vehicles general - pilkch/library GitHub Wiki

Engine

Crank speed: The speed of the engine when the starter motor is cranking it, 200-300 RPM for a good starter motor and battery apparently Idle speed: The speed at which the car is set to idle, 700-1000 RPM, lower for heavier flywheels in trucks and buses, higher for lighter flywheels in racing cars

Four stroke

  1. Intake: This stroke of the piston begins at top dead center (T.D.C.) and ends at bottom dead center (B.D.C.). In this stroke the intake valve must be in the open position while the piston pulls an air-fuel mixture into the cylinder by producing vacuum pressure into the cylinder through its downward motion. The piston is moving down as air is being sucked in by the downward motion against the piston.
  2. Compression: This stroke begins at B.D.C, or just at the end of the suction stroke, and ends at T.D.C. In this stroke the piston compresses the air-fuel mixture in preparation for ignition during the power stroke (below). Both the intake and exhaust valves are closed during this stage.
  3. Combustion: Also known as power or ignition. This is the start of the second revolution of the four stroke cycle. At this point the crankshaft has completed a full 360 degree revolution. While the piston is at T.D.C. (the end of the compression stroke) the compressed air-fuel mixture is ignited by a spark plug (in a gasoline engine) or by heat generated by high compression (diesel engines), forcefully returning the piston to B.D.C. This stroke produces mechanical work from the engine to turn the crankshaft.
  4. Exhaust: Also known as outlet. During the exhaust stroke, the piston, once again, returns from B.D.C. to T.D.C. while the exhaust valve is open. This action expels the spent air-fuel mixture through the exhaust valve.

4 strokes (Piston goes up, down, up, down)
2 revolutions of the crankshaft per cycle

For a 4 cylinder engine each cylinder fires every second revolution, ie. at 800 RPM each cylinder will fire approximately 400 times each minute.
At 8000 RPM each cylinder is firing 4000 times per minute, or approximately 66.6 times per second.

Two stroke

  1. Top dead center (TDC)
  2. Bottom dead center (BDC)

Green: Intake/scavenging
Blue: Exhaust
Orange: Compression
Red: Expansion (power)

We can simplify the cycle to:

  1. Intake/scavenging and Exhaust overlapping
  2. Compression
  3. Combustion

2 strokes (Piston goes up, down)
1 revolution of the crankshaft per cycle

Electric Motor

Voltage (V). The phase-to-phase voltage for a 3-phase supply, or the phase-to-neutral voltage for a single phase supply.
Phase. Whether its a 3-phase or single phase voltage supply.
Rating (P). The power rating of the motor in kW.
Power factor (cosΦ). The rated power factor of the motor. Typically around 0.86.
Efficiency (η). The efficiency of the motor. Typically around 95%. Enter 0.95.

Torque is set by the amps. Power is set by watts (volts times amps.) If you lack the voltage you will still have the low end torque, but you won't have the available rpm range. Voltage allows the motor to continue to take the peak amps to a higher rpm.

Starter Motor

The gear reduction ratio between the starter motor and the flywheel is large, usually between 10: 1 and 15: 1
The components of the breakaway torque consist of the inertia of the engine, which depends upon its size; the friction of the cylinder walls and bearings; and the viscosity of the lubricating oil, which depends partly on the grade of oil and partly on its temperature. A cranking speed of 70 to 100 rev/min is usually sufficient for a petrol engine to draw in an ignitable petrol/air mixture, to start firing and to run up on its own. Diesel engines require a higher cranking speed and a longer starter engagement period.

https://www.uniquecarsandparts.com.au/how_it_works_starter_motor

WOS Performance G/R Starter Motor

http://www.wosperformance.co.uk/products/starter-motors/starter-performance/2-0kw-denso-performance/

Starter Model G/R
Rated Power 2.00kW
Rated Voltage 12.00 V
Free Run Speed 10803 RPM
Free Run Current 87 Amp
Max Of Power 2.23 kW
Max Of Torque 21.23 Nm
Max Of Current 817 Amp
Ripple 18 Amp
Direction CW
Hold Current 12 Amp
Pull Current 30 Amp
Voltage Drop 0.6468 V

42.7Nm required in order to ‘stall’ starter during operation.

