- About Mile per Hour and Revolutions per Minute:
First find the vehicle speed, MPH and the consequent engine RPM
operating range:
1) MPH = TIRE RADIUS ÷ 168 X ENGINE RPM ÷ GEAR RATIO
- Note: Tire Radius is distance, in inches, from center of tire
to ground.
Note: Gear Ratio is Rear Axle ratio multiplied by Transmission
Gear ratio.
Example: What MPH at 6500 RPM with a 4.9 rear axle and 14 inch
radius
tire in 4th (1:1) gear?
MPH = 14 ÷ 168 x 6500 ÷ 4.90 ÷ 1 = 111 MPH
Example: in 3rd gear (1.34)?
MPH = 14 ÷ 168 x 6500 ÷ 4.90 ÷ 1.34 = 83
MPH
- 2) RPM = 168 x GEAR RATIO x MPH ÷ TIRE RADIUS
- Example: For the case in #1, what will be the RPM after shift
from 3rd to 4th gear at 83 MPH?
RPM = 168 x 4.90 x 83 ÷ 14 = 4880 RPM
- 3) GEAR RATIO = TIRE RADIUS x RPM ÷ 168 ÷ MPH
- Example: For the case in #1, what Gear Ratio is required for
120 MPH at 6500 RPM?
GR = 14 x 6500 ÷ 168 ÷ 120 = 4.51
- 4) TIRE RADIUS = 168 x MPH x GEAR RATIO ÷ RPM
- Example: For the case in #1, what tire radius for 110 MPH but
at 6000 RPM with a 4.11 gear?
168 x 110 x 4.11 ÷ 6000 = 12.7 inches
Note: Approximately a 25" diameter tire. Remember that the
tire radius will be less during hard acceleration than when the
vehicle is standing still. Also, radius
will be greater at high speed due to tire expansion from centrifugal
force.
- WHAT HP & TORQUE is needed: Equations #5, #6, and #7
show how to compute the engine horsepower needed for three different
applications.
5) Engine horsepower required to reach MPH in quarter mile (HPq):
HPq = (0.00426 x MPH) x (0.00426 x MPH) x (0.00426 x MPH) x
WEIGHT
- Note: understates HP required at speeds exceeding 100 MPH.
Note: assumes engine HP must be 2 x the HP required at drive
wheels.
Example: What engine HP is required to achieve 110 MPH in a 3200
pound vehicle in 1/4 mile?
HPq = (0.00426 x 110) x (0.00426 x 110) x (0.00426 x 110) x
3200 = 329 engine HP
- 6) Engine horsepower required to sustain MPH on level ground
(HPs):
- HPs = (MPH ÷ 3) + (WEIGHT ÷ 1,000 x MPH ÷ 10)
Note: assumes engine HP must be 2 x the HP required at drive
wheels
Example: What engine HP is required to sustain 75 MPH in a 3600
pound vehicle?
HPs = 75 ÷ 3 + (3600 ÷ 1,000 x 75 ÷10) = 25 +
(3.6 x 7.5) = 52 engine HP
- 7) Engine horsepower required to sustain MPH up a grade of
G% (HPg):
- HPg = HPs + (G ÷ 100 x 0.005 x WEIGHT x MPH)
Note: Assumes engine HP must be 2x HP required at drive wheels,
calculate HPs with #6.
Example: What HP to sustain 75 MPH up a 6 % grade in a 3600 pound
vehicle?
HPg = HPs + (6 ÷ 100 x 0.005 x 3600 x 75) = HPs + 81 =
(3600 ÷ 10,000 + 0.33) x 75 + 81 = 52 + 81 = 133 engine
HP
- 8a) Horsepower = TORQUE x RPM ÷ 5252
8b) Torque = HP x 5252 ÷ RPM
- Horsepower comes from torque. Torque comes from the pressure
of combustion in the cylinder because combustion pressure causes
the piston to turn the
crankshaft which is measured as torque. The trick is to generate
high enough pressure on each stroke and to do it often enough
(RPM) to produce the
horsepower needed.
