmentâ€Â) used. It describes how far you move with each turn of the cranks, which is somewhat more meaningful than “gear inches†which use wheel diameter. Development = wheel circumference × front sprocket/rear sprocket. Of course for a complete picture you also have to take crankarm length into consideration.
– Michael
1 hour ago
@Michael, I'm not sure how meaningful how far I travel for one crank revolution is to me. Wheel diameter is easier to find than circumference (i.e it saves multiplying by À).
– Argenti Apparatus
1 hour ago
@Michael: Crankarm length plays only a part in leverage not in development. It doesn't intervene in the calculation. It's big ring/sprocket multiplied by wheel circumference. Where would you include the crankarm in the calculation?
– Carel
12 mins ago
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
Yes. You can do the math on this: (chainring ÷ sprocket) × wheel size. With a wheel that has an outer diameter at the tire of 26", you'd get 78 gear inches. With a 28" wheel, it would be 84 gear inches, or about 7% higher.
On a multi-gear setup, I've found that a 7% step is just about the smallest step that's really noticeable.
answered 2 hours ago
Adam Rice
4,7881330
add a comment |Â
up vote
2
down vote
Yes. You can do the math on this: (chainring ÷ sprocket) × wheel size. With a wheel that has an outer diameter at the tire of 26", you'd get 78 gear inches. With a 28" wheel, it would be 84 gear inches, or about 7% higher.
On a multi-gear setup, I've found that a 7% step is just about the smallest step that's really noticeable.
answered 2 hours ago
Adam Rice
4,7881330
add a comment |Â
up vote
2
down vote
up vote
2
down vote
Yes. You can do the math on this: (chainring ÷ sprocket) × wheel size. With a wheel that has an outer diameter at the tire of 26", you'd get 78 gear inches. With a 28" wheel, it would be 84 gear inches, or about 7% higher.
On a multi-gear setup, I've found that a 7% step is just about the smallest step that's really noticeable.
answered 2 hours ago
Adam Rice
4,7881330
Yes. You can do the math on this: (chainring ÷ sprocket) × wheel size. With a wheel that has an outer diameter at the tire of 26", you'd get 78 gear inches. With a 28" wheel, it would be 84 gear inches, or about 7% higher.
On a multi-gear setup, I've found that a 7% step is just about the smallest step that's really noticeable.
answered 2 hours ago
Adam Rice
4,7881330
answered 2 hours ago
Adam Rice
4,7881330
answered 2 hours ago
Adam Rice
4,7881330
answered 2 hours ago
Adam Rice
4,7881330
4,7881330
add a comment |Â
add a comment |Â
up vote
2
down vote
A larger wheel (including the tire) just gives a higher effective gear ratio.vBicycle gear ratios are often specified in gear inches, which takes the diameter of the driving wheel into account.
Gear inches = diameter drive wheel (in inches) × size front sprocket / size rear sprocket
[Source: https://www.sheldonbrown.com/gain.html]. Use of inches is traditional, you could of course use metric units.
Obviously, you can calculate the sprocket ratios which give comparable gear inch values for each wheel size.
Also, you can directly calculate the difference in effective gearing by simply dividing one wheel diameter by another. Assuming wheel with tires have an actual diameter of 28" and 26", a 28" wheel gives 28/26 = 1.07 i.e. 7% higher gear ratio over a 26" wheel.
Choosing 26 and 28" wheel diameters for comparison is a little strange. A 559mm rim MTB wheel with a 2" tire would be about 26" in diameter. A 622mm MTB wheel with a 2.25" tire would be about 29".
Assuming you would use similar tire sizes on each size wheel, you can simply divide rim diameters. 622mm / 559mm = 1.11, i.e. the larger wheel gives about a 11% higher effective gear ratio.
answered 2 hours ago
Argenti Apparatus
29.6k23377
Not sure if this is region-specific, but in German discussions I’m often see the term „Entfaltung“ (“developmentâ€Â) used. It describes how far you move with each turn of the cranks, which is somewhat more meaningful than “gear inches†which use wheel diameter. Development = wheel circumference × front sprocket/rear sprocket. Of course for a complete picture you also have to take crankarm length into consideration.
– Michael
1 hour ago
@Michael, I'm not sure how meaningful how far I travel for one crank revolution is to me. Wheel diameter is easier to find than circumference (i.e it saves multiplying by À).
