okay so you have 4 pistons that are 81mm in diameter. in one engine they move 77mm per stroke. in another engine they move 87.20mm per stroke. so lets say you have this formula that i just made up

number of pistons X diameter of piston X stroke X RPM

so lets take a b16 at 3000 RPMS
4 x 81 x 77 x 3000 = 774844000

and now lets take a b18c at 3000 rpms
4 x 81 x 87.20 x 30000 = 84758400

now let's take the figure of the b18c which is 84758400 and divide it by the formula from the b16
84758400 / 4 / 81 / 77 = 3397.402597403 rpms

that means that when a b16 is spinning at roughly 3400 RPMS the pistons are being moved the same amount as a b18c is moving at 3000 rpms. does this make sense? tell me that i'm not the only one that's ever thought of this...

is this the reason that in order to make a b16 have power you must "spin the hell out of it" ?

 

its called rod/stroke ratio. the higher the ratio, the higher up the powerband would be. the lower the ratio, the lower the powerband would be.

and that is what determins piston speed, what your talking about.

the lower the r/s, the higher the piston speed. thats why our 1.54 and 1.58 motors pistons move faster at 8,000 RPMs than an F1 piston moving at 19,000RPMs.

F1 motors have a very short stroke = less piston speed
longer stroke = faster piston speed.

 

r/s ratio has NOTHING to do with average piston speed. r/s has to do with piston ACCELERATION

but yes.. the smaller the engine the higher you have to rev it to get power. small engines dont make alot of tq and power is a function of tq and rpm. 100lb's of tq at 10000 rpms will always be more power than 100lbs of tq at 5000rpms.

 

<TABLE WIDTH="90%" CELLSPACING=0 CELLPADDING=0 ALIGN=CENTER><TR><TD>Quote, originally posted by thebense &raquo;</TD></TR><TR><TD CLASS="quote">okay so you have 4 pistons that are 81mm in diameter. in one engine they move 77mm per stroke. in another engine they move 87.20mm per stroke. so lets say you have this formula that i just made up

number of pistons X diameter of piston X stroke X RPM

so lets take a b16 at 3000 RPMS
4 x 81 x 77 x 3000 = 774844000

and now lets take a b18c at 3000 rpms
4 x 81 x 87.20 x 30000 = 84758400

now let's take the figure of the b18c which is 84758400 and divide it by the formula from the b16
84758400 / 4 / 81 / 77 = 3397.402597403 rpms

that means that when a b16 is spinning at roughly 3400 RPMS the pistons are being moved the same amount as a b18c is moving at 3000 rpms. does this make sense? tell me that i'm not the only one that's ever thought of this...

is this the reason that in order to make a b16 have power you must "spin the hell out of it" ?</TD></TR></TABLE>


If u keep on calculating you'll find that the b16 has a longer dwell time a TDC, and this is why the power is made higher on this motor than say a motor with an 87.2mm stroke or 89mm stroke.

 

<TABLE WIDTH="90%" CELLSPACING=0 CELLPADDING=0 ALIGN=CENTER><TR><TD>Quote, originally posted by Used2beAb16 &raquo;</TD></TR><TR><TD CLASS="quote">If u keep on calculating you'll find that the b16 has a longer dwell time a TDC, and this is why the power is made higher on this motor than say a motor with an 87.2mm stroke or 89mm stroke.</TD></TR></TABLE>

TDC?

 

<TABLE WIDTH="90%" CELLSPACING=0 CELLPADDING=0 ALIGN=CENTER><TR><TD>Quote, originally posted by thebense &raquo;</TD></TR><TR><TD CLASS="quote">TDC?</TD></TR></TABLE>

Top dead center, hes talking about the fraction of a second that the pistons sits still at the top of the exhaust stroke.

<TABLE WIDTH="90%" CELLSPACING=0 CELLPADDING=0 ALIGN=CENTER><TR><TD>Quote, originally posted by JDogg &raquo;</TD></TR><TR><TD CLASS="quote">100lb's of tq at 10000 rpms will always be more power than 100lbs of tq at 5000rpms.</TD></TR></TABLE>

This is only true if the engine is capable of making power that high, but if the motor can make power that high, then your comment is very true. very well put also, this should help a whole bunch of people understand the whole way hondas make power.

 

<TABLE WIDTH="90%" CELLSPACING=0 CELLPADDING=0 ALIGN=CENTER><TR><TD>Quote, originally posted by JDogg &raquo;</TD></TR><TR><TD CLASS="quote">100lb's of tq at 10000 rpms will always be more power than 100lbs of tq at 5000rpms.</TD></TR></TABLE>


but will 100 ft/lb at 10k be more powerful than 200lb ft/lb at 5k? Hmm?

 

what? no one in the class knows?

