Correction - upgrade to Newton's 2nd theorem October 26th 2018 + February 11th 2020 + March 29th 2020
Tights	Said already earlier, that Newton missed vertical systems and moves. He refined Galilei's gravity researches to laws.

I was making ball simulation for ATS Roulette game. Tried to calculate kick off energies for ball. Realized that ball's acceleration - during the moment, the push ends - is zero. 

During the moment ball takes off F_k = ma results to plain zero. This cannot be true. The ball is certain to have kinetic energy, with what it rolls. It cannot roll without kinetic energy.

When I puzzled with the issue, came to the conclusion that kinetic energy of an object is actually speed-weigh system at moment X.

March 29th 2020 - New definitions need tests and verifications.

When I puzzled with power aerodynamic sheets energies got better explanations. Speed get the power of 3 from affected cube. When speed raises the hole object makes in one second, grows linearly. But the box into which growth in tube affects grows exponentially in 3 directions. Current formula uses box affected area. In reality this box is ball. In accurate or divided system you can volumes for affected areas. Use box or ball with speed effect.

Then I resolved acceleration and centrifugal forces. Came to new conclusion, that kinetic, horizontal energy is a match vertical energy. Instead of area, you use mass.

These energies obviously converts into forces with friction factor. Depends on usage. Centrifugal force and momentum are a match to hill force.

You accelerate IMMO from 0 to 50 m/s. First the acceleration takes 5 seconds, then another takes 10 seconds. 5 second acceleration is 10 m/s2 and 10 second system is 5 m/s2. 

When you calculate energies 
- 5 second acceleration force is 1900 kg * 10 m/s2 = 19 kN  * 5 s = 95 kN
- 10 second acceleration force is 1900 kg * 5 m/s2 = 9.5 kN  * 10 s = 95 kN
- Without acceleration, force is 1900 kg * 50 m/s = 95 kN

In roll resistance this kinetic energy grows with speed. In small speeds and vacuums car gets force system, which allows it to move at certain speed. mG force is stable, in low speeds and vacuums, the speed of object is undefined. Gravitational drag comes into formula with weight.

The overall force you need for acceleration is steady. I am not convinced from this, verification is needed. In principle you can use more energy to acceleration than the resulting energy is. You can waste energy to air resistance and friction. Would need a system, where you accelerate weight with calibrated motor and gauge for measuring forces.

During 5 second acceleration you need 19 kW power for 5 second time. During 10 second acceleration you need 9.5 kW additional power for 10 second. In more accurate system you calculate air resistance. Air resistance follows the mass system. Final force is the same. 

When you make calculations, IMMO's engine must produce acceleration power. When you have current and maximum power for used rpm, you can calculate how IMMO reacts to an attempt to increase the speed. The power cap in between current and maximum can be used for acceleration, hill climbing and towing.

After IMMO has reached 50 m/s speed, it's kinetic energy contents is 1900 kg * 50 m/s = 95 kN. This is the force, with what IMMO collides in an accident. Also the force, you start to downgrade, when gear is released. After IMMO stops, the energy has to be zero, speed is zero. Multiplication with zero always results to 0.

In curves, you change direction of IMMO's kinetic energy charge. With turning angle you can calculate the force you need for turning. The resulting force is known as centrifugal force. This centrifugal force is the force, tires has to transport to road with current friction factor. When centrifugal force per tire grows over the mass based friction, car slips. System is obviously valid also for acceleration and braking.

OLD AND OBSOLETE SYSTEMS

With acceleration you calculate speeds at moment X, but actual energy must come from current speed. When freely rotating ball slows down, ma calculation always ends into zero. When you use acceleration for calculating new speed, F_k = mv² results into new kinetic energy for the object.

F_k = Objects kinetic energy at moment X
m = mass * G
v = speed at moment X

Power of two is needed for keeping the old Newton with the formula. This formula cannot be correct. Native unit for Newton is kg m/s². Power of 2 results into m²/s² 

Ooops. ... In the actual calculations for balls I used air's density for correcting the unit. Density is included in full formula for air resistance force. Unit for density is kg/m³. When you for example calculate air resistances for the balls, the formula with Av energy results into incorrect unit. 

Fk = A * v2 * q Fk = kinetic energy at moment X, N - kgm/s2 q  = density of the surrounding gas/liquid, kg/m3 A  = frontal area, m2 v  = speed at moment X, m/s m2 * m2 / s2 and kg / m3 => kg m/s2

- In this formula you do not use mass. Mass affects to the acceleration. You use the mass for calculating changes in speed.
- Dense material results in to bigger kinetic energy. When your roll your balls in thick gas, you need more force for keeping the speed.
- When you have falling object, you calculate gravitational mass-speed system, and add it to your force calculation.
- New formula works also in space, where you do not have gravity. You can use it when you calculate predicted acceleration, braking and steering forces.
- on earth speed is known to operate with curved force system.

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Tuesday 11th February 2020 - Gravitational force for moving object

When I calculated cars fuel energy and used energy systems, there was big cap in between fuel consumption and calculated energies. Possible that G = mG formula is faulty.

When you move on the crust, gravitational drag remains unchanged.
But in principle the moving object creates it's mass system into gravitational every time it moves from point A to B.
When so, you should upgrade the gravitational force with speed.

