so if my fc has 138 torque at 3800 rpm how much torque do i have up hi like 6300 6500 rpm. do you lose alot and its all hp after 3800 or do you only lose a lil? please help

 

you only would lose a tiny bit, the curve on my dyno seemed pretty flat, i have a full exhaust on my NA though

i ended up peaking torque at about 5400 rpm though, and between 4500-6500 the torque was still within 5 ft/lbs of the peak

 

how much did that full exhast help hp wise

 

i didnt have a base line dyno,

but i dynoed 133 hp ant 115 ft/lbs, with an engine with quite alot of miles, over 150K prolly,its a pretty worn engine. compression was around 90 all around.

 

isnt it suposed to have like 120 to be ok and lik140150 great?

 

the factory rated hp is 146 ,

dynoing to the wheels from the factory it would put between 110-120 brand new.

with the low compression my motor had id say it was probably making about 105 if i dynoed it stock.

 

i was talking compression my bad like 120s ok and 140 150 is great 90 realy shity right

 

no 120 is great and 140 or 150 would be a high compression 240sx or something around there.

 

so what normal then lik 110?

 

Torque normally falls off as the engine approaches redline.
This is just the normal dynamics and physics of the internal combustion engine.
There are tricks to keep the torque up - i.e. VDI, switching short / long runners, etc., but torque tends to fall on the top end.

Sorry, but if you want "gobs of torque", you need to go forced induction or more displacement.


-Ted

 

a stock engine making 146 HP at 6500 rpm is making 118 lb-ft of torque at that same 6500 rpm. HP is a function of torque. HP, Torque, and RPM. take any two and you can find out the third

 

So- who's going to make the newest 3.2 Litre Rotary?


LOL

 

Torque??? Whats that?

 

I don'tr think you could find rpm from only tourque and hp numbers could you

 

127HP is the baseline number for a good engine (according to Kahren).

 

Yes, you can. Hp is torque times RPM/5252,
so if H=horespower, T=torque, and R=RPMs, then
H=T*(R/5252)
From this you can get that R=(H/T)*5252

 

^

which is why the torque and hp curves on all dyno graphs should cross at 5252 rpm, because (5252/5252) equals one, and so horsepower will always equal torque at that specific rpm.

 

That is correct IF the torque and hp scales are exactly the same.
Some DynoJet graphs like to scale different torque and hp axes, so be careful about this.


-Ted

 

As long as we're on the subject here, just a little fyi for anyone that doesn't know; Peak Torque ALWAYS occurs at Peak Volumetric Efficiency. Given that, it's easy to understand why Turbo/Supercharging help out Torque alot.

 

alright, that's good to know, so as not become suspicious whenever they don't cross at that point

 

sorry but i don't get it, well it's not that i don't get it just the math isn't working out. Lets go with the new passat it has 276hp at 6200 and 266lb-ft at 2750 so if R=(H/T)*5252 then R=(276/266)*5252 that equals 5449.44 which is niether the rpm of the HP or Torque, Assuming it did work out right would R be the rpm of the HP or torque? even with H=T*(R/5252) Then H=266*(6200/5252) that would mean 314hp unless you ment r is the rpm or the torque in which case H=266*(2750/5252) and that would mean 139.28hp
not trying to be a dick just trying to figure out why it won't work for me