can you say RICE!!
#1
Guest
Posts: n/a
can you say RICE!!
WTF!!
here are some more. btw he seems to think he has 530hp!!! and can do a 9.7 1/4 mile!!
http://www.cardomain.com/member_page...=Honda%20Civic
here are some more. btw he seems to think he has 530hp!!! and can do a 9.7 1/4 mile!!
http://www.cardomain.com/member_page...=Honda%20Civic
#4
Guest
Posts: n/a
I just looked at his mods and he did do an engine swap and has "turbo, supercharger, extrude-honed, nitrous, port and polish, stroke-increase". So i guess maybe he is that fast. BUT STILL FUGLY!!!
When will people learn that a huge spoiler (on a FWD!) and all the other UGLY cosmetic stuff DOES NOT make your car faster.
When will people learn that a huge spoiler (on a FWD!) and all the other UGLY cosmetic stuff DOES NOT make your car faster.
#11
Guest
Posts: n/a
i found this in the 4th gen forum.
it is the funniest website i have ever seen
http://www.riceboypage.com/
it is the funniest website i have ever seen
http://www.riceboypage.com/
#12
there are TONS of ricers around here, one kid has nasty ford festiva with "custom" painted hub caps, a huge apc banner, led squirters and an amazingly big fart can. Another one has a 90-91 celica with a nonfunctional hood scoop, huge fart can, biggest wing I've ever seen in person, painted stock hubcaps and apc seat& steering wheel covers.
I was talking to this kid yesterday and he's like "yeh dude I'm gonna get a sweetass body kit for my metro, paint it jet black, buy some rims(he ment hub caps but didn't know the difference)and add a flame steeringwheel cover".
Atleast it's something to laugh at.
I was talking to this kid yesterday and he's like "yeh dude I'm gonna get a sweetass body kit for my metro, paint it jet black, buy some rims(he ment hub caps but didn't know the difference)and add a flame steeringwheel cover".
Atleast it's something to laugh at.
#16
it is the blown underwater version of a Honda hovercraft/submarine hybrid DX model...note the ventilation/ram air/snorkel vents on teh hood...
did it show a picture of the propeller out back and the bilge pump....hi-performance bilge pump no less...
did it show a picture of the propeller out back and the bilge pump....hi-performance bilge pump no less...
#17
Originally posted by NT2SHBBY
the 530hp comes from the leerjet that pushin behind it
the 530hp comes from the leerjet that pushin behind it
That civic is just, wrong, waaaayyyy to wrong!!! Is it even a civic? Looks like something that came out of the s**t hole
#22
^ just copy and paste the address into the reply box, it'll add the url tags automatically. I'm quite confident that he doesn't have a supercharger and a turbo, but it's been done. It's just amazingly expensive and complicated.
#23
What a bunch of BS. If you do the physics based on the #s (weight of car, power of engine, 0-60/0-147 times) things don't come out right. Even assuming he had 530hp at all times (and all of that at the wheels), it's not enough to accelerate his 2100lb car from 60 to 147 mph in 5.6 sec...
#29
Originally posted by weltonw
What a bunch of BS. If you do the physics based on the #s (weight of car, power of engine, 0-60/0-147 times) things don't come out right. Even assuming he had 530hp at all times (and all of that at the wheels), it's not enough to accelerate his 2100lb car from 60 to 147 mph in 5.6 sec...
What a bunch of BS. If you do the physics based on the #s (weight of car, power of engine, 0-60/0-147 times) things don't come out right. Even assuming he had 530hp at all times (and all of that at the wheels), it's not enough to accelerate his 2100lb car from 60 to 147 mph in 5.6 sec...
#30
The other day I was in town and saw a 2002 Eclipse. It was silver with Blue tint, an Exaust tip that lit up blue, and get this he had a TYPE R badge on each side, u think thats funny he had the GTR badge on the bottom of each tail light on the back lol it also had Greddy stickers on it and tis completely stock
#31
Originally posted by NT2SHBBY
Turbo and SC, eh??? must be runnin 2 separate engines.....
Turbo and SC, eh??? must be runnin 2 separate engines.....
#32
http://cgi.ebay.com/ebaymotors/ws/eB...&category=6256
All he has is intake, exhaust, shifter and a stupid looking body kit bolted on it.
This idiot is a complete liar.
He even has tint on the windshield! What a moron...
All he has is intake, exhaust, shifter and a stupid looking body kit bolted on it.
This idiot is a complete liar.
He even has tint on the windshield! What a moron...
#33
Originally posted by bluemaxx
SHOW YOUR WORK!
SHOW YOUR WORK!
Data
M = 2100lbs = 955 kg
V1 = 60mph = 26.8 meter/s
V2 = 147mph = 65.7 meter/s
T = 5.6s (9.6 - 4.1)
Kinetic energy required to go from 60mph to 147mph
= 1/2 M (V2^2 - V1^2)
= 1.72e6 joules
(ignores rotational energy in wheels)
Continuous power required to do so in 5.6 seconds
= 307000 Watts
= 411 hp
Now approximate air drag...
Drag = 1/2 * Rho * V^2 * A * Cd
Power = Drag*V
where
Rho = air density
V = velocity
A = vehicle frontal area
Cd = drag coefficient
Assume
Rho = 1.275 kg/m^3 = sea level value
A = 2 meters wide x 1.5 meters high (this is reasonable)
Cd = 0.35
Power needed at 60mph = 17hp = P1
Power needed at 147mph = 254hp = P2
Assumption: more of the 5.6 seconds of time will be spent at the higher speeds of the run. Therefore the actual average power needed to overcome drag over the 5.6 seconds is *greater* than the average of P1 and P2, which is 135hp.
