Dynomite

Tech Info - LT5/ZR-1 Technical Calculations

Several posts involving ZR1 and LT5 Technical Calculations follow this first summary post.
This thread is referenced in the first post of Tech Info - LT5 Modifications/Rebuild Tricks (500+hp)

The Calculations Posts in this thread are in DRAFT for now and include the following topics.

Post 2. Tech Info - LT5 Horsepower and Torque Calculations
Post 3. Tech Info - ZR1 Differential Gearing and Vehicle Speed Calculations
Post 4. Tech Info - ZR1 Wind Force, Rolling Resistance, Drivetrain Loss Calculations
Post 5. Tech Info - LT5 Camshaft Timing and Timing Chain Calculations
Post 6. Tech Info - LT5 Pressure Drop In Oil Lines Calculations
Post 7. Tech Info - LT5 RC SL4-205 injectors (500+ hp) Calculations
Post 8. Tech Info - LT5 Summary of Camshaft Timing
Post 9. Tech Info - LT5 Timing Diagrams
Post 10. Tech Info - L98 Frisbee Horsepower
Post 11. Tech Info - LT5 Coolant Flow Calculations
Post 12. Tech Info - Calculating Alternator Pulley Diameter

Additional Calculations Links are provided herein. These Links provide for quick determinations of:
a. 1/4 Mile ET
b. HP
c. MPH
d. RPM
e. Speed
f. Gear Ratios
g. Wheel Offset.

1/4 mile Calculator TIPS
Gear calculator TIPS
Wheel Offset Calculator

Additional posts will be added as they are created.

Last UPDATE of post 1 Sep, 2012

Dynomite

Tech Info - LT5 Horsepower and Torque Calculations

One horsepower is defined as Omega T or rotation speed (radians per second) multiplied by torque (foot-pounds)

One horsepower is defined as 550 foot-pounds per second (33,000 foot-pounds per minute). Engine speed is revolutions per minute (rpm). The units of torque are foot-pounds.

A radian of a circle is a ratio of the length of an arc divided by the length of a radius, so the units of length cancel out and you're left with a dimensionless measure called a radian.

One revolution is 360 degrees of a circle. The circumference of a circle is 2 x pi x radius (there are 2pi radians in a revolution where pi is 3.14 which is the ratio of circumference of circle to radius). To convert revolutions per minute to radians per minute, you multiply revolutions per minute (rpm) by 2pi which equals 6.28 radians per minute per rpm (revolution per minute). (6.28 radians per minute)/(60 seconds per minute) equals 0.10472 radians per second per rpm (revolution per minute).

So....one rpm is 6.28 raidans per minute or .1047 radians per second.

If we use rpm (converted to radians per minute) in the Horespower equation (one horsepower = 33,000 ft-lb/min) we have to divide by 6.28 radians per minute per rpm (revolution per minute).

If we use one horsepower = 550 ft-lbs/sec we have to divide by the conversion 6.28 radians per minute/60 seconds per minute equals .10472 radians per second per rpm (revolution per minute).

Divide the 550 ft-lbs/sec by the 0.10472 radians/second per rpm, we get 550/0.10472, which equals 5,252 ft-lbs-rpm (Conversion factor).

Horsepower = Torque (ft-lbs) x engine speed (rpm)/5,252 ft-lbs-rpm.
OR.....Torque (ft-lbs) = 5,252 ft-lbs-rpm x Horsepower/engine speed (rpm).

You should be able to construct any Torque/rpm curve from any Horsepower/rpm curve and vice versa.

Torque = 5252 x Horsepower/rpm at any specific rpm.
Horsepower = rpm x Torque/5252 at any specific rpm.

If you are talking crankshaft, stay at the crankshaft for both. If you are talking rear wheel, stay at the rear wheels for both.

Rotating Horsepower Calculators

Last UPDATE of post 2 Nov, 2012

Dynomite

Tech Info - ZR1 Differential Gearing and Vehicle Speed Calculations

1. ZF-S6-40 Transmission gear ratios. To convert transmission output rpm to engine rpm.
Gear ratios - 1st-2.68, 2nd-1.80, 3rd-1.29, 4th-1.0, 5th-0.75, 6th-0.50, Reverse-2.50
Transmission Output rpm = Engine rpm/transmission gear ratio.
For Engine Speed of 7,000 rpm and third gear.....
Transmission Output rpm = 7,000/1.29 = 5,426rpm

2. Differential Gear Ratios and Tire rph (Revolutions Per Hour). To convert transmission output rpm to engine rpm.
Two typical Differential Gear Ratios are 3.45 and 4.10.
Tire rpm = Transmission Output rpm/Differential Gear Ratio.
Tire Revolutions per hour = Tire rpm x 60min/hr
For Engine Speed of 7,000 rpm in third gear with 3.45 Differential......
Tire Revolutions per hour = (7,000/1.29/3.45) x 60min/hr = 94,371rph

3. Tire circumferences.
Stock 315x17x35 Tires. 25.68 tire diameter x 3.14 = 80.64inch circumference.

a. It is assumed the tire radius is measured from center of tire to ground or deflected (Loaded) radius of tire to calculate tire diameter used in the speed calculations.
b. It is assumed centrifugal force does not increase tire loaded diameter at high rpms.

