engine/car formulas
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engine/car formulas
alright i figured this would be a good thread to start..its so every one in here can just post formulas they know for figureing engine/car stuff out.
1.how to figure out your cubic inches .
bore x bore x stroke x .785 x number of cylinders= your cubic inches.
example
4.00 x 4.00 x 3.48 x .785 x 8 = 349.67 witch is a chevy 350
2.how to figure out your cam over lap
int duration + exh duration / by 4 - lobe x 2 = valve over lap
example
285 + 292 / 4 - 112 x 2 = 64.5
3.calculating how much cfm your engine will use(for carb selection)
engine cubic inches x max rpm / 3456 = cfm
example
355 x 8500 / 3456 = 873
4.et calculating stuff
your 1/8 mile time x 1.5832 = 1/4 time
your 1/4 mile time / 1.5832 = 1/8 time
TUEBO STUFF
5.on a turbo map ever 10lbs =144.72 cfm
6.figuring out pressure ratio from boost
atmospheric pressure at sea level is 14.7 psi so
14.7 + amount of boost / 14.7 = pressure ratio
example
14.7 + 34 / 14.7 = 3.3
7.figureing out your engine volumetric flow or EVF in cfm.
step 1
engine ci / 1428 =
step 2
engine rpm / 2 =
step 3
take step 1 answer and times it by step 2s answer witch will give u your evf.
8.figuring out your engines airflow in lb/min
step 1
your psi u plan on running x your evf x 29 =
step 2
10.73 x 460 =
step 3
take step 1s answer and div it by step 2s witch then u get your engines airflow in lb/mins
1.how to figure out your cubic inches .
bore x bore x stroke x .785 x number of cylinders= your cubic inches.
example
4.00 x 4.00 x 3.48 x .785 x 8 = 349.67 witch is a chevy 350
2.how to figure out your cam over lap
int duration + exh duration / by 4 - lobe x 2 = valve over lap
example
285 + 292 / 4 - 112 x 2 = 64.5
3.calculating how much cfm your engine will use(for carb selection)
engine cubic inches x max rpm / 3456 = cfm
example
355 x 8500 / 3456 = 873
4.et calculating stuff
your 1/8 mile time x 1.5832 = 1/4 time
your 1/4 mile time / 1.5832 = 1/8 time
TUEBO STUFF
5.on a turbo map ever 10lbs =144.72 cfm
6.figuring out pressure ratio from boost
atmospheric pressure at sea level is 14.7 psi so
14.7 + amount of boost / 14.7 = pressure ratio
example
14.7 + 34 / 14.7 = 3.3
7.figureing out your engine volumetric flow or EVF in cfm.
step 1
engine ci / 1428 =
step 2
engine rpm / 2 =
step 3
take step 1 answer and times it by step 2s answer witch will give u your evf.
8.figuring out your engines airflow in lb/min
step 1
your psi u plan on running x your evf x 29 =
step 2
10.73 x 460 =
step 3
take step 1s answer and div it by step 2s witch then u get your engines airflow in lb/mins
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Brake Math
-it's all about the square inch area of the master cylinder pushing against the square inch area of the caliper piston.
-Then you have a ratio between the pedal to the pivot length, and from the pivot to master cylinder rod length.
-If you had a 2" diameter piston in one caliper you have 3.1416 square inches of area.
-If you have a master cylinder that is 1" diameter you have .7854 square inches of area.
-So you take 3.1416, divide by .7854 and you get a hydraulic ratio of 4:1.
-This means that if you could get a pressure at the master cylinder rod of 180 pounds you get 180 times 4 or 720 pounds of force at the caliper piston.
-You can get the 180 pounds at the master cylinder pushrod by having a 9" long brake pedal arm hanging down, with the top end mounted say about 3" above the master cylinder piston, a pivot for the master cylinder rod clevis at 3" from the top, and then 6 inches from that pivot down to a foot pedal.
-the 6" divided by the 3" is a 2:1 ratio meaning that if you moved the foot pedal two inches you'd push the master cylinder rod 1".
-So to get 180 pounds at the master cylinder rod you'd need to push 90 pounds of push from your foot on the brake pedal.
-it's all about the square inch area of the master cylinder pushing against the square inch area of the caliper piston.
-Then you have a ratio between the pedal to the pivot length, and from the pivot to master cylinder rod length.