– Argenti Apparatus
1 hour ago
@Michael: Crankarm length plays only a part in leverage not in development. It doesn't intervene in the calculation. It's big ring/sprocket multiplied by wheel circumference. Where would you include the crankarm in the calculation?
– Carel
12 mins ago
add a comment |Â
up vote
2
down vote
A larger wheel (including the tire) just gives a higher effective gear ratio.vBicycle gear ratios are often specified in gear inches, which takes the diameter of the driving wheel into account.
Gear inches = diameter drive wheel (in inches) × size front sprocket / size rear sprocket
[Source: https://www.sheldonbrown.com/gain.html]. Use of inches is traditional, you could of course use metric units.
Obviously, you can calculate the sprocket ratios which give comparable gear inch values for each wheel size.
Also, you can directly calculate the difference in effective gearing by simply dividing one wheel diameter by another. Assuming wheel with tires have an actual diameter of 28" and 26", a 28" wheel gives 28/26 = 1.07 i.e. 7% higher gear ratio over a 26" wheel.
Choosing 26 and 28" wheel diameters for comparison is a little strange. A 559mm rim MTB wheel with a 2" tire would be about 26" in diameter. A 622mm MTB wheel with a 2.25" tire would be about 29".
Assuming you would use similar tire sizes on each size wheel, you can simply divide rim diameters. 622mm / 559mm = 1.11, i.e. the larger wheel gives about a 11% higher effective gear ratio.
answered 2 hours ago
Argenti Apparatus
29.6k23377
Not sure if this is region-specific, but in German discussions I’m often see the term „Entfaltung“ (“developmentâ€Â) used. It describes how far you move with each turn of the cranks, which is somewhat more meaningful than “gear inches†which use wheel diameter. Development = wheel circumference × front sprocket/rear sprocket. Of course for a complete picture you also have to take crankarm length into consideration.
– Michael
1 hour ago
@Michael, I'm not sure how meaningful how far I travel for one crank revolution is to me. Wheel diameter is easier to find than circumference (i.e it saves multiplying by À).
– Argenti Apparatus
1 hour ago
@Michael: Crankarm length plays only a part in leverage not in development. It doesn't intervene in the calculation. It's big ring/sprocket multiplied by wheel circumference. Where would you include the crankarm in the calculation?
– Carel
12 mins ago
add a comment |Â
up vote
2
down vote
up vote
2
down vote
A larger wheel (including the tire) just gives a higher effective gear ratio.vBicycle gear ratios are often specified in gear inches, which takes the diameter of the driving wheel into account.
Gear inches = diameter drive wheel (in inches) × size front sprocket / size rear sprocket
[Source: https://www.sheldonbrown.com/gain.html]. Use of inches is traditional, you could of course use metric units.
Obviously, you can calculate the sprocket ratios which give comparable gear inch values for each wheel size.
Also, you can directly calculate the difference in effective gearing by simply dividing one wheel diameter by another. Assuming wheel with tires have an actual diameter of 28" and 26", a 28" wheel gives 28/26 = 1.07 i.e. 7% higher gear ratio over a 26" wheel.
Choosing 26 and 28" wheel diameters for comparison is a little strange. A 559mm rim MTB wheel with a 2" tire would be about 26" in diameter. A 622mm MTB wheel with a 2.25" tire would be about 29".
Assuming you would use similar tire sizes on each size wheel, you can simply divide rim diameters. 622mm / 559mm = 1.11, i.e. the larger wheel gives about a 11% higher effective gear ratio.
answered 2 hours ago
Argenti Apparatus
29.6k23377
A larger wheel (including the tire) just gives a higher effective gear ratio.vBicycle gear ratios are often specified in gear inches, which takes the diameter of the driving wheel into account.
Gear inches = diameter drive wheel (in inches) × size front sprocket / size rear sprocket
[Source: https://www.sheldonbrown.com/gain.html]. Use of inches is traditional, you could of course use metric units.
Obviously, you can calculate the sprocket ratios which give comparable gear inch values for each wheel size.
Also, you can directly calculate the difference in effective gearing by simply dividing one wheel diameter by another. Assuming wheel with tires have an actual diameter of 28" and 26", a 28" wheel gives 28/26 = 1.07 i.e. 7% higher gear ratio over a 26" wheel.