4. Vehicle Speed at 7,000rpm 3rd Gear (Fly Wheel RPM).
Vehicle Speed in Miles Per Hour (mph) = (Tire Circumference in ft x Tire revolutions/hour)/ 5,280ft/mile

(7,000rpm/1.29/3.45) = Wheel rpm =1572.85rpm = 1572.85rpm x 60 = 94,371 revolutions/hour
80.64/12 = wheel circumference in ft = 6.72ft
94,371 revolutions/hour x 6.72ft/revolution/ 5,280 ft/mile = 120.11 miles/hour

OR
One Mile = 5,280ft
(1572.85rpm x 6.72ft x 60 min/hr)/ 5,280ft = 120.11 mph

5. Vehicle speed with tire loaded deflection of 1/4 inch at Engine rpm (7,000 rpm).
Assuming third gear (Transmission 1.29 gears and Differential 3.45 gears).

Stock 315x17x35 Tires. 25.68 tire diameter x 3.14 = 80.64inch circumference.
Deflected Tire Diameter 25.68in - .5in = 25.18 inch tire effective deflected tire diameter.
New wheel circumference in ft = 6.59
Vehicle Speed at engine speed of 7,000rpm = (1572.85rpm x 6.59ft x 60 min/hr)/ 5,280 ft = 117.78mph


Last UPDATE of post 3 Aug, 2014

Dynomite

Tech Info - ZR1 Wind Force, Rolling Resistance, Drivetrain Loss Calculations

(100 mph, 50 deg F and Sea Level).

Starting from scratch............Coefficient of drag (Cd) depends somewhat on the Rynolds Number (for this discussion we shall consider Cd constant or 0.34). Reynolds Number Calculator (just to show what is involved as there is no chart showing Cd vrs Reynolds Number for ZR1 vehicles) Reynolds Number Calculator

For the air resistance or drag (D) we have:

D = (1/2)* rho*(V^2)*A*Cd. (* or x are used interchangeably as multiplying)

Rho is density of the fluid, air in this case, which is approximated as 0.0024 slugs/cubic foot (depends on temperature and elevation - 50 deg F and Sea Level) Air Density Calculator
One slug per cubic foot (slug/ft^3) = 32.17 lbm/cubic foot (divide lbm/ft^3 in chart by 32.17 to get slugs per cubic foot).

V is the velocity, in ft/sec, relative to the air.
V = (100 miles/60 minutes)
V = ((100 miles * 5280 ft/1 mile) / (60 min * 60 sec/1 min)) = 146 ft/sec
A is your frontal area. Let's assume car is 4 feet tall and 6 foot wide as an average, for an area of about 24 square feet.

Cd is the Drag Coefficient. For a 1990 Chevrolet Corvette ZR1 lets use a Cd of 0.34.

1. Wind Force.
So the air Drag on the ZR1 (at 100 mph) at sea level on a cool 50 deg F day would be (1/2)*0.0024*(146^2)*24*.34 = 208 slugs*ft/sec^2 or 208 pounds.

2. Energy Loss.
But your real question is ft# or total energy per a mile (5280 ft) for example. So in a mile we have 208 lbs * 5280 ft or 1,098,240 ft # .

But.........what you really want is Horsepower (Hp) or Drag times velocity

33,000 ft-lbs/min (550 ft-lbs/sec) is one horsepower........

3. Horsepower Loss.
I think then we would have 208 lbs x 146 ft/sec or about 30,368 ft-lb/sec or about 55 Hp (Drag x velocity) at 100 mph (sea level, 50 deg F).

or ft-lb (Energy) per unit of time. Lets say the mile is accomplished in 36 seconds at 100 mph.
This is the same as 1,098,240 ft-lbs/36 seconds or close to 30,368 ft-lbs/sec or 55 Hp (energy/unit of time).