-If you had a 2" diameter piston in one caliper you have 3.1416 square inches of area.
-If you have a master cylinder that is 1" diameter you have .7854 square inches of area.
-So you take 3.1416, divide by .7854 and you get a hydraulic ratio of 4:1.
-This means that if you could get a pressure at the master cylinder rod of 180 pounds you get 180 times 4 or 720 pounds of force at the caliper piston.
-You can get the 180 pounds at the master cylinder pushrod by having a 9" long brake pedal arm hanging down, with the top end mounted say about 3" above the master cylinder piston, a pivot for the master cylinder rod clevis at 3" from the top, and then 6 inches from that pivot down to a foot pedal.
-the 6" divided by the 3" is a 2:1 ratio meaning that if you moved the foot pedal two inches you'd push the master cylinder rod 1".
-So to get 180 pounds at the master cylinder rod you'd need to push 90 pounds of push from your foot on the brake pedal.
Formulas for Various Computations
Key;
* Pi = 3.14159
* F = Degrees Farehenheit
* C = Degrees Celsius (Centigrade)
* R = radius ( When preceeded by degrees F or degrees C, it refers to Absolute temperature [Rankin])
* D = Diameter
Weights:
* 1 Gallon of Oil = 7.0 lbs.
* 1 Gallon of Water = 8.354 lbs. @ 60 deg.F.
* 1 Gallon Gasoline = 6.2 lbs @ 60 deg.F.
* 1 Gallon Alcohol = 6.616 lbs @ 60 deg.F.
* 1 Gallon Nitro = 9.505 lbs @ 60 deg.F.
Conversions;
* Cubic Inches to Gallons = Divide by 231
* Gallons to Cubic Inches = Multiply by 231
* Cubic Inches to Ounces = Multiply by .554
* Ounces to Cubic Inches = Divide by .554
*
* Cubic Centimeters to Cubic Inches = Multiply by .06102
* Cubic Inches to Cubic Centimeters = Multiply by 16.39
Circles:
* Area of a Circle = Pi x R ^2 (3.14159 x R^2)
* Area of a Circle = .7854 * D
*
* Circumference of a Circle = 2Pi x R (6.2832 x R)
* Circumference of a Circle = Pi x D
Temperature:
* Absolute Zero = -459.7 deg. F.
* Absolute Zero = -273.2 deg. C.
*
* Absolute Temperature deg. F. = deg. F. + 459.7 ( May be rounded to -460)
* Absolute Temperature deg. C. = deg. C. + -273.2 (May be rounded to -273)
Jet Change for Fuel Injection
* New Air Density - A
* Old Air Density - B
* Current Jet.... - C
* Formula; (A-B)/ (B) x 100 = % Change in Density
* Sqrt(((C/2)^2(PI)(-(A-B)/(B))/(C/2)^2(PI))/(PI))x2 = New Jet Size.
Fuel Delivery in GPM @ RPM
Lbs.Fuel @ RPM
--------------
Weight of 1 Gallon Fuel.
Lbs Fuel @ RPM
Lbs. Air@RPM
------------
Desired A/F Ratio
Lbs Air@RPM
CFM @ RPM x Weight of 1 cu.Ft Air
Weight of 1 Cu.Ft. Air
1.326 x Barometer
Rankine= Temp Deg F +460
Key;
* Pi = 3.14159
* F = Degrees Farehenheit
* C = Degrees Celsius (Centigrade)
* R = radius ( When preceeded by degrees F or degrees C, it refers to Absolute temperature [Rankin])
* D = Diameter
Weights:
* 1 Gallon of Oil = 7.0 lbs.
* 1 Gallon of Water = 8.354 lbs. @ 60 deg.F.
* 1 Gallon Gasoline = 6.2 lbs @ 60 deg.F.
* 1 Gallon Alcohol = 6.616 lbs @ 60 deg.F.
* 1 Gallon Nitro = 9.505 lbs @ 60 deg.F.
Conversions;
* Cubic Inches to Gallons = Divide by 231
* Gallons to Cubic Inches = Multiply by 231
* Cubic Inches to Ounces = Multiply by .554
* Ounces to Cubic Inches = Divide by .554
*
* Cubic Centimeters to Cubic Inches = Multiply by .06102
* Cubic Inches to Cubic Centimeters = Multiply by 16.39
Circles:
* Area of a Circle = Pi x R ^2 (3.14159 x R^2)
* Area of a Circle = .7854 * D
*
* Circumference of a Circle = 2Pi x R (6.2832 x R)
* Circumference of a Circle = Pi x D
Temperature:
* Absolute Zero = -459.7 deg. F.
