stroker kits for the b16a
Thread Starter
Registered User
Joined: Jul 2002
Posts: 24
Likes: 0
stroker kits for the b16a
are stroker kits worth the money and time? i know some companys have the 1.8 and a 2.0 stroker kits for the b16a, but is it really worth it? someone mentioned just getting the crank from a 1.8 or even getting a b18c5 short block and putting the b16a head on it. im not to educated in the whole stroker/crank area so any info would be helpful. thanx for your time.
Guest
Posts: n/a
RE: Stroker kit
The Stroked D16 kit is $3800 from Crower, I hear horrid stories of all stroker motors including 302->347 kits as being unreliable and things BREAKING. Considering Crower's kits are billit... I wouldn't go with one.
Tetris on a stick
Joined: Apr 2001
Posts: 389
Likes: 0
to quote the scc guide to honda building... b16s should never be stroked.
notice how they did not finish the stroker motor article in tuner/turbo...
notice how they did not finish the stroker motor article in tuner/turbo...
__________________
Eat a live toad every morning, and that will be the worst thing to happen every day.
Eat a live toad every morning, and that will be the worst thing to happen every day.
Import Tuner
Joined: May 2002
Posts: 70
Likes: 0
In my opinion, I think it would be alot cheaper to just go and find a 1.8 or 2.0L B-Series block and work with that. Stroker kits arent cheap and in a b16 they can be lethal.
__________________
03' BMW 330XI
03' BMW 330XI
insert witty retort here
Joined: Aug 2002
Posts: 850
Likes: 0
besides, why mess with the B16A's 1.74 rod/stroke ratio? 
if you want a stroked B16A, just get a 92-93 integra GS-R engine, thats all the B17A1 is, a B16A stroked to 81.4mm (1.7L). it shares the same head & block as the B16A2, but with shorter rods and a longer stroke.
if you want a stroked B16A, just get a 92-93 integra GS-R engine, thats all the B17A1 is, a B16A stroked to 81.4mm (1.7L). it shares the same head & block as the B16A2, but with shorter rods and a longer stroke.
__________________
Shawn
-----
The answer to your question is sqrt(pi)
Shawn
-----
The answer to your question is sqrt(pi)