Just trying to make this whole discussion more universal being able to now calculate ft-lbs of energy expended in any distance for any velocity and any car (any temperature and any altitude) and from there calculate hp expended

With head lights up or down see Wind Tunnel Test

4. Tire Rolling Resistance.
Now....lets calculate tire rolling resistance Horsepower (RR Hp) .
Lets say rolling resistance coefficient (RRc) of .02 (depends on tire type and tire pressure velocity, altitude, temperature independant).
Rolling resistance R = Weight x RRc = 3,300 lbs x .02 = 66 lbs. At 146 ft/sec (100 mph) the RR Hp would be:
RR Hp = (146 ft/sec x 66 lbs)/550 ft-lbs/sec) = 17.5 Hp

The total Horsepower loss due to aerodynamics (I should say poor aerodynamics) and tire rolling resistance at 100 mph is then 55 Hp + 17.5 Hp or 72.5 Hp. This can then be compared to the Hp at the rear wheels since we have not accounted for drivetrain losses. Drivetrain losses (Transmission 1.5%, Drive shaft .5%, Differential 7% or greater) might add up to 10% or more of Hp output. The drivetrain losses estimated here are very conservative.

Lets try 185 mph just for kicks (seal level, 50 deg F).
V = 271 ft/sec
Drag due to air resistance would be 720 lbs. That would be around 354 Hp at 185 mph. Add in Rolling Resistance that would be 372 hp total......we are now very close (or exceed) 375 hp at the flywheel if you include drivetrain loss (not calculated).

Lets now try 180 mph and include drivetrain losses (sea level and 50 deg F)
V = 264 ft/sec
Drag due to air resistance would be 682 lbs. That would be around 327 Hp at 180 mph. Add in Rolling Resistance that would be 344 hp total.

5. Drivetrain Loss.
Now lets add Drivetrain losses (1.5% transmission, .5% drive shaft, 7% Differential) and we get an additional hp loss of 31 hp for a total hp loss of 375 hp......we are now at 375 hp at the flywheel when we include Drivetrain losses (which drivetrain losses are very conservative estimates).

If we use 3.45:1 differential gears and .5:1 Transmission 6th gear overdrive with 12.5 inch loaded radius rear wheels we are at 4,200 rpm (to make sure we have the 375 hp at the flywheel from the horsepower curve we would need appropriate differential gears) SPEED vrs RPM Calculator

Keep your Head Lights DOWN and transmission hot.

Last UPDATE of post 4 July, 2017

Dynomite

1. Tech Info - LT5 Camshaft Timing Calculations

I set the exhaust at 108 deg because of the SW headers and SW 3 inch exhaust having less backpressure and intake at 114 deg.

This timing accomodates the dynamics of the air flow (effects on cam timing) resulting from porting (plenum/injector housing/heads) and headers/3 inch exhaust/no CATS.

There are 20 teeth on the cam sprocket or 18 degrees per tooth. There are 14 holes in sprocket or 25.7 degrees if you rotate vernier plate without removing pin from hole in vernier plate and pin next hole in sprocket (15 holes in vernier plate and 14 holes in sprocket). There is 1.6 deg on the camshaft sprocket if you rotate vernier washer to align with next pin hole in sprocket (360/15/15). So if you went 11 holes that would be like one tooth on cam sprocket or 17.6 degrees on the camshaft sprocket retard or advance. The smallest amount of retard or advance you could set would be 1.6 deg on the camshaft sprocket by pinning next aligned holes (3.2 deg on the crankshaft front pully).

Marc Haibeck suggests there is about 3 deg of play (6 deg measured at the crankshaft) between the sprocket timing plate flat (vernier plate) and camshaft flat (90'-92'). As one tightens the camshaft bolt you have to watch not to turn the camshaft in the timing plate (vernier plate). It helps to hold a 19 mm open end wrench on the rear of each cam during tightening of the camshaft bolt.

This 3 deg of play (at the camshaft) actually allows you to set the camshaft timing at any angle within the 1.6 deg (at the camshaft) allowed by the pin holes in the vernier plate and sprocket.

This camshaft play is 3 deg (90-92 having single flats) and is reduced to 1.5 deg on 93s' (having double flats).

See LT5 Camshaft Specifications and Camshaft Timing Tricks for supplemental information.

See Tech Info - LT5 Summary of Camshaft Timing from Start to Finish for supplemental information.

2. Tech Info - LT5 Timing Chain Calculations

Timing chain breakage although very rare has been blamed on repeated engine pulses or fatigue (dead injector or bad valves for example). The timing chain calculations in this post debunk all such theories on timing chain failure.


LT5 Cranskshaft sprocket has 20 teeth.
LT5 Idler sprocket on primary side has 42 teeth
LT5 Idler sprocket on secondary side has 21 teeth
Each LT5 camshaft sprocket has 20 teeth
LT5 Primary chain has 54 pins
LT5 Secondary chain on LH has 102 pins
L T5 Secondary chain on RH has 94 pins

So...we know the camshaft sprockets rotate half as fast as the crankshaft sprocket.