Example: What torque is required to generate 329 HP at 6000 RPM?
T = 329 x 5252 ÷ 6000 = 288 foot pounds @ 6000 RPM
Example: What torque is required for 296 HP at 4880 RPM?
T = 296 x 5252 ÷ 4880 = 319 foot pounds @ 4880
- About Cubic Inches, Volumetric Efficiency,
Combustion Efficiency and CFM:
9) CID = NUMBER OF CYLINDERS x SWEPT VOLUME
- CID = N x 0.7854 x bore x bore x stroke (all in inches)
Example: What is CID of a V8 with a "30 over", 4 inch
bore and
3.48 inch stroke?
CID = 8 x 0.7854 x 4.030 x 4.030 x 3.48 = 355 cu. inches
- 10) VE = VOLUMETRIC EFFICIENCY = Engine Actual Air Intake
÷ CID:
- If VE is less than 1 (or 100%) the amount and quality of charge
in the cylinder
is reduced so less torque is produced. VE above 100% is a supercharging
effect and more torque is produced.
Power Level: Stock Performer Torker II Perf. RPM Victor
Jr. Victor
Peak VE% 60-80 75-90 90-100 95-105 105-115 110-122
- 11) CE = COMBUSTION EFFICIENCY = How well the energy in
the fuel is
converted into crankshaft torque. Affected by; air/fuel ratio,
ignition timing, charge mixing and other factors.
Condition Best Power Best Economy Lean Misfire
Air/Fuel Ratio 12-12.5 14.5-15.5 17
- 12) CFM = CUBIC FEET PER MINUTE
- A measure of air flow
into and out of an engine (CFM = CID x RPM x VE ÷ 3464).
Example: What CFM is consumed by a 355 CID engine at 4478 RPM
if VE = 105% (1.05)?
CFM = 355 x 4478 x 1.05 ÷ 3464 = 482 CFM
Example: What CFM by the same engine at 6400 RPM if VE
has fallen to 95% (0.9)?
CFM = 355 x 6400 x 0.95 ÷ 3464 = 623 CFM
- About Compression Ratio, Cubic Inches and Horsepower:
13) CR = COMPRESSION RATIO = CYL. VOLUME @ BDC ÷ CYLINDER
VOLUME @ TDC
- = 1 + (SWEPT VOLUME ÷ VOL @ TDC)
= 1+ (0.7854 x BORE x BORE x STROKE) ÷ (CCV + HGV + PDV)
CCV = Combustion Chamber Volume, in cubic inches
Note: if volume is given in cc’s then ÷ 16.4 to get
cubic inches.
HGV = Head Gasket Volume, in cubic inches,
= Head gasket compressed thickness x 0.7854 x bore x bore
PDV = (Piston Deck Volume) + (Piston Dome Effective Volume)
= (0.7854 x bore x bore x deck to piston distance) + (volume
of piston
depressions - volume of piston bumps)
Example: What is CR of the engine in #9 if heads have 72 cc chamber,
head gasket is compressed to 0.040 inch and flat top pistons give
0.025 deck clearance at TDC?
CCV = 72 ÷ 16.4 = 4.39 cubic inches
HGV = 0.040 x 0.7854 x 4.030 x 4.030 = 0.51 c.i.
PDV = 0.025 x 0.7854 x 4.030 x 4.030+ 0- 0 = 0.32 c.i.
CR = 1+ (0.7854 x 4.030 x 4.030 x 3.48 ÷ (4.39 + 0.51+
0.32) = 1+ (44.39 ÷ 5.22) = 9.5 CR
- 14) HP = Atmos. Press. x CR x VE x CID x RPM ÷ 5252 ÷
150.8
- Example: What HP from a 350 CID Torker-level engine @ 6000 RPM
at sea level?
