Front Tire for the drag bike - Stock 85 Rim

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I have it narrowed down to the following tires that all weigh around 10lbs and are all V-rated tires.

Michelin M50E $100
Pirelli Sport Demon $100
Dunlop GT501 - $115
Bridgeston Battlax BT45 - $95

Maybe I should set up a poll and let everyone here decide.

Thanks,

Jeff
 
to be honest Jeff, I think you should bin the front wheel and use the one from yer mountain bike - thas wot most drag racers do don't they? ;-)
 
I had the GT501's on my Nighthawk S and they are nice too! They are a new design too.

Chris
 
Please tell me this will fit..... cause I just purghased it. They go for around $200 new, so I don't think I did too bad, but we'll see:

http://cgi.ebay.com/ebaymotors/ws/eBayISAPI.dll?ViewItem&item=320329872773&sspagename=STRK%3AMEWNX%3AIT&viewitem=

Here are the specs.... is our rim width 2.5"? I guess I should have done a little more research first.

Tire Size:23.50 x 4.50-18
Rim Diameter (in):18 in.
Sidewall Style:BlackwallTire
Construction:Bias-ply
Directional:No
Tire Diameter (in):23.30 in.
Section Width (in):3.30 in.
Tread Width (in):4.30 in.
Minimum Recommended Rim Width (in):2.50 in.
Maximum Recommended Rim Width (in):2.50 in.
Asymmetrical Tread Pattern:No
DOT-Approved:No
Tube Required:Yes
Tire Compound:Hard
Quantity:Sold individually.
 
Jeff,Most drag bikes use a very small and light wieght front end and wheel off of a small bike.





I was wondering what you all thought about the best front tire for my drag bike project. I think for now I am going to use the stock 85 front wheel/rim and was planning on running a 100/90-18 tire. I think that equates to a 3.25/4.00-18 tire (or tyre depending where you live).

Here are some options I found on eBay:
KENDA Challenger Tire K-657F 100-90-18 = $55
DUNLOP D404 SIZE 100/90-18 FRONT TIRE = $73
Metzeler Marathon ME880 100/90/18 H = $130
Bridgestone Battlax BT-45V 100/90 V-18 = $93

I'm sure there are more (cheng shin), just wanted to see what you all thought and if you have any imput. Of course I want something lightweight, but I also don't want it flying apart on me.
Thanks,
Jeff
 
Jeff,Most drag bikes use a very small and light wieght front end and wheel off of a small bike.

Xcatly wot I said about 20 posts ago - use the wheel from yer mountain bike!! :rofl_200::biglaugh::rofl_200::biglaugh::rofl_200:

Once U have that blower going the front's never gonna touch the ground anyway..:clapping::clapping::clapping::clapping:
 
Depends on what calls your building your bike to race...

Bracket Bike, I want a good grip tire on the front.... Ever watch down at the finish line watching bracket racers?? Sand Bagging hitting the brakes at the last minute to prevent a break out?? I have seen 3 people go down due to this,, all were running skinny drag tires...
How much weight are you going to save?? 3-5-7 lbs? For crying out loud your on your Tank, a VMAX so having a meaty front tire on the front helps even with the weight transfer after crossing the finish line..

I ran my OEN GSXR 3.0 in wheel with a Z rated radial and glad a few times bearing hard on the breaks to prevent that .0001 breakout.

Unless your building a top gas + bike I wouldnt put a skinny on a dragbike even more so with a raked front end
 
Lankee, since you like my numbers and calculations so much, I decided to model up a tire in my CAD program and calculated the numbers with 2 different densities. Now I know the rubber densities won't be twice as dense, and the changes are more in the thickness of the tires, but this will at leas show you how the rotational inertia will change with a 9.3 lb tire vs. a 4.6 lb tire. These are just a little lighter than a stock tire and the tire I purchased. The 'X' axis would be the axil.


LIGHT TIRE
VOLUME = 8.4822438e+06 MM^3
SURFACE AREA = 2.7036586e+06 MM^2
DENSITY = 2.5000000e-07 KILOGRAM / MM^3
MASS = 2.1205609e+00 KILOGRAM

CENTER OF GRAVITY with respect to _TIRE_TEST coordinate frame:
X Y Z 0.0000000e+00 0.0000000e+00 0.0000000e+00 MM

INERTIA with respect to _TIRE_TEST coordinate frame: (KILOGRAM * MM^2)
INERTIA TENSOR:
Ixx Ixy Ixz 5.3283652e+05 0.0000000e+00 0.0000000e+00
Iyx Iyy Iyz 0.0000000e+00 2.7832740e+05 0.0000000e+00
Izx Izy Izz 0.0000000e+00 0.0000000e+00 2.7832753e+05

INERTIA at CENTER OF GRAVITY with respect to _TIRE_TEST coordinate frame: (KILOGRAM * MM^2)
INERTIA TENSOR:
Ixx Ixy Ixz 5.3283652e+05 0.0000000e+00 0.0000000e+00
Iyx Iyy Iyz 0.0000000e+00 2.7832740e+05 0.0000000e+00
Izx Izy Izz 0.0000000e+00 0.0000000e+00 2.7832753e+05