Lets calculate......Crankshaft sprocket rotates 360 deg or 20 teeth. Idler sprocket primary rotates 20/42 of 360 deg or 20 teeth. Idler sprocket on secondary side rotates 20/42 of 360 deg or 10 teeth. Camshaft sprocket rotates 180 deg or 10 teeth. Exactly 180 deg like it is suppose to.

Now lets see how much of total length the RH and LH secondary chains rotate. Crankshaft sprocket rotated 20 teeth or 360 degrees. Primary chain has 54 pins so it rotated 20/54ths of its length. Camshaft sprocket rotated 180 deg or 10 teeth so for RH chain with 94 pins that is 10/94ths of its length and for LH chain with 102 pins that is 10/102nds of its length.

Lets put this another way.......each time the crankshaft sprocket rotates 360 degees, the RH chain moves 10/94ths of its length and the LH chain moves 10/102nds of its length. Or if the RH chain were to rotate its full length, the crankshaft sprocket would rotate 9.4 times and if the LH chain were to rotate its full length, the crankshaft sprocket would rotate 10.2 times.

Now...let us look at the analysis in above posts wherein a miss fire could cause chain failure.

The bad cylinder miss fires once every two revolutions of the crankshaft. That little tug on the chain would occur on the RH chain 4.7 times as it rotated its full length. That little tug would occur on the LH chain 5.1 times as it rotated its full length. And the little tug from a bad injector would occur at different locations on each chain every revolution of the crankshaft since these (4.7 and 5.1) are not whole numbers.

Primary drive crankshaft sprocket................................Primary chain


Idler sprocket........................................ ...........Secondary chain (LH top, RH bottm)


One of four camshaft sprockets


Last UPDATE of post 5 July, 2012

Dynomite

Tech Info - LT5 Pressure Drop In Oil Lines Calculations

Pressure drop in oil lines of different internal diameters and lengths

Flow rate in gallons per minute (gal/min) Q (lets say 3 gal/min at 2,300 engine rpm).
Inside diameter of pipe or hose in inches (in) D (lets say .75 in)
Kinematic viscosity of fluid (at operating temperature) in square in/sec) kv (lets say .1309 square in/sec)
Density of the fluid in pounds per cubic inch (lbs/in cubed) ρ (lets say .0318 lbs/in cubed)
Length of the pipe, tube or hose in inches (in) L (lets say 36 inches or two hoses 18 inches long each)

Kinematic Viscosity (kv)
AMSOIL 10W-30 kv is .1024 square in/sec @ 104 deg F (66.1 cSt)
kv is .0181 square in/sec @ 212 deg F (11.7 cSt)
AMSOIL 10W-40 kv is .1309 square in/sec @ 104 deg F (84.5 cSt) (worst case with cold oil)
kv is .0223 square in/sec @ 212 deg F (14.4 cSt)
(1 cSt = .001549 square in/sec)

v = (Q x 3.85)/(D x D x .785) = 26.11 in/sec
Where
v = velocity in inches per second (in/sec)
Q = flow rate in US gallons per minute (gpm)
D = inside diameter of pipe or hose in inches (in)

Re = (v x D) /(kv) = 149
where
Re = Reynolds Number
v = velocity in inches per second (in/sec)
D = inside diameter of pipe or hose in inches (inches)
kv = kinematic viscosity of fluid (at operating temperature) in square inches per sec (in squared/sec)

f = 64/Re = 64/149 = .429
Where
f = friction factor
Re = Reynolds Number < 2300

Pressure drop in oil line
Δp = v x v x f x L x ρ)/(2xDxg) = (26.1 x 26.1 x .429 x 36 x .0318)/(1.5 x 386)

Where
Δp = pressure drop in lbs/square inch (psi)
v = velocity inches per second (in/sec)
f = friction factor
L = length of pipe or hose inches (in)
ρ = density of the fluid in lbs per cubic inch (.0314-.0322 lbs/in cubed for hydraulic oil)
D = inside diameter of pipe or hose inches (in)
g = 386 in/sec squared (Gravitational acceleration)

= .577psi (using .75 inch ID hose at 2,300 engine rpm, cold start).
= 2.79 psi (using .5 inch ID hose at 2,300 engine rpm, cold start ).
= 1.7 psi (using .75 inch ID hose at 7,000 engine rpm, cold start).
= 8.8 psi (using .5 inch ID hose at 7,000 engine rpm, cold start).
= .29 psi (using .75 inch ID hose at 7,000 engine rpm, hot).
= 1.5 psi (using .5 inch ID hose at 7,000 engine rpm, hot).
Oil pump flow is 1.3 gpm per engine 1,000 rpm using the gerotor pump in the LT5 engine.