* Absolute Zero = -273.2 deg. C.
*
* Absolute Temperature deg. F. = deg. F. + 459.7 ( May be rounded to -460)
* Absolute Temperature deg. C. = deg. C. + -273.2 (May be rounded to -273)
Jet Change for Fuel Injection
* New Air Density - A
* Old Air Density - B
* Current Jet.... - C
* Formula; (A-B)/ (B) x 100 = % Change in Density
* Sqrt(((C/2)^2(PI)(-(A-B)/(B))/(C/2)^2(PI))/(PI))x2 = New Jet Size.
Fuel Delivery in GPM @ RPM
Lbs.Fuel @ RPM
--------------
Weight of 1 Gallon Fuel.
Lbs Fuel @ RPM
Lbs. Air@RPM
------------
Desired A/F Ratio
Lbs Air@RPM
CFM @ RPM x Weight of 1 cu.Ft Air
Weight of 1 Cu.Ft. Air
1.326 x Barometer
Rankine= Temp Deg F +460
Piston speed:
stroke in inches x rpm/6=fpm
Blower Capacities:
Sizes and capacity are:-
6:71 small diameter
Rotor dia=5.505", length=14.975", displacement per full turn of rotor=339CI.
6:71 big diameter
Rotor dia=5.778", length=14.975", displacement per full turn of rotor=411CI.
8:71
Rotor dia=5.778", length=15.905", displacement per full turn of rotor=436CI.
10:71
Rotor dia=5.778", length=17.000", displacement per full turn of rotor=466CI.
14:71
Rotor dia=5.778", length=19.000", displacement per full turn of rotor=521CI
This is theoretical displacement.
Retro or high helix will change this considerably.
A worn blower will reduce this a little.
A very good tight new blower and a Teflon stripped blower should be about the same
Supercharger plenum size:
surmising an engine size of 202 6 cyl (you put your own numbers in here) Inlet air density ratio of 1.55.
Eg; 202/6= 33.6666 x 1.55= 52.183333. Air cushion of 3 times cylinder volume is needed.
So 52.183333 x 3= 156.54999 cubic inches for CC's times this figure by 16.387 = 2565.3848
Air Density Index
ADI = 100 * [ (PBAR-PWV) / 29.92 ] * [ 519.67 / ( 459.67 + T ) ]
PBAR = Barometric Pressure (inches Hg)
PWV = Water Vapor Pressure (inches Hg)
T = Temperature (deg Fahrenheit)
stroke in inches x rpm/6=fpm
Blower Capacities:
Sizes and capacity are:-
6:71 small diameter
Rotor dia=5.505", length=14.975", displacement per full turn of rotor=339CI.
6:71 big diameter
Rotor dia=5.778", length=14.975", displacement per full turn of rotor=411CI.
8:71
Rotor dia=5.778", length=15.905", displacement per full turn of rotor=436CI.
10:71
Rotor dia=5.778", length=17.000", displacement per full turn of rotor=466CI.
14:71
Rotor dia=5.778", length=19.000", displacement per full turn of rotor=521CI
This is theoretical displacement.
Retro or high helix will change this considerably.
A worn blower will reduce this a little.
A very good tight new blower and a Teflon stripped blower should be about the same
Supercharger plenum size:
surmising an engine size of 202 6 cyl (you put your own numbers in here) Inlet air density ratio of 1.55.
Eg; 202/6= 33.6666 x 1.55= 52.183333. Air cushion of 3 times cylinder volume is needed.
So 52.183333 x 3= 156.54999 cubic inches for CC's times this figure by 16.387 = 2565.3848
Air Density Index
ADI = 100 * [ (PBAR-PWV) / 29.92 ] * [ 519.67 / ( 459.67 + T ) ]
PBAR = Barometric Pressure (inches Hg)
PWV = Water Vapor Pressure (inches Hg)
T = Temperature (deg Fahrenheit)
Plus if anybody needs em ive got everything from piston kinematic programs, to enderle and hilborn conversion charts, to nitro mixing and tuning charts, Dynamic CR calcs, valvetrain timing/geometry .....yadayadayada