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PostPosted: Thu Jun 21, 2012 4:25 am 
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Joined: Mon Oct 15, 2007 9:28 pm
Posts: 1132
I am trying to figure out reach difference between a flat bar with 9 degree bend and flat bar with 11 degree bend. I am not up on my trig.
Bar is 640mm or 320mm per side.


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PostPosted: Thu Jun 21, 2012 7:18 am 
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Joined: Mon Aug 15, 2011 4:48 am
Posts: 419
11 degree has a negitive reach from bar centre of 62mm.

9 degree has a negitive reach from bar centre of 50mm.

So the 9 degree will make your reach 12mm longer, basically

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PostPosted: Thu Jun 21, 2012 7:24 am 
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Joined: Thu Oct 13, 2011 5:25 pm
Posts: 168
Sin(theta)= opposite over hypotnuse. Hypotnuse = 320 mm, assuming the length of the bar is measured according to the bend

so:

320*((sin11deg)-(sin9deg)) will get you pretty close. This is the amount in milimeters the bars will be closer to the saddle in one dimension. This will be a good approximation. To know exactly how much closer to your hands the bars would be, we would need to know the length from the saddle to the center of where the bars are being clamped (to solve in two dimensions),and even more precisely, the drop (to solve in three dimensions).

Using google's calculator
0.0343745303 * 320 = 10.9998497

So, roughly 1 cm closer.

If the length is measured in the bar's x direction (i.e. the 9 and 11 degree bars have the same 'effective' width) , tan(theta)= opposite over adjacent

so: 320*((tan11)-(tan9))

yields
11.518678mm

The difference accounting for more dimensions would be pretty much negligible, so I'd count on 1 cm closer to you.


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PostPosted: Thu Jun 21, 2012 11:46 am 
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Joined: Mon Aug 15, 2011 4:48 am
Posts: 419
yeah your talking 1cm ish, depending on your grips stem, clamp diametre etc. very small differerence

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