These oil flow rates through the oil filter housing are assuming the Oil Pressure Regulation Valve is not opening . An open Oil Pressure Regulation Valve would let some oil bypass back to the oil pan without going through the oil filter housing while maintaining 50 psi - 60 psi Oil Pressure.

As can be seen, one should use .75 inch ID hose for this application at the higher rpms and cold starts.





Background Tech Info – Pressure Drop In Oil Lines for the LT5

Oil Pressure Restriction

The largest restriction in oil flow (what you see on the oil pressure gauge) is the head loss once the oil reaches the main bearings and camshafts. That is the largest restriction that maintains oil pressure at the gauge.

If the oil filter becomes clogged, the oil filter bypass valve will open reducing the excessive oil filter head loss restriction if that restriction becomes appreciable. The oil flow is determined by the total head loss after the pump. Keep in mind the Oil Pressure Regulation Valve (OPRV) located just after oil pump opens at a pressure of 50-60 psi. This OPRV is prolly always open a bit (especially on a cold start). The oil pump has more volume capability than that volume of oil flowing through the oil filter housing when the oil is of high viscosity, or the oil pump is operating at high rpm.

As oil heats up and becomes less viscous the oil pressure drops a bit with the oil flow through the oil filter housing prolly reaching full volumetric capacity of the oil pump with the total head loss decreasing below the 50-60 psi OPRV setting (especially at low oil pump rpms). As you increase engine rpm and oil pump rpm, the volumetric capacity of the oil pump will increase prolly above a 50-60 psi head loss limit and the OPRV will again open. This all depends on temperatures of oil and resulting oil viscosity as well as initial oil viscosity. This also depends on oil viscosity breakdown with use and condition of main bearings and camshafts journals/cam covers.

Lets say I add head loss (5 psi) by adding more restrictive oil lines. Keep in mind the total head loss the pump will see is 50 psi (50-60 psi) or it will bypass flow through the OPRV. Now if I add 5 psi at the oil lines, the head loss at the main bearings and camshafts has to decrease 5 psi so the total head loss remains matched to the allowable 50 psi. The only way that will happen is if I decrease oil flow to the main bearings and camshafts. So I decrease oil flow a bit and decrease head loss in my oil lines as well as the main bearings with a bit more oil going through the OPRV. It is an iterative process as you can see because when I decreased oil flow through the oil filter housing I went a bit below my assumed 5 psi head loss in my new oil lines.

You start out with about 50-60 psi just after the pump. So......I wanted to maintain the original oil flow with minimal oil flow reduction due to additional head loss in hoses between the oil filter adapter and relocated dual oil filters.

I run oil filter relocation on a Toyota Tacoma which I had to do because of the near impossibility of removing that oil filter without removing the engine pan beneath it (and what a mess as that engine pan got its fair share of oil every oil change). I tipped that oil filter upright at the new location just behind the radiator at about the height of the radiator drain plug. When I remove that relocated oil filter, I get just a bit of oil down the sides of the oil filter as I spin it off (all of which drips into the oil drain pan). That works great and I did some calculations on that system also determining the oil flow was nearly unaltered as the head loss in the added oil filter relocation hoses I used was minimal. The key is to use larger SS braided hose (most use 8AN and better oil filter relocation systems use 10AN (I am using 12AN).

Last UPDATE of post 6 July, 2012

Dynomite

Tech Info - RC SL4-205 injectors (500+ hp) Calculations

I use RC SL4-205 injectors (500+ hp). You would use RC 225s for 650 hp or more.

Some Injector Calculatons
Assuming .5 lbs fuel per hour per Hp (BSFC or Brake Specific Fuel Consumption for non-turbocharged engines)....

Lbs fuel per hour per injector = (Hp x BSFC)/(Number of Injectors x .80) where 80% is duty cycle.

(500 x .50)/(16 x .80) = 19.53 lbs/hr (per injector) or 205.06 cc/min

RC 205 Injectors are rated at 43.5 psi fuel pressure.
Hp rating for RC 205s would be....

(20 lbs/hr x .80% duty)/.50 BSFC = 32 hp/injector or 512 hp for sixteen injectors.

Flow Rate - CC's:
205 CC's / MIN @ 43.5 PSI
Flow Rate - LB's:
20 LBS / HR @ 43.5 PSI
Resistance:
12.5 Ohms @ 68 F
Voltage:
8-15 Volts, nominal 13.5 Volts
Amperage:
1.0 Amps
Pressure:
Min 30 PSIG / Max 100 PSIG

RC Injectors

Last UPDATE of post 7 July, 2012

Dynomite

Tech Info - LT5 Summary of Camshaft Timing

See:
LT5 Camshaft Specifications and Camshaft General Timing (pinning)
LT5 Camshaft Specific Timing (degree wheel)
LT5 Camshaft Timing Additional Tricks

1. Camshaft General Timing by Pinning.

a. Set the camshaft timing relatively close using the pinning method (using a 15/64 drill bit as a pin).
b. Use Cylinder #1 for drivers side bank and Cylinder #6 for passenger side bank.
c. Install Timing Chain temporary manual tensioners.