PRINCIPAL MOMENTS OF INERTIA: (KILOGRAM * MM^2)
I1 I2 I3 2.7832726e+05 2.7832767e+05 5.3283652e+05



STOCK TIRE
VOLUME = 8.4822438e+06 MM^3
SURFACE AREA = 2.7036586e+06 MM^2
DENSITY = 5.0000000e-07 KILOGRAM / MM^3
MASS = 4.2411219e+00 KILOGRAM

CENTER OF GRAVITY with respect to _TIRE_TEST coordinate frame:
X Y Z 0.0000000e+00 0.0000000e+00 0.0000000e+00 MM

INERTIA with respect to _TIRE_TEST coordinate frame: (KILOGRAM * MM^2)
INERTIA TENSOR:
Ixx Ixy Ixz 1.0656730e+06 0.0000000e+00 0.0000000e+00
Iyx Iyy Iyz 0.0000000e+00 5.5665481e+05 0.0000000e+00
Izx Izy Izz 0.0000000e+00 0.0000000e+00 5.5665506e+05

INERTIA at CENTER OF GRAVITY with respect to _TIRE_TEST coordinate frame: (KILOGRAM * MM^2)
INERTIA TENSOR:
Ixx Ixy Ixz 1.0656730e+06 0.0000000e+00 0.0000000e+00
Iyx Iyy Iyz 0.0000000e+00 5.5665481e+05 0.0000000e+00
Izx Izy Izz 0.0000000e+00 0.0000000e+00 5.5665506e+05

PRINCIPAL MOMENTS OF INERTIA: (KILOGRAM * MM^2)
I1 I2 I3 5.5665452e+05 5.5665535e+05 1.0656730e+06


I'm sure you know all about rotational inertia and some people claim that 1lb of rotational mass ='s 5 to 7 lbs of static weight, so by me saving anywhere from 5 to 7 lbs of rotational mass, that ='s quite a few burritos.

If anyone else wants to explain this more feel free, I am digging through some old info and will try to correlate this back to a HP savings, but I might give up before I do.

Jeff
 
Nope I dont have a clue to what you just typed Jeff.. But I know that the back tire spins forward but if you back up then the front tire will spin backwards..
Put a good front tire on and save your brain to focus on riding



Lankee, since you like my numbers and calculations so much, I decided to model up a tire in my CAD program and calculated the numbers with 2 different densities. Now I know the rubber densities won't be twice as dense, and the changes are more in the thickness of the tires, but this will at leas show you how the rotational inertia will change with a 9.3 lb tire vs. a 4.6 lb tire. These are just a little lighter than a stock tire and the tire I purchased. The 'X' axis would be the axil.


LIGHT TIRE
VOLUME = 8.4822438e+06 MM^3
SURFACE AREA = 2.7036586e+06 MM^2
DENSITY = 2.5000000e-07 KILOGRAM / MM^3
MASS = 2.1205609e+00 KILOGRAM
CENTER OF GRAVITY with respect to _TIRE_TEST coordinate frame:
X Y Z 0.0000000e+00 0.0000000e+00 0.0000000e+00 MM
INERTIA with respect to _TIRE_TEST coordinate frame: (KILOGRAM * MM^2)
INERTIA TENSOR:
Ixx Ixy Ixz 5.3283652e+05 0.0000000e+00 0.0000000e+00
Iyx Iyy Iyz 0.0000000e+00 2.7832740e+05 0.0000000e+00
Izx Izy Izz 0.0000000e+00 0.0000000e+00 2.7832753e+05
INERTIA at CENTER OF GRAVITY with respect to _TIRE_TEST coordinate frame: (KILOGRAM * MM^2)
INERTIA TENSOR:
Ixx Ixy Ixz 5.3283652e+05 0.0000000e+00 0.0000000e+00
Iyx Iyy Iyz 0.0000000e+00 2.7832740e+05 0.0000000e+00
Izx Izy Izz 0.0000000e+00 0.0000000e+00 2.7832753e+05
PRINCIPAL MOMENTS OF INERTIA: (KILOGRAM * MM^2)
I1 I2 I3 2.7832726e+05 2.7832767e+05 5.3283652e+05


