will i need to move the front mech...

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S-Express

Guest
Depends how much clearance there already is.
 

Sharky

Guru
Location
Kent
I knew there was a reason for learning geometry at school now after 48 years I've found it.

Cirmcumference = 2 pie R
so R = Circumference / 2 pie
so 53*0.5 / 2* pie - 52 *0.5 /2* pie = "not a lot"
 
Location
Loch side.
Geometry is very clear in this. It says maybe.
Precise science aside, surely, if you have enough tech savvy to remove and replace a chainring, you have enough skills to loosen a FD and move it up a fraction?
 

fossyant

Ride It Like You Stole It!
Location
South Manchester
You'll likely have some adjustment with a new cranks for shifting but you might just get away with not moving the mech up if you have enough gap.
 

ianrauk

Tattooed Beat Messiah
Location
Rides Ti2
I knew there was a reason for learning geometry at school now after 48 years I've found it.

Cirmcumference = 2 pie R
so R = Circumference / 2 pie
so 53*0.5 / 2* pie - 52 *0.5 /2* pie = "not a lot"
30635546.jpg
 

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Ajax Bay

Guru
Location
East Devon
I knew there was a reason for learning geometry at school now after 48 years I've found it.
Cirmcumference = 2 pie R
so R = Circumference / 2 pie
so 53*0.5 / 2* pie - 52 *0.5 /2* pie = "not a lot"
There was a reason, @Sharky - so why didn't you? Number of teeth is not 'r'.

Radius of a '53' is 108mm. Radius of a 52 is therefore 108 x52 /53 (= ~106mm).
Hence @User 's 2mm recommendation, with the probably / "if the mech is in the right position to start with" caveat.
 

Dogtrousers

Kilometre nibbler
There was a reason, @Sharky - so why didn't you? Number of teeth is not 'r'.

Radius of a '53' is 108mm. Radius of a 52 is therefore 108 x52 /53 (= ~106mm).
Hence @User 's 2mm recommendation, with the probably / "if the mech is in the right position to start with" caveat.
@Sharky 's forumula for the radius of a 53 is pretty good I thought: 53*0.5 / 2* pie gives 4.21" or 107.1 mm. The overall delta formula 53*0.5 / 2* pie - 52 *0.5 /2* pie = "not a lot" gives a delta of 0.08" or 2mm - in line with what @User suggested.
 
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