Use Jeffvette Billet Aluminum Camshaft Retainers which will remain in place after camshaft timing is complete.

To start the cam timing process rotate the crankshaft to the 51 deg mark on the front pulley aligned with "0" on timing plate (degree wheel installed on front crankshaft pulley).
Keep in mind that this discussion is for the Drivers side Heads. For the Passenger side Heads use cylinder number 6 instead of cylinder number 1 when setting the camshaft timing using the degree wheel with the method of finding maximum cam lift. TDC for cylinder number 6 will be 360 deg from cylinder number 1. 114 deg ATDC IN and 110 deg BTDC EX will be related to the TDC of cylinder number 6..

Install all cams in the neutral position (no lifters being compressed) using the Billet Aluminum (oiled) Camshaft Retainers with Torx bolts tightened (89 in-lbs) with loctite 262.

Looking at the front of the engine toward the rear so clockwise is normal engine rotation from that perspective.

Rotate exhaust camshafts to insert a 15/64 drill bit as a pin into the pinning hole in exhaust camshafts (hole in front camshaft Retainer). Pin counterclockwise one pin hole to allow for 110 deg BTDC EX. Rotate intake camshafts to insert 15/64 drill bit as a pin into the pinning hole in intake camshafts (hole in front camshaft Retainer) as that should be close to 114 deg ATDC IN.

The Pinning Method will get you close to these timing numbers (110 deg BTDC EX and 114 deg ATDC IN). Fine tuning the camshaft timing for a ported engine (for example), use the degree wheel and a dial gauge on the lifters for the additional fine camshaft timing adjustments. See step 5c of LT5 Camshaft Timing Additional Tricks

Rotate the camshaft sprockets counterclockwise against the chain, and then set vernier plate pin in "next" hole clockwise where the pin might fit. Then tighten a bit the camshaft bolts such that the "flat" of the camshaft is counterclockwise tight against the flat of the vernier plate which was rotated clockwise (holding a 19 mm box wrench on the rear end of the camshaft).

Use Loctite 262 on the camshaft bolt, and lubricate both sides of the washer. Tighten to 19 ft lbs and then mark a straight line and then proceed to tighten (Stretch) the bolt another 80-85 degrees (while holding a 19 mm box wrench on the other end of the camshaft).

After the camshaft timing is set, Remove the manual chain tensioners and install the factory hydraulic chain tensioners set in travel position. Give the chain tensioners a tap to activate each tensioner.

Last UPDATE of post 8 Feb, 2013

Dynomite

Tech Info - LT5 Timing Diagrams




Last UPDATE of post 9 Aug, 2012

Dynomite

Tech Info - L98 Frisbee Horsepower

There has been lots of discussion regarding elimination of the Frisbee (The flat plate bolted to front of water pump pully).
Calculations of Horsepower gained by elimination of the Frisbee are included herein.

1. Background.
Originally Posted by rodj
From the CF Tech pages

"During a Q & A session at Corvettes at Carlisle 2001, I posed the question to Gordon Kellebrew as to the purpose of the "frisbee". His response was that this disk was added by GM in 1988 to absorb vibrations when the A/C compressor turned On/Off. He said that they later found that this caused the serpentine belt to chirp upon sudden changes in engine speed and so it eventually removed from engines in later years."

2. Frisbee physical characteristics.
Frisbee measurements are 8.5 inches (.71 ft) in diameter (.35 ft radius) and 3.6 lbs. Weight is concentrated a bit on outer ring which I will not consider in these calculations.

3. Moment of Inertia.
Moment of Inertia I = 1/2 M x R x R
where M is mass in slugs (Weight/g) and R is radius in feet and I is moment of inertia in lbs divided by ft /sec^2 multiplied by ft^2or
lbs sec^2 ft (where ^ indicates power of or 2 is squared). g is acceleration of gravity or 32 ft/sec^2

4. Angular acceleration.
Angular acceleration is in Radians per sec^2 and there are 2 pi radians per revolution or 6.28 radians per revolution.

Our Frisbee weighs 3.6 lbs and is .71 foot diameter.
Moment of Inertia (I) would be (3.6 lbs/(2)(32) ft/sec^2) x .35 ft x .35 ft or .0069 lbs ft sec^2

Lets say our Frisbee rotates from 0 to 3,000 rpm in 1 second reving the engine in neutral.
3,000 rpm would be 18,840 radians per minute or 314 radians per second accelerated in 1 second or 314 radians per sec^2.