 

<TABLE WIDTH="90%" CELLSPACING=0 CELLPADDING=0 ALIGN=CENTER><TR><TD>Quote, originally posted by LsVtec92Hatch &raquo;</TD></TR><TR><TD CLASS="quote">


but will 100 ft/lb at 10k be more powerful than 200lb ft/lb at 5k? Hmm?</TD></TR></TABLE>

Torque * RPM

Horsepower = ------------

5252

there ya go... any time you have the hp peak after 5252 the hp will be higher than the tq number

100 lbs at 10000 rpms = 190 hp

200 lbs at 5000 rpms = 190hp

now use some mechanical advantage through gearing to make both those engines (one that revs to 10000 and one that revs to 5000 ) top out at the same mph.. the one with 100ft lbs at 10000 is gona go with the same acceleration as the other one

 

<TABLE WIDTH="90%" CELLSPACING=0 CELLPADDING=0 ALIGN=CENTER><TR><TD>Quote, originally posted by JDogg &raquo;</TD></TR><TR><TD CLASS="quote"> Torque * RPM

Horsepower = ------------

5252

there ya go... any time you have the hp peak after 5252 the hp will be higher than the tq number

100 lbs at 10000 rpms = 190 hp

200 lbs at 5000 rpms = 190hp

now use some mechanical advantage through gearing to make both those engines (one that revs to 10000 and one that revs to 5000 ) top out at the same mph.. the one with 100ft lbs at 10000 is gona go with the same acceleration as the other one </TD></TR></TABLE>

but doesn't the torque curve usually lose it's peak before 10k?

 

<TABLE WIDTH="90%" CELLSPACING=0 CELLPADDING=0 ALIGN=CENTER><TR><TD>Quote, originally posted by thebense &raquo;</TD></TR><TR><TD CLASS="quote">

but doesn't the torque curve usually lose it's peak before 10k?</TD></TR></TABLE>


only if you want it to

 

and that's what cams do, right?

 

it takes a little more than that.. cams are one piece of the puzzle, you have think about head work, intake manifold, header, the bottom end, etc

 

so i guess that blows the idea of me trying to gain the highest trap speed by peaking my cammed b16 at around 8000 rpms and bringing all the power in as early as possible.

Because now it peaks at about 9000 or so. Makes sense.

 

you have to gear the car to match the power and then drive in the power... thats the hardest part of a honda... staying in the power the whole way, including coming off the line.... 2step at 8000 rpms

 

<TABLE WIDTH="90%" CELLSPACING=0 CELLPADDING=0 ALIGN=CENTER><TR><TD>Quote, originally posted by JDogg &raquo;</TD></TR><TR><TD CLASS="quote">you have to gear the car to match the power and then drive in the power... thats the hardest part of a honda... staying in the power the whole way, including coming off the line.... 2step at 8000 rpms </TD></TR></TABLE>

See thats the hardest part about driving the a b16 to get the same traps as a b18 with the same power. The curve of course is flatter with a b18 displacement. All my power is from 6k to 9k because of cam gear tuning. And launching at 4000-5000 rpms.

Can you increase trap speeds by moving all the power from 5k to 8k with cam gear tuning??

Because i dont think the hp and tq #s would be as high peaking from 5k -8k as it would between 6k-9k, know what i mean??

 

Cams are not the be all end all when it comes to where your engine likes to make its power. Everything plays a part. All the way from your intake to your selection of header, to your stroke and r/s ratio. Let me put it this way...

You stated the cams in your b16 like to make the most power from 6-9k. If i stuck those same cams in my 1.8 and tuned them, they would still make more power between 6-9k than your 1.6l, because the 1.8 is 'breathing deeper'.

 

<TABLE WIDTH="90%" CELLSPACING=0 CELLPADDING=0 ALIGN=CENTER><TR><TD>Quote, originally posted by LsVtec92Hatch &raquo;</TD></TR><TR><TD CLASS="quote">Cams are not the be all end all when it comes to where your engine likes to make its power. Everything plays a part. All the way from your intake to your selection of header, to your stroke and r/s ratio. Let me put it this way...

You stated the cams in your b16 like to make the most power from 6-9k. If i stuck those same cams in my 1.8 and tuned them, they would still make more power between 6-9k than your 1.6l, because the 1.8 is 'breathing deeper'. </TD></TR></TABLE>

actually i was referring to shift points in relation to peak power and peak tq.

Most of my power is between 6000 and 9000 and peaking around 9000 rpms, because i tuned the cams that way. Cam tuning just shifts the power curve.

if i were to tune a b18 with cams and all the right combination of bolt ons, i would tune the cams to make the flattest curve up to 8300 or so and peak at that point. The longer the stronger, the lower the peak tq will be in rpms, compared to a shorter stroke, so it would be stupid to tune for peak power up to 9000 rpms with a b18, unless you have a ported head with some higher compression.