Lets say that car weighs is 100 kg. With 9.81 gravitational drag mass is 981 kgm/s².
- when car moves 10 m/s, mass grows to 9 810 kgm²/s³

If car's top projection is 2 x 5 meters the drag area is 10 m². Otherwise fine, but units don't match.
- when car moves 10 m/s, mass grows with power of two in to 98 100 kgm³/s⁴

No. You must add the speed into area. When car moves 10 m/s, the new area is 5m * 10 / 5 =  10 x 2 meters = 20 m².
- the total area for mass creation is 30 m².

When 100 kg weight is for the car, the mass must be divided with car's area : 100 kg / 10 m2 = 10 kg/m2.
- when multiply area with square meter weight, you get weight for cars mass creation, it is 300 kg.

With these the Fg = 300 kg * 9.81 m/s2.

Fg = mx * At * G A = width * length : area of top projection Av = velocity * width : area from speed At = A + Av : total top projection mx = m / A : mass per square meter Formula is not verified. But it can and will be checked.

Force for rolling resistance comes, when you multiply the force with friction factor.

Wednesday 12th February 2020 - Air resistance

When I calculated cars fuel energy and used energy systems, there was big cap in between fuel consumption and calculated energies. Possible that G = mG formula is faulty.

When vehicle moves, the volume of the air, you must move is dependent on your speed.
Gases viscosity or capability to move Gas atoms is around 50%.

When so, body moves 50% from the atoms and the another half converts into pressure.

Most from  air resistance comes from the compression work. Required compression force grows quickly when created compression ratio and pressure grows.

Front's tilt angle reduces the amount of gas vehicle moves at the time. First the vehicle moves and compresses the front most projection.
After that vehicle moves the differences in the preceding and current side projection.

From the tables below you see how the compression ration grows with speed. And also how tilt angle reduces the pressure against.
The pressure on one square meter plate converts directly in bars. 25 m3 / second means 25 bar pressure at the plate.
- typical pressure in tires is 2 bars, petrol engine 9 bars, diesel 20 bars.
-when you put a pressure sensor to car's hood, the measured pressure is a match to the compress column in the tables.
- these values are calculated estimations. Values can and will be checked.

Square meter plate in upright position and 90 degree tilt

Speed		Tilt		Volume		Viscosity		Move		Compress
0	0	90	deg		m3/s	50	%	0	m3/s	0	m3/s	0		0
1	m/s	90	deg	1	m3/s	50	%	0,5	m3/s	0,5	m3/s	3,6	km/h	2,25	mph
10	m/s	90	deg	10	m3/s	50	%	5	m3/s	5	m3/s	36	km/h	22,49	mph
20	m/s	90	deg	20	m3/s	50	%	10	m3/s	10	m3/s	72	km/h	44,97	mph
30	m/s	90	deg	30	m3/s	50	%	15	m3/s	15	m3/s	108	km/h	67,46	mph
40	m/s	90	deg	40	m3/s	50	%	20	m3/s	20	m3/s	144	km/h	89,94	mph
50	m/s	90	deg	50	m3/s	50	%	25	m3/s	25	m3/s	180	km/h	112,43	mph

Square meter plate with 45 degree tilt

Speed		Tilt		Volume		Viscosity		Move		Compress
0	m/s	45	deg	0	m3/s	50	%	0	m3/s	0,00	m3/s	0		0
1	m/s	45	deg	0,5	m3/s	50	%	0,25	m3/s	0,25	m3/s	3,6	km/h	2,25	mph
10	m/s	45	deg	5	m3/s	50	%	2,50	m3/s	2,50	m3/s	36	km/h	22,49	mph
20	m/s	45	deg	10	m3/s	50	%	5,00	m3/s	5,00	m3/s	72	km/h	44,97	mph
30	m/s	45	deg	15	m3/s	50	%	7,50	m3/s	7,50	m3/s	108	km/h	67,46	mph
40	m/s	45	deg	20	m3/s	50	%	10,00	m3/s	10,00	m3/s	144	km/h	89,94	mph
50	m/s	45	deg	25	m3/s	50	%	12,50	m3/s	12,5	m3/s	180	km/h	112,43	mph

Square meter plate with 30 degree tilt

Speed		Tilt		Volume		Viscosity		Move		Compress
0	m/s	30	deg	0,00	m3/s	50	%	0,00	m3/s	0,00	m3/s	0		0
1	m/s	30	deg	0,33	m3/s	50	%	0,17	m3/s	0,17	m3/s	3,6	km/h	2,25	mph
10	m/s	30	deg	3,33	m3/s	50	%	1,67	m3/s	1,67	m3/s	36	km/h	22,49	mph
20	m/s	30	deg	6,67	m3/s	50	%	3,33	m3/s	3,33	m3/s	72	km/h	44,97	mph
30	m/s	30	deg	10,00	m3/s	50	%	5,00	m3/s	5,00	m3/s	108	km/h	67,46	mph
40	m/s	30	deg	13,33	m3/s	50	%	6,67	m3/s	6,67	m3/s	144	km/h	89,94	mph
50	m/s	30	deg	16,67	m3/s	50	%	8,33	m3/s	8,33	m3/s	180	km/h	112,43	mph