Jump to content
SAU Community

Recommended Posts

Close! it's 6,000 because you can't rev an rb30 beyond that no matter you do with them

all things being equal, you can always rev a motor with a shorter stroke more no matter how you build them.

That rb 30 z is a drag car, would be totally useless around a track and irrelevant to the ops aim. To get any torque or power out of that thing it would have to be revving at 5000rpm and have nos to get it on boost just to get it off the line.

all things being equal, you can always rev a motor with a shorter stroke more no matter how you build them.

That's fine, but for the average joe blow on a "normal budget", why would you want to rev the thing past 8000 RPM.

That's fine, but for the average joe blow on a "normal budget", why would you want to rev the thing past 8000 RPM.

You dont need to.who said you did. but it would get there at the same time a 30 will get to 6500. 26s spin up a lot faster because of the shorter stroke thats my point

So if you have a rolling run from say 4500 RPM in 4th, both engines make the same peak power but the RB30 has more torque through the mid range will still accelerate the car slower than the "rev happy" RB 26?

So if you have a rolling run from say 4500 RPM in 4th, both engines make the same peak power but the RB30 has more torque through the mid range will still accelerate the car slower than the "rev happy" RB 26?

4500rpm in 4th is same speed as you would be changing from 2nd to 3rd in a RB26 :pirate:

You dont need to.who said you did. but it would get there at the same time a 30 will get to 6500. 26s spin up a lot faster because of the shorter stroke thats my point

In neutral maybe.

Not when a torquier engine is pushing 1500kgs.

So if you have a rolling run from say 4500 RPM in 4th, both engines make the same peak power but the RB30 has more torque through the mid range will still accelerate the car slower than the "rev happy" RB 26?

Rb30 would kill it of course with the extra torque, but put both in there sweet spot the 26 cover a greater rev range in the same time if you had identical power and torque figures, which of course you dont. be interested to see this on a graph though.

If you think torque isn't important go back to the graphs and look at the torque readings when a turbo engine comes on boost . Homologation specials like GTRs are not IMO a great example of what production turbo engines/cars are like - because its reason for being was a narrow focus special . Typical production road cars make more torque (same engine/capacity) in turbo form than the NA versions do and surprisingly , for some , at lower revs than the NA ones often do .

Gootta run trains to move , cheers A .

  • Like 1

And who is saying torque is not important.....geez

I am going to take my 70 kw alfa33 out for a trashing now just to prove to myself that...don't know but I am taking it out anyway....I know carbies are better than FI. Hopefully I can find a vl.

Create an account or sign in to comment

You need to be a member in order to leave a comment

Create an account

Sign up for a new account in our community. It's easy!

Register a new account

Sign in

Already have an account? Sign in here.

Sign In Now



  • Similar Content

  • Latest Posts

    • Yeah, that's fine**. But the numbers you came up with are just wrong. Try it for yourself. Put in any voltage from the possible range and see what result you get. You get nonsense. ** When I say "fine", I mean, it's still shit. The very simple linear formula (slope & intercept) is shit for a sensor with a non-linear response. This is the curve, from your data above. Look at the CURVE! It's only really linear between about 30 and 90 °C. And if you used only that range to define a curve, it would be great. But you would go more and more wrong as you went to higher temps. And that is why the slope & intercept found when you use 50 and 150 as the end points is so bad halfway between those points. The real curve is a long way below the linear curve which just zips straight between the end points, like this one. You could probably use the same slope and a lower intercept, to move that straight line down, and spread the error out. But you would 5-10°C off in a lot of places. You'd need to say what temperature range you really wanted to be most right - say, 100 to 130, and plop the line closest to teh real curve in that region, which would make it quite wrong down at the lower temperatures. Let me just say that HPTuners are not being realistic in only allowing for a simple linear curve. 
    • I feel I should re-iterate. The above picture is the only option available in the software and the blurb from HP Tuners I quoted earlier is the only way to add data to it and that's the description they offer as to how to figure it out. The only fields available is the blank box after (Input/ ) and the box right before = Output. Those are the only numbers that can be entered.
    • No, your formula is arse backwards. Mine is totally different to yours, and is the one I said was bang on at 50 and 150. I'll put your data into Excel (actually it already is, chart it and fit a linear fit to it, aiming to make it evenly wrong across the whole span. But not now. Other things to do first.
    • God damnit. The only option I actually have in the software is the one that is screenshotted. I am glad that I at least got it right... for those two points. Would it actually change anything if I chose/used 80C and 120C as the two points instead? My brain wants to imagine the formula put into HPtuners would be the same equation, otherwise none of this makes sense to me, unless: 1) The formula you put into VCM Scanner/HPTuners is always linear 2) The two points/input pairs are only arbitrary to choose (as the documentation implies) IF the actual scaling of the sensor is linear. then 3) If the scaling is not linear, the two points you choose matter a great deal, because the formula will draw a line between those two points only.
    • Nah, that is hella wrong. If I do a simple linear between 150°C (0.407v) and 50°C (2.98v) I get the formula Temperature = -38.8651*voltage + 165.8181 It is perfectly correct at 50 and 150, but it is as much as 20° out in the region of 110°C, because the actual data is significantly non-linear there. It is no more than 4° out down at the lowest temperatures, but is is seriously shit almost everywhere. I cannot believe that the instruction is to do a 2 point linear fit. I would say the method I used previously would have to be better.
×
×
  • Create New...