2.0kW Power

3500 RPM Speed

12 V Voltage

21 Nm Torque

Compression Stroke

Calculating how much energy is required to compress the air in each cylinder and how much it heats up as a result.

https://physics.stackexchange.com/a/163761
https://en.wikipedia.org/wiki/Adiabatic_process#Example_of_adiabatic_compression https://what-when-how.com/automobile/requirements-of-the-starting-system-automobile/

Car engines compress air adiabatically. The work W to compress a gas adiabatically is given by:

W = K(V1 − γf − V1 − γi) / 1 − γ

where γ = CPCV = specific heat at constant pressure CP over specific heat at constant volume CV for the gas K = PiVγi Pi = initial pressure before compression Vi = initial volume before compression Vf

= final volume after compression

This calculator uses the equation above to give you the work done compressing the gas. For the scenario you describe of compressing 0.26 L air starting at STP (20°C, 101.3 kPa) to one eighth that volume the result is approximately 90 Joules. That is roughly the work required to lift 10 kg or 20 lbs by 1 meter— certainly on the same order of magnitude as turning a hand-crank to start an engine:

https://physics.stackexchange.com/questions/163752/how-much-energy-is-required-to-compress-one-engine-cylinder-full-of-air

MAP and MAF Sensors

MAP

Simplify to just "reading = manifold pressure at the cylinder during the intake phase"?

𝑊cyl = 𝜂v * 𝑛e / 2 * Vd * 𝑃 / 𝑅𝑇

𝑊cyl is the mean value of air introduced into the cylinders 𝜂e is the volumetric efficiency of the engine condition 𝑉d is the engine displacement 𝑛e is the engine speed 𝑃 is the manifold pressure 𝑇 is the intake manifold temperature 𝜂v is the volumetric efficiency under different engine condition

MAF

Simplify to just "reading = get intake air pressure and temperature"?

Air flow rate is estimated in the intake manifold directly with a hot wire anemometer mass air flow sensor. Slower response than MAFless.

Exhaust O2 Sensor

...

Fuel injection

Note: The right pedal is not the gas pedal; it is the air pedal. The throttle pedal determines the air, and in turn, the air mass determines the fuel mass. The same is true for carburetors, only carburetors were volume, not mass based devices. With some recent systems, the right pedal isn't even an "air pedal"... it has evolved to a "power demand pedal" - it isn't connected to the throttle at all, it signals the CPU how far the driver has depressed the pedal, and the CPU determines how far to open the throttle using an electric motor. This has many benefits some of which include: controlling emissions during transients, cruise control, traction control, engine start/cranking, driveline clunk, idle speed control, air conditioning load compensation, etc.

The three elemental ingredients for combustion are fuel, air and ignition. However; complete combustion can only occur if the air and fuel is present in the exact stoichiometric ratio, which allows all the carbon and hydrogen from the fuel to combine with all the oxygen in the air, with no undesirable polluting leftovers.

To achieve stoichiometry, the air mass flow into the engine is measured and combined with the fact that the stoichiometric air/fuel ratio is 14.64:1 (by weight) for gasoline. The required fuel mass that must be injected into the engine is then translated to the required pulse width for the fuel injector.

Deviations from stoichiometry are required during non-standard operating conditions such as heavy load, or cold operation, in which case, the mixture ratio can range from 10:1 to 18:1 (for gasoline).

Sample pulsewidth calculations

Note: These calculations are based on a 4-stroke-cycle, 5.0L, V-8, gasoline engine. The variables used are real data.

Calculate injector pulsewidth from airflow

First the CPU determines the air mass flow rate from the sensors - lb-air/min. (The various methods to determine airflow are beyond the scope of this topic. See MAF sensor, or MAP sensor.)

        (lb-air/min) × (min/rev) × (rev/4-intake-stroke) = (lb-air/intake-stroke) = (air-charge)

            - min/rev is the reciprocal of engine speed (RPM) – minutes cancel.
            - rev/4-intake-stroke for an 8 cylinder 4-stroke-cycle engine.

        (lb-air/intake-stroke) × (fuel/air) = (lb-fuel/intake-stroke)

            - fuel/air is the desired mixture ratio, usually stoichiometric, but often different depending on operating conditions.

        (lb-fuel/intake-stroke) × (1/injector-size) = (pulsewidth/intake-stroke)

            - injector-size is the flow capacity of the injector, which in this example is 24-lbs/hour if the fuel pressure across the injector is 40 psi.

Combining the above three terms . . . 

        (lbs-air/min) × (min/rev) × (rev/4-intake-stroke) × (fuel/air) × (1/injector-size) = (pulsewidth/intake-stroke)

Substituting real variables for the 5.0L engine at idle.

        (0.55 lb-air/min) × (min/700 rev) × (rev/4-intake-stroke) × (1/14.64) × (h/24-lb) × (3,600,000 ms/h) = (2.0 ms/intake-stroke)

Substituting real variables for the 5.0 L engine at maximum power.