HP = 14.7 x 9.5 x 0.95 x 350 x 6000 ÷ 5252 ÷ 150.8
= 352 HP
Note: Torker VE is typically 100% at Torque peak RPM but 95%
at HP peak RPM.
Example: Effect of a carb restrictor plate that causes 1.5 PSI
additional manifold vacuum?
HP = 14.7 x 9.5 x 0.9 x 350 x 6000 ÷ 5252 ÷ 150.8
= 336 HP
- 15) CID = HP x 5252 x 150.8 ÷ Atmos. Press. ÷ CR
÷ VE ÷ RPM
- Example: What CID required for 352 HP from a Performer-level
engine?
CID = 352 x 5252 x 150.8 ÷ 14.7 ÷ 8.5 ÷ 0.85
÷ 5000 = 525 c.i.
Note: Performer is 8.5 CR & HP peak is 5000 RPM @ 85% VE
CFM RULES...
CFM and Carburetors:
- Carburetors are rated by CFM (cubic
feet per minute)
capacity. 4V carburetors are rated at 1.5 inches (Hg) of pressure
drop (manifold vacuum) and 2V carburetors at 3 inches (Hg). Rule:
For maximum performance, select a carburetor that is rated higher
than the engine CFM requirement. Use 110% to 130% higher on single-plane
manifolds . Example: If the engine needs 590 CFM, select a carburetor
rated in the range of 650 to 770 CFM for a single-plane manifold.
A 750 would be right. An 850 probably would cause driveability
problems at lower RPM. A 1050 probably would cause actual loss
of HP below 4500 RPM. For dual-plane manifolds use 120% to 150
% higher.
- CFM and Manifolds:
- Manifolds must be sized to match the
application. Because manifolds are made for specific engines,
select manifolds based on the RPM range.
- CFM and Camshafts:
- With the proper carburetor and manifold
it is possible to select a cam that loses 5% to15% of the potential
HP. These losses come from the wrong lift and duration which try
to create air flow that does not match the air flow characteristics
of the carburetor, manifold, head and exhaust so volumetric efficiency
is reduced. An increase in camshaft lobe duration of 10 degrees
will move the HP peak up 500 RPM but watch out, it may lose too
much HP at lower RPM.
- CFM and Cylinder Heads:
- Usually, cylinder heads are the
limiting component in the whole air flow chain. That is why installing
only a large carburetor or a long cam in a stock engine does not
work. When it is not possible to replace the cylinder heads because
of cost, a better matching carburetor, manifold, cam and exhaust
can increase HP of most stock engines by 10 to 15 points. To break
100% Volumetric Efficiency, however, better cylinder heads or
OEM “HO” level engines are usually needed.
- CFM and Exhaust:
- An engine must exhaust burned gases before
it can intake the next fresh charge. Cast iron, log style manifolds hamper
the exhaust process. Tube style exhaust systems are preferred. But headers
are often too big; especially for Performer and Performer RPM levels.
Improving an engine’s Volumetric Efficiency depends on high exhaust
gas velocity to scavenge the cylinder but this will not happen if the
exhaust valve dumps into a big header pipe. On the newer computer
controlled vehicles it is also important to ensure that all emissions
control devices, and especially the O2 sensor, still work as intended.
- CFM and Engine Control:
- Spark timing must be matched to
Volumetric Efficiency because VE indicates the quantity and quality
of charge in each cylinder on each stroke of the engine. Different
engine families require distinctly different spark advance profiles.
And even engines of equal CID but different CR require their own
unique spark advance profiles. Rule: Expect 0.1% to 0.5% loss
in Torque for each 1 degree error in spark timing advanced or
retarded from best timing. Also, detonation will occur with spark
advanced only 3 degrees to 5 degrees over best timing and detonation
will cause 1% to 10% torque loss, immediately, and engine damage
if allowed to persist.