So we need 411hp for the kinetic energy and at least 135hp for the drag, and that already puts us over...
#35
Originally posted by weltonw
No prob - here it is in metric units. Tried to keep it simple.
Data
M = 2100lbs = 955 kg
V1 = 60mph = 26.8 meter/s
V2 = 147mph = 65.7 meter/s
T = 5.6s (9.6 - 4.1)
Kinetic energy required to go from 60mph to 147mph
= 1/2 M (V2^2 - V1^2)
= 1.72e6 joules
(ignores rotational energy in wheels)
Continuous power required to do so in 5.6 seconds
= 307000 Watts
= 411 hp
Now approximate air drag...
Drag = 1/2 * Rho * V^2 * A * Cd
Power = Drag*V
where
Rho = air density
V = velocity
A = vehicle frontal area
Cd = drag coefficient
Assume
Rho = 1.275 kg/m^3 = sea level value
A = 2 meters wide x 1.5 meters high (this is reasonable)
Cd = 0.35
Power needed at 60mph = 17hp = P1
Power needed at 147mph = 254hp = P2
Assumption: more of the 5.6 seconds of time will be spent at the higher speeds of the run. Therefore the actual average power needed to overcome drag over the 5.6 seconds is *greater* than the average of P1 and P2, which is 135hp.
So we need 411hp for the kinetic energy and at least 135hp for the drag, and that already puts us over...
No prob - here it is in metric units. Tried to keep it simple.
Data
M = 2100lbs = 955 kg
V1 = 60mph = 26.8 meter/s
V2 = 147mph = 65.7 meter/s
T = 5.6s (9.6 - 4.1)
Kinetic energy required to go from 60mph to 147mph
= 1/2 M (V2^2 - V1^2)
= 1.72e6 joules
(ignores rotational energy in wheels)
Continuous power required to do so in 5.6 seconds
= 307000 Watts
= 411 hp
Now approximate air drag...
Drag = 1/2 * Rho * V^2 * A * Cd
Power = Drag*V
where
Rho = air density
V = velocity
A = vehicle frontal area
Cd = drag coefficient
Assume
Rho = 1.275 kg/m^3 = sea level value
A = 2 meters wide x 1.5 meters high (this is reasonable)
Cd = 0.35
Power needed at 60mph = 17hp = P1
Power needed at 147mph = 254hp = P2
Assumption: more of the 5.6 seconds of time will be spent at the higher speeds of the run. Therefore the actual average power needed to overcome drag over the 5.6 seconds is *greater* than the average of P1 and P2, which is 135hp.
So we need 411hp for the kinetic energy and at least 135hp for the drag, and that already puts us over...
#36
Originally posted by weltonw
No prob - here it is in metric units. Tried to keep it simple.
Data
M = 2100lbs = 955 kg
V1 = 60mph = 26.8 meter/s
V2 = 147mph = 65.7 meter/s
T = 5.6s (9.6 - 4.1)
Kinetic energy required to go from 60mph to 147mph
= 1/2 M (V2^2 - V1^2)
= 1.72e6 joules
(ignores rotational energy in wheels)
Continuous power required to do so in 5.6 seconds
= 307000 Watts
= 411 hp
Now approximate air drag...
Drag = 1/2 * Rho * V^2 * A * Cd
Power = Drag*V
where
Rho = air density
V = velocity
A = vehicle frontal area
Cd = drag coefficient
Assume
Rho = 1.275 kg/m^3 = sea level value
A = 2 meters wide x 1.5 meters high (this is reasonable)
Cd = 0.35
Power needed at 60mph = 17hp = P1
Power needed at 147mph = 254hp = P2
Assumption: more of the 5.6 seconds of time will be spent at the higher speeds of the run. Therefore the actual average power needed to overcome drag over the 5.6 seconds is *greater* than the average of P1 and P2, which is 135hp.
So we need 411hp for the kinetic energy and at least 135hp for the drag, and that already puts us over...
No prob - here it is in metric units. Tried to keep it simple.
Data
M = 2100lbs = 955 kg
V1 = 60mph = 26.8 meter/s
V2 = 147mph = 65.7 meter/s
T = 5.6s (9.6 - 4.1)
Kinetic energy required to go from 60mph to 147mph
= 1/2 M (V2^2 - V1^2)
= 1.72e6 joules
(ignores rotational energy in wheels)
Continuous power required to do so in 5.6 seconds
= 307000 Watts
= 411 hp
Now approximate air drag...
Drag = 1/2 * Rho * V^2 * A * Cd
Power = Drag*V
where
Rho = air density
V = velocity
A = vehicle frontal area
Cd = drag coefficient
Assume
Rho = 1.275 kg/m^3 = sea level value
A = 2 meters wide x 1.5 meters high (this is reasonable)
Cd = 0.35
Power needed at 60mph = 17hp = P1
Power needed at 147mph = 254hp = P2
Assumption: more of the 5.6 seconds of time will be spent at the higher speeds of the run. Therefore the actual average power needed to overcome drag over the 5.6 seconds is *greater* than the average of P1 and P2, which is 135hp.
So we need 411hp for the kinetic energy and at least 135hp for the drag, and that already puts us over...
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01-22-2001 09:35 PM