        (28 lb-air/min) × (min/5500 rev) × (rev/4-intake-stroke) × (1/11.00) × (h/24-lb) × (3,600,000 ms/h) = (17.3 ms/intake-stroke)

    Injector pulsewidth typically ranges from 2 ms/engine-cycle at idle, to 20 ms/engine-cycle at wide-open throttle. The pulsewidth accuracy is approximately 0.01 ms; injectors are very precise devices.

Calculate fuel-flow rate from pulsewidth

        (Fuel flow rate) ≈ (pulsewidth) × (engine speed) × (number of fuel injectors)

            Looking at it another way:

        (Fuel flow rate) ≈ (throttle position) × (rpm) × (cylinders)

            Looking at it another way:

        (Fuel flow rate) ≈ (air-charge) × (fuel/air) × (rpm) × (cylinders)

Substituting real variables for the 5.0 L engine at idle.

        (Fuel flow rate) = (2.0 ms/intake-stroke) × (hour/3,600,000 ms) × (24 lb-fuel/hour) × (4-intake-stroke/rev) × (700 rev/min) × (60 min/h) = (2.24 lb/h)

Substituting real variables for the 5.0L engine at maximum power.

        (Fuel flow rate) = (17.3 ms/intake-stroke) × (hour/3,600,000-ms) × (24 lb/h fuel) × (4-intake-stroke/rev) × (5500-rev/min) × (60-min/hour) = (152 lb/h)

    The fuel consumption rate is 68 times greater at maximum engine output than at idle. This dynamic range of fuel flow is typical of a naturally aspirated passenger car engine. The dynamic range is greater on a supercharged or turbocharged engine. It is interesting to note that 15 gallons of gasoline will be consumed in 37 minutes if maximum output is sustained. On the other hand, this engine could continuously idle for almost 42 hours on the same 15 gallons.

https://wikicars.org/en/Fuel_injection

Higher air flow = more fuel
Lower temperature = more fuel Higher engine speed = more fuel

Turbos, Turbines, Torque Converters

https://www.nuclear-power.net/nuclear-engineering/fluid-dynamics/centrifugal-pumps/eulers-turbomachine-equations/

Engine to Drivetrain to Wheels

Clutch

In operation, a clutch can have 4 operating states:

  • open, zero torque is transmitted between the input and the output shafts
  • slipping, some amount of torque is transmitted between input and output shafts; the speed difference between input and output shaft is significant (e.g. 500 rpm)
  • micro-slipping, almost all of the input torque is transmitted through the clutch; the speed difference between input and output shaft is very small, around 5-10 rpm
  • closed (clamped, locked-up), there is no slip between input and output shaft, all the input torque is transmitted through the clutch

Torque Converter

  • The stall speed is the engine RPM at which the torque converter transfers the power of the engine to the transmission.
  • Lowering the stall speed reduces the transmission’s temperature. When the impeller (input side) of the torque converter is spinning quickly, while the turbine (output side) is spinning slowly or not at all. The motion energy of the impeller is being converted into heat energy, most of which is passed on to the transmission fluid. The higher the stall speed, the more heat will be generated which is bad for the transmission. With a lower stall speed, less time elapses before the motion energy of the impeller is converted to motion energy to drive the turbine, so the transmission runs cooler and lives longer.

https://official.bankspower.com/tech_article/understanding-stall-speed/

A torque converter has three stages of operation:

  • Stall. The prime mover is applying power to the impeller but the turbine cannot rotate. For example, in an automobile, this stage of operation would occur when the driver has placed the transmission in gear but is preventing the vehicle from moving by continuing to apply the brakes. At stall, the torque converter can produce maximum torque multiplication if sufficient input power is applied (the resulting multiplication is called the stall ratio). The stall phase actually lasts for a brief period when the load (e.g., vehicle) initially starts to move, as there will be a very large difference between pump and turbine speed.
  • Acceleration. The load is accelerating but there still is a relatively large difference between impeller and turbine speed. Under this condition, the converter will produce torque multiplication that is less than what could be achieved under stall conditions. The amount of multiplication will depend upon the actual difference between pump and turbine speed, as well as various other design factors.
  • Coupling. The turbine has reached approximately 90 percent of the speed of the impeller. Torque multiplication has essentially ceased and the torque converter is behaving in a manner similar to a simple fluid coupling. In modern automotive applications, it is usually at this stage of operation where the lock-up clutch is applied, a procedure that tends to improve fuel efficiency.

https://en.wikipedia.org/wiki/Torque_converter#Operational_phases

... insert curves image here

h (m) = is P(out)/P(in). This is a function of fluid viscosity, fin design in the turbine and impeller units, T(out), T(in) and other variables. Torque converters run at efficiencies anywhere from 0-95% depending on w(in), w(out), and T(in) and T(out). For example, when a car is stopped at a traffic light, the engine still applies power to the input shaft, but the brakes and transmission prevent the output shaft from rotating. Since P(out) = T(out) * w(out), and w(out) equals zero, P(out) equals zero. Therefore, the efficiency equals zero.