5. Torque.
Torque = Moment of Inertia multiplied by angular acceleration or .0069 lbs ft sec^2 multiplied by 314 radians per sec^2
Torque equals 2.166 ft lbs which is constant as the Frisbee accelerates assuming the angular acceleration is constant.

6. Horsepower to accelerate the Frispee in rotation (Angular Acceleration).
To get HP which is variable as the Frisbee accelerates angularily.....we will look at 3,000 rpm. 3,000 rpm is 314 radians per second as above. There is 550 ft lbs per second in one HP.

HP = angular velocity multiplied by Torque.
We have 314 radians per second multiplied by 2.166 ft lbs or 680 ft lbs per second which is approximately 1.23 hp (680/550) which is the horsepower generated at the angular velocity of 3,000 rpm as the Frisbee is still accelerating angularily.

7. Comments and additional considerations.
So the only way you get that high an angular acceleration of the Frisbee is reving the engine in neutral. Also keep in mind you have several other pullies with angular acceleration at different rpms depending on their diameters. And keep in mind a couple of those pulleys (Air, Power Steering, Alternator and Water Pump) may have a torque resistance because they are doing work during constant rpms.

Also keep in mind you have ONE BIG Pulley (flywheel) which sees the same kind of angular acceleration as the rather heavy Harmonic Balancer on the front of the crankshaft. A bit different angular acceleration than the Frisbee because the pully diameters are different than the primary driving pully (Harmonic Balancer).

8. Summary of Frisbee Horsepower waste.
So for first go around spinning the Frisbee from zero to 3,000 rpm or 314 radians per second in one second with a required torque of 2.166 ft lbs we would have 680 ft lbs per second or 1.23 hp which is the horsepower generated at the angular velocity of 3,000 rpm as the Frisbee is still accelerating angularily

9. Test Ride and confirmation of possible Frisbee elimination issues.
Eliminated Frisbee and Eliminated Air Induction Pump.
A test ride in a 90' L98 going from zero to 100 mph with Air Compressor on and off several times I found as the Air Compressor engaged a little bit of surge hardly noticeable.

No other indications regarding water pump vibration, pulley chirps when Air Compressor engaged and disengaged were noticed during the test drive up to speeds of 100 mph.

The Air Pump had been eliminated several months before Frisbee elimination and although the Air Pump elimination was not the focus herein, this road test does confirm Frisbee and Air Pump elimination function well together without any associated engine issues. TPiS Air Pump Elimination Kit

1990 Corvette (L98) Performance
1990 Corvette (L98) Modifications




Last UPDATE of post 10 Sep, 2012

Dynomite

Tech Info - LT5 Coolant Flow Calculations

An attempt will be made here to show that the coolant flow velocity in the coolant pipes that may contain air will self evacuate air due to the high flow rate of coolant. In other words a bubble of air cannot travel backwards in the flow at the coolant and will be pushed with the coolant to the radiator. Once Air is evacuated to the top of the radiator the air remains on top and the accumulated air evacuates through the vent on the passenger top side of the radiator back to the pressurized coolant reservoir sitting at the highest point in the coolant system just in front of the passenger side under the hood.

1. Coolant Flow.
Centrifugal Water Pump is assumed to deliver flow in direct proportion to its rotation speed. The calculatons assume the thermostast has thermally opened allowing free flow of water through the radiator.

When first starting the engine, coolant recirculates through the engine only and even though air is self evacuated as described below, the evacuated air does NOT get to the top of the radiator untill the engine/coolant is warmed to operating temperature with thermostat open for radiator flow.

2. Water Pump Flow Characteristics.
1. 11 gpm at 900 rpm
2. 61 gpm at 5,000 rpm
3. 85 gpm at 7,000 rpm


Thermostat/Cooling Discussion

3. Thermostat Bypass to protect radiator.
Coolant pressure (as the water pump spins faster the coolant pressure increases) forces the bypass to open at around 5,000 rpm so max flow through radiator is a bit over 61 gpm (lets assume 65 gpm). The 65 gpm maximum flow through the radiator keeps the radiator at a stable pressure.

The two IH coolant manifolds have pipe inside diameters of 1.25 inch or a cross sectional area of 1.3 square inches each.
A water pump flow rate of 65 gpm would be 15,014 cubic inches per minute or 250 cubic inches per second flow rate or 125 cubic inches per second in each IH coolant manifold pipe. The coolant velocity in each IH coolant manifold pipe would then be 125 cubic inches per second/1.3 square inches = 96 inches per second or 8 ft per second.

Before the thermostat opens back to engine at 5,000 rpm all the coolant flow moves through the single pipe back to the top of the radiator. This single pipe has a cross sectional area of 1.3 square inches also. The Coolant velocity in that single pipe would be 250 cubic inches per second/1.3 square inches = 192 inches per second or 16 ft per second.

