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Discussion Starter #1
What is the weight difference between titanium and stainless headers? Is it really worth the extra money to get the titanium systems over the s/s? I know its a lot lighter weight but its also a nice penny more so i wasnt sure if it is worth it
 

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Let me ask you another question, will 5 or even 10 pounds be noticeable to you? I really doubt it.

I don't know how much savings it will be, but I would just get stainless if I were you. It also may depend on what brand for how much less it will weigh.
 

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No way dude hes probably a 10 top AMA racer and it would make a HUGE difference pay the extra money!!
 

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^^^ exactly. unless you were nicky hayden, i doubt you would notice the effects of 5-10 pound loss. do you notice the weight difference between a full tank of gass and an empty one??? now imgine that weight down low, im going to say that a full Ti system is worth its weight in feathers, unless your sponsored by a company that will give you one for free.
 

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If your racing and 1/10's of seconds are important it may matter to you. For the street it's a waste of money performance wise.
 

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Everything else equal you can get a better performance gain (weight loss) by taking a dump before you ride.

It's a noticable difference in your hand when you hold one vs. the other, but not terribly noticable (if at all) on the bike.

R
 

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....IMO you don't need ANY performance modifications or weight savings if you ride on the street...WASTE OF MONEY.

I ride my bike on the track and I the only performance modification I have would have to be my racing tires. Otherwise my bike is pretty much stock, minus the lights, passenger pegs, and mirrors. Now I DO notice the difference but that is only because I'm pushing my bike hard on the track. If you're pushing your bike on the street to the point in which you can actually tell the difference with any weight savings...your endangering yourself and others.
 

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Discussion Starter #8
Everything else equal you can get a better performance gain (weight loss) by taking a dump before you ride.

It's a noticable difference in your hand when you hold one vs. the other, but not terribly noticable (if at all) on the bike.

R
thanks. thats what i figured.. i wasnt sure.
 

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10 lbs is a lot if you are riding on the track. more than an accelration improvement, you feel it when you flick side to side hard! that combined with all the other weight savings from taking lights and other junk off- you might save close to 20 lbs- a big difference.for the street, don't bother unless you want some pose factor.
 

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its easy to calc.... try to find out the TI and than find the density....its online...... ie) .29, take the over shape of the header to be a few cylinders and find the volume.... multiply ur density by the volume and ull get the weight. Its probably 5-7 lbs..... your better off loosing weight yourself than spending a $$$$$$$$
 

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on the track... as light as possible is the way to go... on the street... letting everyone know you have a full ti system is the way to go...
 

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nice one nick>>>>>>

carbon is purdy too
 

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i heard TI isnt much more lightweight than s/s... it will disperse heat better than s/s. i could be wrong tho
 

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Ti-is lighter
 

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I can save you from endless discussion with the power of Newton. No not those yummy fig snacks, I'm talking Mr. Isaac. You know, the apple story. Well you see, it turns out that the acceleration of your bike is the quotient of the net force on your bike over the mass, or for sake of argument, weight of your bike. So if you've all passed highschool math, you will quickly see that by reducing the denominator of that fraction (weight) you will increase the result (acceleration) [Ref: a=F/m]. The question is, by how much???

Well let's transform that 5 or 10 pounds quoted earlier into a relative quantity called percentage (so fun!) Because the snow is melting here I'll be optimistic and say 10 pounds reduction in weight. Assuming you own the 2005 model, your dry weight according to Sport Rider (http://www.sportrider.com/bikes/2005/146_05_honda_cbr600rr_specs/) is 361 pounds. I'll also assume you weight 150lb, making it 511lbs total. So, to transform 511 to 501 what is the percentage we need to apply? Simple, 501/511 (rocket science!) = 0.9804 (98.04%). Now let's carry it over to our original Newtonian equation a=F/m. By applying this percentage, we get a=F/(.9804*m). Then by the power of multiplication, we transfer it over to the acceleration (hold on to your hats!) (1/.9804)*a=F/m ---> (1.01996)*a=F/m. Wow!!!! an increase of 1.996% (or 2%) in acceleration! worth every penny!!!!! Isn't science so cool!

Here is a picture of Newton, to which I pleasure myself every night.



Sexy man.
 

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I can save you from endless discussion with the power of Newton. No not those yummy fig snacks, I'm talking Mr. Isaac. You know, the apple story. Well you see, it turns out that the acceleration of your bike is the quotient of the net force on your bike over the mass, or for sake of argument, weight of your bike. So if you've all passed highschool math, you will quickly see that by reducing the denominator of that fraction (weight) you will increase the result (acceleration) [Ref: a=F/m]. The question is, by how much???

Well let's transform that 5 or 10 pounds quoted earlier into a relative quantity called percentage (so fun!) Because the snow is melting here I'll be optimistic and say 10 pounds reduction in weight. Assuming you own the 2005 model, your dry weight according to Sport Rider (http://www.sportrider.com/bikes/2005/146_05_honda_cbr600rr_specs/) is 361 pounds. I'll also assume you weight 150lb, making it 511lbs total. So, to transform 511 to 501 what is the percentage we need to apply? Simple, 501/511 (rocket science!) = 0.9804 (98.04%). Now let's carry it over to our original Newtonian equation a=F/m. By applying this percentage, we get a=F/(.9804*m). Then by the power of multiplication, we transfer it over to the acceleration (hold on to your hats!) (1/.9804)*a=F/m ---> (1.01996)*a=F/m. Wow!!!! an increase of 1.996% (or 2%) in acceleration! worth every penny!!!!! Isn't science so cool!

Here is a picture of Newton, to which I pleasure myself every night.



Sexy man.
lol freakin hilarious. NICE ONE!
 

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Dude, you don't multiply by the reciprocal to cancel it out on the right side, you would have .9084^2 * a in the denominator then.

But then it seems to reduce the acceleration. So theoretically, you are correct, but algabreically, I'm not seeing it.
 

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Dude, you don't multiply by the reciprocal to cancel it out on the right side, you would have .9084^2 * a in the denominator then.

But then it seems to reduce the acceleration. So theoretically, you are correct, but algabreically, I'm not seeing it.
dude. you're kidding me.

IF A = B / C
THEN (1/D)*A = (1/D)*(B/C) = B/(C*D)
 
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