3

On my bike, on the left shifter, It is labeled low to high and has lots of dots between them.

I don't know how to shift to one individual gear on the front derailleur. It seems that If I shift it once it kind of moves and you have to shift several times to move to another ring.

Can I shift more than one gear at a time, so I can move to another ring faster?

enter image description here

Improve this question
2
  • 2
    This really depends on the shifter and the derailer. With some setups you need to do about 3 clicks per shift, especially for the front, while with others it's one click. In some cases the 3 clicks is intentional, to allow some "tuning" of the front derailer based on the rear derailer's setting.
    – Daniel R Hicks
    Commented Mar 24, 2020 at 2:48
  • 4
    They're called Revoshifts or Grip Shifters... and they're not exactly precision components. At that price point there's little chance of accuracy, so your left shifter (front mech) is really just a big friction shifter.
    – Criggie ♦
    Commented Mar 24, 2020 at 4:05

1 Answer 1

Reset to default
5

As comments to the question point out, your front shifter is a grip shift. It operates as a friction shifter, whether intentionally or not. This means its operation is more "analog" than "digital", with no one-to-one correspondence between grip's positions and front chainrings. Rather, there is a spectrum of derailleur states, some of which do cause the chain to change its position, while the remaining intermediate positions only cause the chain to rub against one ring or another.

You are certainly allowed/supposed to rotate the shifter grip faster to its final position. Make sure you find positions where the chain does not rub against anything. As with any front derailleur, you are recommended to somewhat reduce pedaling pressure while making a front shift, to prevent excessive loads on the chain while it is skewed between the chainrings.

Improve this answer

Not the answer you're looking for? Browse other questions tagged or ask your own question.