Choosing 26 and 28" wheel diameters for comparison is a little strange. A 559mm rim MTB wheel with a 2" tire would be about 26" in diameter. A 622mm MTB wheel with a 2.25" tire would be about 29".
Assuming you would use similar tire sizes on each size wheel, you can simply divide rim diameters. 622mm / 559mm = 1.11, i.e. the larger wheel gives about a 11% higher effective gear ratio.
answered 2 hours ago
Argenti Apparatus
29.6k23377
edited 1 hour ago
answered 2 hours ago
Argenti Apparatus
29.6k23377
answered 2 hours ago
Argenti Apparatus
29.6k23377
answered 2 hours ago
Argenti Apparatus
29.6k23377
29.6k23377
Not sure if this is region-specific, but in German discussions I’m often see the term „Entfaltung“ (“developmentâ€Â) used. It describes how far you move with each turn of the cranks, which is somewhat more meaningful than “gear inches†which use wheel diameter. Development = wheel circumference × front sprocket/rear sprocket. Of course for a complete picture you also have to take crankarm length into consideration.
– Michael
1 hour ago
@Michael, I'm not sure how meaningful how far I travel for one crank revolution is to me. Wheel diameter is easier to find than circumference (i.e it saves multiplying by À).
– Argenti Apparatus
1 hour ago
@Michael: Crankarm length plays only a part in leverage not in development. It doesn't intervene in the calculation. It's big ring/sprocket multiplied by wheel circumference. Where would you include the crankarm in the calculation?
– Carel
12 mins ago
add a comment |Â
Not sure if this is region-specific, but in German discussions I’m often see the term „Entfaltung“ (“developmentâ€Â) used. It describes how far you move with each turn of the cranks, which is somewhat more meaningful than “gear inches†which use wheel diameter. Development = wheel circumference × front sprocket/rear sprocket. Of course for a complete picture you also have to take crankarm length into consideration.
– Michael
1 hour ago
@Michael, I'm not sure how meaningful how far I travel for one crank revolution is to me. Wheel diameter is easier to find than circumference (i.e it saves multiplying by À).
– Argenti Apparatus
1 hour ago
@Michael: Crankarm length plays only a part in leverage not in development. It doesn't intervene in the calculation. It's big ring/sprocket multiplied by wheel circumference. Where would you include the crankarm in the calculation?
– Carel
12 mins ago
Not sure if this is region-specific, but in German discussions I’m often see the term „Entfaltung“ (“developmentâ€Â) used. It describes how far you move with each turn of the cranks, which is somewhat more meaningful than “gear inches†which use wheel diameter. Development = wheel circumference × front sprocket/rear sprocket. Of course for a complete picture you also have to take crankarm length into consideration.
– Michael
1 hour ago
Not sure if this is region-specific, but in German discussions I’m often see the term „Entfaltung“ (“developmentâ€Â) used. It describes how far you move with each turn of the cranks, which is somewhat more meaningful than “gear inches†which use wheel diameter. Development = wheel circumference × front sprocket/rear sprocket. Of course for a complete picture you also have to take crankarm length into consideration.
– Michael
1 hour ago
@Michael, I'm not sure how meaningful how far I travel for one crank revolution is to me. Wheel diameter is easier to find than circumference (i.e it saves multiplying by À).
– Argenti Apparatus
1 hour ago
@Michael, I'm not sure how meaningful how far I travel for one crank revolution is to me. Wheel diameter is easier to find than circumference (i.e it saves multiplying by À).
– Argenti Apparatus
1 hour ago
@Michael: Crankarm length plays only a part in leverage not in development. It doesn't intervene in the calculation. It's big ring/sprocket multiplied by wheel circumference. Where would you include the crankarm in the calculation?
– Carel
12 mins ago
@Michael: Crankarm length plays only a part in leverage not in development. It doesn't intervene in the calculation. It's big ring/sprocket multiplied by wheel circumference. Where would you include the crankarm in the calculation?
– Carel
12 mins ago
add a comment |Â
Ari is a new contributor. Be nice, and check out our Code of Conduct.
Ari is a new contributor. Be nice, and check out our Code of Conduct.
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Do the math. Circumference = 2 pi r.
– Daniel R Hicks
2 hours ago