drive ratio blowers
 

Is there a formula to figure what size pulleys to use on blowers
to achieve different boost levels?
The blowershop.com goes into combos using underdriven ratios
aswell as overdriven. How do you go about figuring your size pulleys?

Example.
I have a 540 8.1 static compression
Merlin Fully ported Cast iron heads.
Crane 139741 cam
Lee intercooler
Mooneyham 10-71
The blowers came with 8 mm pulleys #49 and # 52
The chart on theblowershop.com this would give 6.12%
bottom pulley being the 52
Is the size of the belt a limiting factor?
How does one figure the boost other then trial and error?

I would like to run the highest level possible on 93 octane.
possibly 9 lbs?
Thanks in advance.

Please respond
Gerry

caveman

I am no expert but many things change actual boost and final compression ratio. Cam - Exhaust - Blower setup/clearance - heads and carburetors. I would say you are close. My little experience tells me to start low boost and work your way up. Not much help I know. Try it and go from there. Sound like you are close.

I run a 525ci - 1071 at close to 6% under and it runs 5PSI with no intercooler. 9psi at 6% over just swap the pulleys. :D 8 to 1 at 9psi intercooled with 93 octane should not be a problem as long as you watch your timing advance total. Just my .02.

I agree with paradigm...WOT rpm matters too.
I run 10-71's on 557inch motors with a 6%OD (53/50) combo, no innercooler and make 7psi at 5100rpm.
If i turn the motors/blowers faster the boost will increase.

Thanks for the responses.
I was thinking volumetric efficiency ( Big Words)
plays a role. Like cam lift profile flowing specs on heads.
I am trying to get a idea to get close. One engine is going to the dyno. I would like to have it close that day.

Maybe some one has a combo similar to the above and good give me there pulley specs. and WOT rpm.

Wot in this case would be 5400 rpm max. (hydraulic Roller)

What about bolting one to a engine cradle and run it.
install boost gauge and whack it. I will use the existing pulleys:)

Back4more, What static compression are you running and what type of cam, carbs etc . And did you ever dyno the motors , to see what type of power you got out of them. I'm allways interested to see what everybody is doing with there motor combo's. I'm in the process of building new power for my boat and it will be similar set-up. Mine will be 572's w/10-71. Thanks, Tom

There's no cast-iron formula.
A roots blower will theoretically pump a given volume of air per revolution at atmospheric pressure. At low rpm, leakage around the rotor strips will reduce this from theoretical, and as boost level increases, you will get a reduction as well from seal leakage.

While percent of change seems to give you a good rule of thumb to follow, Paradigm's experience shows otherwise. Example: at 6% under he saw 5 # boost. At 6% over, he saw 9 # boost. The math doesn't give you this but real world does.

Here's what the math shows:
6% under (0.94) versus 6% over (1.06).
The difference is (.94-1.06=.12). The percent of change is (0.12/0.94 = 13.7%). This means that his blowers will "try" to pump 13.7% more charge into the motors.

This does NOT equate to a 13.7% increase in boost. If his motor was capable of flowing a HUGE amount of increase, this may be partially so, but his motor is already showing resistance to the flow at 6% under or else he would not see that much positive manifold pressure. SO: 13.7% increase in attempted flow increases the back pressure in a nonlinear fashion, and results in 9 psi on his combo.

What we gain from this is that doing math on blower speeds is a crapshoot. There is probably a dreamwheel out there that gives the ratios and observed results to be able to slide-rule your way to a good guess.

If you need a guess worthy formula, go with Paradigm's example and we derive:

13.7% increase in blower speed gives an 80% increase in boost pressure.

The graph curve is most likely logarithmic (a parabola), so let's just go with this for a starter:

1.137 squared times a constant = 1.8
1.293 C = 1.8
C=1.392

Therefore,
(Percent of Blower Speed Increase)squared x 1.392 = (New percent of boost)


If you have 50% underdriven now at 3 pounds of boost, and you go to 1:1, then your new blower speed is 200%.

(2.00)squared x 1.392 = 5.568
New boost will be 3 x 5.568 = 16.7 pounds of boost.

Dunno if this looks close or not.

Thanks again everyone for the responses.

I feel that the two pulleys supplied are not going to
achieve enough boost. If I compare the chart on the
blowershop.com it will give 6.12% overdriven.
The pulleys are three teeth apart. 49, 52 If I use Paradigm's example his also when overdriven gives 6.00% (9lbs boost) three teeth apart. I must figure in the pressure drop across the Intercooler. I estimate
7.5lbs. I Figure approx 10.00% overdriven. Change the
bottom pulley to a 54 tooth. I Guess the thing to do would be to bring a assortment of pulleys to the Dyno.

What do you guys think?

If my earlier guess at a formula is anywhere close, then going from 6% over thru an intercooler (guessing 7.5 pounds) to 10% over:

That's a 3.7% increase in blower speed.
1.037 squared times 1.392 = 1.497 times the previous boost pressure.
7.5 times 1.497 = 11.22 pounds of boost.

Dunno if it is anywhere close or not.
Guess I'm not helping much.

Tom,
I have 8:1CR in DART Big M's 4.500X4.375
Isky Hyd. rollers...248*/258*@.050 with .608/.617 lift.
Canfield 350cc Alum. heads 2.30/1.88 valves.
Holley 750's
Dyno'd at 850HP@5500(6.5psi 51/51 pulleys) with a dead flat TQ curve of 830 from 4500-5300.
Hope that helps.
Good luck with the build...let me know how they do, PM me if you can.
Mark

Mark, Thanks for the info , that combo that you have sounds great. Tom

ps . The motors look great too.