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Road Cycling It is by riding a bicycle that you learn the contours of a country best, since you have to sweat up the hills and coast down them. Thus you remember them as they actually are, while in a motor car only a high hill impresses you, and you have no such accurate remembrance of country you have driven through as you gain by riding a bicycle. -- Ernest Hemingway

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Old 01-10-14, 06:13 PM   #1
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Raise saddle after moving it forward?

I keep looking for the answer to this question, but all I can find is the reverse: the need to move a saddle forward to maintain knee position after raising it. That is not what I am asking. I want to know how much do I have to raise my saddle after moving it forward (for whatever reason). I'm thinking the reach to the crank will have been shortened as the saddle moves more directly over it, not much but some. Anyone have a formula for this? I moved my saddle forward 1 cm.

I did a really quick and dirty approximation that suggests the amount the saddle has to be raised x is about y(cos theta) where y is how much I move the saddle forward and theta is the seat tube angle. Assuming 73 deg I get roughly 0.3 of the distance moved forward (or backward and lowering the saddle). So for my 1 cm change, I would need to raise the saddle about 3 mm. Not only is that a very small amount, but my seat post is stuck, so I will likely ignore it for now. At least until I decide if I like the new knee position. I know the saddle height will affect this, but I will work it out as I go.

Nevertheless, I would be interested in hearing what other folks come up with.

Thanks.

Robert
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Old 01-10-14, 06:44 PM   #2
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I do saddle adjustment by feel.
I do use the rail #'s as a reference, but mostly it's a lot of times in and out of the garage with a hex wrench.


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Old 01-10-14, 07:05 PM   #3
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id start with a baseline height using one of the popular formulas. i use the lemond formula (inseam x .883) to determine initial seat height (distance from top of seat to middle of bottom bracket). get that set first then do the fore/aft set up. feel pressure in front of your knees? move the seat back. feel like you are sitting on the narrow end of a 2 x 4? move the seat forward. make sure seat is level always. tah dah....
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Old 01-10-14, 07:05 PM   #4
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I thought the OP was the authority on bike related matters.
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Old 01-10-14, 07:06 PM   #5
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Are we supposed to be applying math to this question? If so, then of course I am of zero assistance.

However, I will rehash my Retul fit experience, wherein my saddle was moved one inch higher and almost an inch forward, resulting in optimized measured power output (still quite low by 41 standards). Therefore, according to my scientific sample size of one...ipso facto the ratio is 1/1....
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Old 01-10-14, 07:07 PM   #6
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I thought the OP was the authority on bike related matters.
He used to be, but IMHO has measurably mellowed like fine wine...
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Old 01-10-14, 07:09 PM   #7
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Make the adjustment then post it in the "Hot r Not" thread. They'll tell you exactly the adjustment you'll need to make.
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Old 01-10-14, 07:09 PM   #8
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Rather than get out my TI-30, I just use a tape measure and measure from the center of the BB spindle to the center of the saddle rail. As I move the saddle forward, I have to raise it up so that it's the same distance.
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Old 01-10-14, 07:10 PM   #9
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I thought the OP was the authority on bike related matters.
Testy, eh?
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Old 01-10-14, 07:11 PM   #10
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Not really. Just was surprised is all.
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Old 01-10-14, 07:17 PM   #11
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Rather than get out my TI-30, I just use a tape measure and measure from the center of the BB spindle to the center of the saddle rail. As I move the saddle forward, I have to raise it up so that it's the same distance.
Sure, the empirical method should have been obvious to me. I was just wondering whether there was an established formula for this that soneone could supply
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Old 01-10-14, 07:22 PM   #12
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"established formula" = x+y-r/2
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Old 01-10-14, 07:26 PM   #13
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"established formula" = x+y-r/2
That's what I'm talkin' about. I can't wait to check this against the measurements. I'm guessing you will be anxious to know the results.
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Old 01-10-14, 07:34 PM   #14
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I thought the OP was the authority on bike related matters.
Actually I admit to entering discussions with inferior knowledge, expressing what I think is correct and then using the rebuttals as a learning tool. Kind of a screwed up Socratic method.
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Old 01-10-14, 07:38 PM   #15
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STA is less relevant, as seatposts have offset. I would look at it as c=√(a+b). C is a constant, your leg extension, which does not need to be solved, as it will be the same before and after the change. If a is vertical and b is horizontal from BB, you now have √(a+b)=√((a+∆v)+(b-∆h)). Square both sides, use foil, and solve for ∆v.
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Old 01-10-14, 07:42 PM   #16
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Actually I admit to entering discussions with inferior knowledge, expressing what I think is correct and then using the rebuttals as a learning tool. Kind of a screwed up Socratic method.
This makes you a 'made member' of the 41.

Congratulations.

Now lets do the ceremonial cleaning of the chain.
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Old 01-10-14, 07:49 PM   #17
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STA is less relevant, as seatposts have offset. I would look at it as c=√(a+b). C is a constant, your leg extension, which does not need to be solved, as it will be the same before and after the change. If a is vertical and b is horizontal from BB, you now have √(a+b)=√((a+∆v)+(b-∆h)). Square both sides, use foil, and solve for ∆v.
That looks completely correct. Is it not true that the trig approach I used inherently incorporates the formula you provided? You're right about the lack of need for using the STA, but I wonder if the use of it, just as an approximation doesn't simplify the whole process.
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Old 01-10-14, 07:54 PM   #18
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If you are going to use angles, I recommend switching to radians. If I get a choice, I love talking about pi rotating around my crank.
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Old 01-10-14, 07:56 PM   #19
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If you are going to use angles, I recommend switching to radians. If I get a choice, I love talking about pi rotating around my crank.
Perhaps pi likes it too.
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Old 01-10-14, 08:02 PM   #20
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Sure, the empirical method should have been obvious to me. I was just wondering whether there was an established formula for this that soneone could supply
I nominate this statement to be the new motto of the Forty-One.
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Old 01-10-14, 08:09 PM   #21
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I nominate this statement to be the new motto of the Forty-One.
I'll second the motion provided you support my campaign to have "Iron Alloy Appreciation Week". I am still working on a pithy slogan like "Steal a moment for Iron Alloy" or "The irony of Iron Alloy" or "Iron Alloy: it is your father's alloy" or "Available for Centuries for centuries".
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Old 01-10-14, 08:13 PM   #22
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That looks completely correct. Is it not true that the trig approach I used inherently incorporates the formula you provided? You're right about the lack of need for using the STA, but I wonder if the use of it, just as an approximation doesn't simplify the whole process.
Yes you could solve with angles. I wouldn't because you don't care what the starting or ending angle are, but you will need to solve them, and use them in a subsequent equation. They are intermediate calculations to move from ∆v to ∆h. If you can make a formula that uses only ∆v &∆h, why not solve direct?
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Old 01-10-14, 08:18 PM   #23
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Wear two three pairs of shorts.
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Old 01-10-14, 08:21 PM   #24
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I'll second the motion provided you support my campaign to have "Iron Alloy Appreciation Week". I am still working on a pithy slogan like "Steal a moment for Iron Alloy" or "The irony of Iron Alloy" or "Iron Alloy: it is your father's alloy" or "Available for Centuries for centuries".
May I suggest, "Iron alloy, a better use for coke."
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Old 01-10-14, 08:23 PM   #25
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Wear two three pairs of shorts.
Now you see? That is truly creative. Especially appropriate if I can't get my seat post unstuck. I love it. We have a winner. No one else need apply.
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