When a car is traveling at highway speeds, the turbine is rotating nearly as fast as the impeller. Recalling that they are attached to the output shaft and the input shaft respectfully, then P(in) » P(out) and therefore efficiency is rather high.

https://web.mit.edu/2.972/www/reports/torque_converter/torque_converter.htm

Differential

On corners the inside wheel travels a shorter path than the outside wheel, causing scrubbing. An open diff solves this, allowing the inside and outside wheels to travel at different speeds.

But, if one driven wheel has low traction such as hitting mud, gravel, snow, ice it will begin to spin when torque is applied, and the vehicle will be less controllable and the vehicle could get bogged.
A locking differential solves this by locking the driven wheels to the same speed.
However, when the traction improves, on bitumen for example, a locked diff will promote understeer, and the tires will scrub again, so it needs to be turned off.

Differential locks:
Simple dog clutch on one of the half-shafts. When engaged, this effectively locks both half-shafts together and prevents the differential action.
Multi-plate limited slip differential. Uses a set of multi-plate clutches to lock up the differential gears.

A multi-plate limited slip differential allows a certain amount of slip, but as it increases, the clutch pack gradually engages until both half-shafts are effectively locked together, and there is no differential action. For acceleration we are trying to send torque to wheels with grip, for deceleration (In 1.5 and 2 way LSDs) we are trying to send torque to the non-locked up wheel.
1 way: Lock up for accelerating
1.5 way: Lock up for accelerating and decelerating but at different proportions
2 way: Lock up for accelerating and decelerating

Open: Each wheel always gets 50% of the torque whether it is spinning or not

Locking diff with one wheel off the ground, stationary, torque is sent to the wheel on the ground, and both tires remain stationary until the torque is adequate enough to either spin the tire or move the car forward.

Slipping wheels

If we have 1 wheel on bitumen and one wheel on snow, maximum of 1000N and 300N in the example above.
For a locking diff:
0 to 600N: 50% to each wheel
600 to 1300N: Resistance in bitumen contact means 300N is applied to the snow and the rest is applied to the bitumen wheel
1300N and above: Both wheels may spin

For an open diff:
0 to 600N: 50% to each wheel
600N and above: 300N to bitumen wheel, remainder to snow wheel

https://www.youtube.com/watch?v=_HOa0aRZYpw&list=PL2ir4svMoaYim-RSNWEh-aIfdcM6plSly&index=3

Wheels, Tires, Driving

The maximum force you can apply to a driven wheel is:
F = mN

F: Output force
m: Friction coefficient
N: Normal force that we are pushing down into the ground with

Bitumen example:
m = 1
N = 1000N
F = 1000N

Snow example:
m = 0.3
N = 1000N
F = 300N

NOTE: For a sliding/spinning wheel the frictional coefficient goes down

Downforce and Lift

Street cars are generally neutral until high speeds (150KPH+) when the body of the car can generate either lift of downforce.

Calculating drag:

Fdrag = 0.5CdAV²
where:
Fdrag - Aerodynamic drag
Cd- Coefficient of drag
D- Air density
A- Frontal area
V- Object velocity

Cd is the coefficient of drag determined by the exact shape of the car and its angle of attack.

Calculating downforce (Essentially the same equation but using the coefficient of lift/downforce instead of the coefficient of friction):

Fdown = 0.5ClAV²
where:
Fdown - Aerodynamic downforce
Cl- Coefficient of lift
D- Air density
A- Frontal area
V- Object velocity

In the case of a modern Formula 1 car, the lift-to-drag ratio Cl/Cd has a typical value of, say, 2.5, so downforce dominates performance.

2017 Formula 1 cars
At ~193KPH produces enough downforce to drive upside down (Just over 722kg)
3000Kg of downforce at 300KPH apparently? So around 29419N?

I think we can just create wings with position on the car, Cd, Cl and A values, then work out the drag and downforce for each one and apply it as a force at that position?

The radiator also affects the airflow, creating drag, slowing down, mixing and creating dirty air. It can be tilted forwards or backwards to force more air through the radiator.

Suspension