At an engine rpm of 2500 the coolant flow would be half and the coolant velocity in the coolant pipes would be half of that calculated above.

At idle (lets say 900 rpm) the coolant flow would be 11 gpm and coolant velocity in the two IH coolant manifold pipes would be 1.4 ft per second and for the single return pipe to the top side of the radiator the coolant velocity would be 2.8 ft per second.

4. Coolant System Self Air Evacuation.
What is significant here in these calculations is the coolant velocity and the speed at which a bubble of air (air pocket) would have to go against the flow to not be evacuated. It is impossible to imagine a bubble of air flowing by gravity up a slight incline from the top of the radiator to the top of the IH coolant manifolds at a speed of 1.4 ft per second. A bubble of air in a vertical test tube would not rise that fast in a column of water.

From these calculations it is concluded that once the water pump is void of air (either it is or isn't since a partially loaded with water impeller would move that water and entrapped air outward sucking in more water), any air sitting elsewhere in the system will be forced to the top of the radiator by the coolant velocity.

5. The Air Locked Water Pump.
One has only to assure oneself that the water pump does get flooded with water which may take several pumping cycles adding water each time to the top radiator hoses and allowing water to seep through a closed thermostat back upwards to the water pump filling the water pump void.

See for the exception of the Air Locked Water Pump Filling With Coolant and the Air Locked Water Pump

An anology might help explain. Take an empty quart jar and tip it upside down and lower it into a bucket of water. The jar remains full of air even though the water surrounding the jar is much higher. Now tip the jar a bit and see some bubbles rise from the bottom end. You will now see the water level in the jar has come up a bit but there is still a lot of air inside the jar. Your water pump empeller is in that jar.

Last UPDATE of post 11 Sep, 2013

Dynomite

Calculating Alternator Pulley Diameter.

Starting with diameter of belt surface on Harmonic Balancer to convert rpm to surpentine belt speed we can calculate Alternator rpm depending on Alternator Pulley diameter.

The Harmonic Balancer diameter is 7-1/8 inch to outer lip (90' ZR1). The surpentine belt riding surface is about 1/8 in less radius so surpentine belt riding surface diameter would be 6-7/8 inches (6.875 inches).

If the alternator puts out 150 amp max and 120 amp at idle....the question is at what rpms is alternator spinning when the engine is idling at 850 rpm with a surpentine belt speed of 18,349 inches per minute. (6.875 inches x 3.14) in per rev x 850 rpm = 18,349 inches per minute.

My alternator pully is about 68mm (2.7 inches) diameter (I just bought two new alternator pulleys yesturday).



At a Surpentine Belt speed (idle) of 18,349 inches per minute, that alternator would be spinning at (18,349 inches per minute)/(2.7 inches x3.14) = 2,164 rpm (alternator spin rate) at idle (engine idle speed 850 rpm).

Or...just divide the Harmonic Balancer Diameter by the Alternator Pulley Diameter and multiply by engine idle speed to get Alternator Idle speed.

So the question might be at what Alternator idle speed the alternator still puts out the current needed to keep up with ZR1 current usage at idle (assuming you are not sitting there with head lights on at idle). Size that alternator pulley accoringly and then see what the maximum alternator rpm might be at 7,000 rpm engine speed (making sure you are not overspinning the alternator).

If you are overspinning the alternator and do that often, get a larger alternator pulley and suffer the consequences when sitting at a stop light with lights on loosing battery charge as your alternator cannot keep up with current load.

After figuring out the alternator pulley size required, get a surpentine belt length accordingly.

Last UPDATE of post 12 Nov, 2014

Dynomite

Additional Tech Info - LT5/ZR-1 Technical Calculations

Last UPDATE of post 13 Aug, 2012

Dynomite

Additional Tech Info - LT5/ZR-1 Technical Calculations

Last UPDATE of post 14 Aug, 2012

Dynomite

Additional Tech Info - LT5/ZR-1 Technical Calculations

Last UPDATE of post 15 Aug, 2012

Dynomite

Additional Tech Info - LT5/ZR-1 Technical Calculations

Last UPDATE of post 16 Aug, 2012

Dynomite

Additional Tech Info - LT5/ZR-1 Technical Calculations

Last UPDATE of post 17 Aug, 2012

Dynomite

Additional Tech Info - LT5/ZR-1 Technical Calculations

Last UPDATE of post 18 Aug, 2012

Dynomite

Additional Tech Info - LT5/ZR-1 Technical Calculations

Last UPDATE of post 19 Aug, 2012

ghlkal

Wow, what a wealth of info and a great read

Thanks for putting this together Cliff



Gary