 

<TABLE WIDTH="90%" CELLSPACING=0 CELLPADDING=0 ALIGN=CENTER><TR><TD>Quote, originally posted by JDogg &raquo;</TD></TR><TR><TD CLASS="quote">r/s ratio has NOTHING to do with average piston speed. r/s has to do with piston ACCELERATION

but yes.. the smaller the engine the higher you have to rev it to get power. small engines dont make alot of tq and power is a function of tq and rpm. 100lb's of tq at 10000 rpms will always be more power than 100lbs of tq at 5000rpms.</TD></TR></TABLE>

Actually r/s ratio doesn't tell you much at all about acceleration either. The acceleration is mostly dictated by the stroke. Your rod would have to be extremely long (like 12+") to really start to have a significant impact on the acceleration.

 

<TABLE WIDTH="90%" CELLSPACING=0 CELLPADDING=0 ALIGN=CENTER><TR><TD>Quote, originally posted by thebense &raquo;</TD></TR><TR><TD CLASS="quote">that means that when a b16 is spinning at roughly 3400 RPMS the pistons are being moved the same amount as a b18c is moving at 3000 rpms. does this make sense? tell me that i'm not the only one that's ever thought of this...

is this the reason that in order to make a b16 have power you must "spin the hell out of it" ?</TD></TR></TABLE>

The pistons aren't being moved the same amount, but I think you have the right idea...they move roughly the same amount of air. That's most of what determines how much power an engine makes: how much air it moves. And no, you're not the first who has thought of this. It's pretty much the first thing you realize when you learn about engines....

 

<TABLE WIDTH="90%" CELLSPACING=0 CELLPADDING=0 ALIGN=CENTER><TR><TD>Quote, originally posted by kommon_sense &raquo;</TD></TR><TR><TD CLASS="quote">
Actually r/s ratio doesn't tell you much at all about acceleration either. The acceleration is mostly dictated by the stroke. Your rod would have to be extremely long (like 12+") to really start to have a significant impact on the acceleration.</TD></TR></TABLE>

This is true. The r/s does have an impact on acceleration, but it's very little. A b18c piston pulls about 6400g's at 10000rpm, for example. A b18a would pull about 6600g's at the same rpm.

 

<TABLE WIDTH="90%" CELLSPACING=0 CELLPADDING=0 ALIGN=CENTER><TR><TD>Quote, originally posted by thebense &raquo;</TD></TR><TR><TD CLASS="quote">okay so you have 4 pistons that are 81mm in diameter. in one engine they move 77mm per stroke. in another engine they move 87.20mm per stroke. so lets say you have this formula that i just made up


number of pistons X diameter of piston X stroke X RPM

so lets take a b16 at 3000 RPMS
4 x 81 x 77 x 3000 = 774844000

and now lets take a b18c at 3000 rpms
4 x 81 x 87.20 x 30000 = 84758400

now let's take the figure of the b18c which is 84758400 and divide it by the formula from the b16
84758400 / 4 / 81 / 77 = 3397.402597403 rpms

that means that when a b16 is spinning at roughly 3400 RPMS the pistons are being moved the same amount as a b18c is moving at 3000 rpms. does this make sense? tell me that i'm not the only one that's ever thought of this...

is this the reason that in order to make a b16 have power you must "spin the hell out of it" ?</TD></TR></TABLE>

one big problem with your calculations. The diameter of the piston doesn't tell you anything. You need the area of the bore.

bore area = pi * (81/2)^2

However for what you were trying to do thats still ok.

What you ended up showing was that the b16 has to spin @3400rpm to move the same amount of air that the b18 can move @3000rpm...

b16@3400rpm = 95.28cfm
b18@3000rpm = 95.21cfm

The shorter stroke means that you are gonna have to spin @ higher rpms than a comparable engine with a larger stroke to move the same amount of air.

 

<TABLE WIDTH="90%" CELLSPACING=0 CELLPADDING=0 ALIGN=CENTER><TR><TD>Quote, originally posted by Lsos &raquo;</TD></TR><TR><TD CLASS="quote">This is true. The r/s does have an impact on acceleration, but it's very little. A b18c piston pulls about 6400g's at 10000rpm, for example. A b18a would pull about 6600g's at the same rpm.</TD></TR></TABLE>

Agree. However think about this... Lets say that you have 2 engines. Engine A has stroke of 1mm and rod length of 2mm, engine B has stroke of 100mm and rod length of 200mm. Both engines have a r/s of 2, but INCREDIBLY different piston speeds and acceleration. r/s ratio doesn't really tell you much about what is going on inside the engine.

 

Good example. That would show that that both those setups, which have the same r/s ratio, would have a proportionate piston dwell time at TDC, which would mean it would do more work in less time than a shittier r/s ratio.

 

any of you guys know of a site that explains engine theory and rod/stroke ratio?