STOCK TIRE
VOLUME = 8.4822438e+06 MM^3
SURFACE AREA = 2.7036586e+06 MM^2
DENSITY = 5.0000000e-07 KILOGRAM / MM^3
MASS = 4.2411219e+00 KILOGRAM
CENTER OF GRAVITY with respect to _TIRE_TEST coordinate frame:
X Y Z 0.0000000e+00 0.0000000e+00 0.0000000e+00 MM
INERTIA with respect to _TIRE_TEST coordinate frame: (KILOGRAM * MM^2)
INERTIA TENSOR:
Ixx Ixy Ixz 1.0656730e+06 0.0000000e+00 0.0000000e+00
Iyx Iyy Iyz 0.0000000e+00 5.5665481e+05 0.0000000e+00
Izx Izy Izz 0.0000000e+00 0.0000000e+00 5.5665506e+05
INERTIA at CENTER OF GRAVITY with respect to _TIRE_TEST coordinate frame: (KILOGRAM * MM^2)
INERTIA TENSOR:
Ixx Ixy Ixz 1.0656730e+06 0.0000000e+00 0.0000000e+00
Iyx Iyy Iyz 0.0000000e+00 5.5665481e+05 0.0000000e+00
Izx Izy Izz 0.0000000e+00 0.0000000e+00 5.5665506e+05
PRINCIPAL MOMENTS OF INERTIA: (KILOGRAM * MM^2)
I1 I2 I3 5.5665452e+05 5.5665535e+05 1.0656730e+06

I'm sure you know all about rotational inertia and some people claim that 1lb of rotational mass ='s 5 to 7 lbs of static weight, so by me saving anywhere from 5 to 7 lbs of rotational mass, that ='s quite a few burritos.

If anyone else wants to explain this more feel free, I am digging through some old info and will try to correlate this back to a HP savings, but I might give up before I do.

Jeff
 
Lankee, since you like my numbers and calculations so much, I decided to model up a tire in my CAD program and calculated the numbers with 2 different densities. Now I know the rubber densities won't be twice as dense, and the changes are more in the thickness of the tires, but this will at leas show you how the rotational inertia will change with a 9.3 lb tire vs. a 4.6 lb tire. These are just a little lighter than a stock tire and the tire I purchased. The 'X' axis would be the axil.


LIGHT TIRE
VOLUME = 8.4822438e+06 MM^3
SURFACE AREA = 2.7036586e+06 MM^2
DENSITY = 2.5000000e-07 KILOGRAM / MM^3
MASS = 2.1205609e+00 KILOGRAM

CENTER OF GRAVITY with respect to _TIRE_TEST coordinate frame:
X Y Z 0.0000000e+00 0.0000000e+00 0.0000000e+00 MM

INERTIA with respect to _TIRE_TEST coordinate frame: (KILOGRAM * MM^2)
INERTIA TENSOR:
Ixx Ixy Ixz 5.3283652e+05 0.0000000e+00 0.0000000e+00
Iyx Iyy Iyz 0.0000000e+00 2.7832740e+05 0.0000000e+00
Izx Izy Izz 0.0000000e+00 0.0000000e+00 2.7832753e+05

INERTIA at CENTER OF GRAVITY with respect to _TIRE_TEST coordinate frame: (KILOGRAM * MM^2)
INERTIA TENSOR:
Ixx Ixy Ixz 5.3283652e+05 0.0000000e+00 0.0000000e+00
Iyx Iyy Iyz 0.0000000e+00 2.7832740e+05 0.0000000e+00
Izx Izy Izz 0.0000000e+00 0.0000000e+00 2.7832753e+05

PRINCIPAL MOMENTS OF INERTIA: (KILOGRAM * MM^2)
I1 I2 I3 2.7832726e+05 2.7832767e+05 5.3283652e+05



STOCK TIRE
VOLUME = 8.4822438e+06 MM^3
SURFACE AREA = 2.7036586e+06 MM^2
DENSITY = 5.0000000e-07 KILOGRAM / MM^3
MASS = 4.2411219e+00 KILOGRAM

CENTER OF GRAVITY with respect to _TIRE_TEST coordinate frame:
X Y Z 0.0000000e+00 0.0000000e+00 0.0000000e+00 MM

INERTIA with respect to _TIRE_TEST coordinate frame: (KILOGRAM * MM^2)
INERTIA TENSOR:
Ixx Ixy Ixz 1.0656730e+06 0.0000000e+00 0.0000000e+00
Iyx Iyy Iyz 0.0000000e+00 5.5665481e+05 0.0000000e+00
Izx Izy Izz 0.0000000e+00 0.0000000e+00 5.5665506e+05

INERTIA at CENTER OF GRAVITY with respect to _TIRE_TEST coordinate frame: (KILOGRAM * MM^2)
INERTIA TENSOR:
Ixx Ixy Ixz 1.0656730e+06 0.0000000e+00 0.0000000e+00
Iyx Iyy Iyz 0.0000000e+00 5.5665481e+05 0.0000000e+00
Izx Izy Izz 0.0000000e+00 0.0000000e+00 5.5665506e+05

PRINCIPAL MOMENTS OF INERTIA: (KILOGRAM * MM^2)
I1 I2 I3 5.5665452e+05 5.5665535e+05 1.0656730e+06


I'm sure you know all about rotational inertia and some people claim that 1lb of rotational mass ='s 5 to 7 lbs of static weight, so by me saving anywhere from 5 to 7 lbs of rotational mass, that ='s quite a few burritos.

If anyone else wants to explain this more feel free, I am digging through some old info and will try to correlate this back to a HP savings, but I might give up before I do.

Jeff

:rofl_200:Burritos...The rest of the math..lost
 
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