Power

Power, in physics is actually a specific thing. Energy is the main currency but as everything exists within a time frame, we measure power, as it is the amount of energy per second. The unit of power is the watt and it measures joules per second.

The formulas for power are:
Power (watts) = work (energy in joules) / time (in seconds)

Additionally it is:
Power = force * velocity
I personally find this a bit abstract and so better explanations will follow.

The watt is a derived unit of power in the International System of Units. People generally think of it as an electrical rating, and was a measure of lightbulbs. You may even know the electrical equation:
watts = volts * amps
W = V * A

Electrical companies that supply our homes tend to measure electricity in kilowatt hours. As a kilo is 1000 and there are 3600 seconds in an hour, the kilowatt hour is 3.6Mj. I’m not going to go into much detail on this but the point is that it is a unit of power. We are used to electrical power as it’s so easily converted to mechanical power and is largely generated from mechanical power. Typically chemical power such as a gas mix like HydrOx or nuclear power is used to put energy into a liquid, typically water. That heat energy pushes a turbine that spins a magnet in a coil, generating electrical power. That power is transported down wires and goes into a resistive metal coil in a kettle that heats up and boils water again. Of course energy escapes via things like resistance, sound and light, as no system is 100% effective in conversion.

For now though, I’m thinking about rocket engines so the power equations I’m thinking of are:
W = J / s = (N * M) / s = (kg * m²) / s³

A newton metre is another measure of energy, it is the energy required to apply the force of 1N for 1m. It is very handy but if used as specifically to the physical mechanics of torque it doesn’t directly relate to other expressions of energy.
People can confuse types of energy. A hot atom does not have more newton metres than a cold one, but it does have more joules of energy.
As Force = Mass * Acceleration, then the force of newtons can be broken out into kilograms and acceleration. Also acceleration is the change in velocity over time so the units get broken down further.
If velocity is distance per second, acceleration is the change in velocity i.e. distance per second² then power expresses the increase in acceleration or distance per second³. That feels a bit weird to be in the third dimension of velocity, especially as we are in the second dimension of distance, which is area right? So is a watt is a measure of increasing the acceleration of a kilogram that is multiplied by an area? That doesn’t make sense. Area is not a factor here. Like the newton metre itself it is dimensionally equivalent, but you can’t think of it as area. It would be better to say they are derivatively different. So let’s simplify it.

Let’s just look at the application of power for exactly 1 second.
1W * 1S = 1J
As we’re looking at a watt per second we can dispense with the per second bit of the equation. We are now just looking at energy:
J = N * M = (kg * m²) / s²

1 watt of power is output of 1 joule per second.
Power = Energy / time
A joule is the energy required to apply the force of 1 newton across 1 metre.
Energy = Force * distance
1 newton is the force needed to accelerate 1 kilogram at a rate of 1 metre per 1 second per 1 second.
Force = mass * acceleration.
Starting as stationary, accelerating at 1 meter per second per second, in the first second you will only travelled only 0.5 meters
s=ut+0.5att
This is because, at the beginning of the second, you had no velocity, so even time was passing, you’ve not moved. As you’ve defined the acceleration as 1m/s/s, where your final velocity was 1m/s/s and your start velocity was zero, you are going to average a velocity of 0.5m/s and therefore travel 0.5m in the second.
s=((v-u)/2)t
This the mathematical difference between a calculated acceleration which is an average, and the calculus acceleration which is actually your real acceleration as you recognise it over longer distances. Both are correct but mean different things. See Velocity.
Energy at 1m/s/s is therefore 0.5J. If you up the acceleration to 2m/s/s the energy is 2J and if you accelerate at 3m/s/s you require the energy of 4.5J. The reason why is that your anchor point is the second. In 1 second, at higher acceleration, you must firstly apply more force to accelerate at a faster rate, you are pushing harder. Additionally, as you are accelerating faster, you cover more distance in a second. The double whammy is why you see the energy increase rapidly compared to the acceleration. If you are looking to cover the same distance at different accelerations, then you would require less time and use the same energy.

Energy

The golden rule of energy is that it cannot be destroyed. It can be transferred or transformed but it cannot be destroyed. If your equations show energy disappearing or just appearing, then you are doing something wrong or you don’t understand what is going on correctly. This is a law of nature and irrefutable.

Newton’s Laws of Motion

When viewed in an inertial reference frame, an object either remains at rest or continues to move at a constant velocity, unless acted upon by an external force.
The vector sum of the forces ‘F’ on an object is equal to the mass ‘m’ of that object multiplied by the acceleration vector ‘a’ of the object. F = ma. Note that 1 newton is the force it takes to change the velocity of 1 kilogram by 1 meter per second in 1 second.
When one body exerts a force on a second body, the second body simultaneously exerts a force equal in magnitude and opposite in direction on the first body.

If you are floating along in space with a ball in your hand, then within that inertial frame of reference you are still. You throw the ball. In the moment that you throw the ball you and the ball are accelerating, as the process that changes your velocity. As soon as you let go of the ball you both stop accelerating and continue in opposite directions relative to your mass. If you are 100 times the balls mass then the ball will travel away from where you were at 100 times the velocity you travel away from you were. As you have stopped applying force to the ball you are no longer accelerating so you and the ball will maintain your velocity. It will be constant.

Relativity

A quick word on relativity as it’s the thing following Newton and is the work of Einstein. If a gun fries a bullet at 1,200m/s can you catch it? If you try it here on earth it will blow your hand off and possibly your arm too! However, what if this was all in space and you are also traveling at 1,200m/s, could you grab the bullet? Yes, you could, because it is stationary to you. So the real question is in fact, ‘if a gun fries a bullet that leaves at a speed of 1,200m/s relative to the gun. If you are stationary relative to the gun, can you catch it? No. Normally on Earth, all speeds are relatively similar so you can simplify the question by assuming all items are stationary to each other. It’s known as a frame of reference. Back to the gun example where you, the gun and the bullet are within the same frame of reference. The fact the earth is spinning at 465m/s, traveling at 30,000m/s around the sun while the sun itself is moving through the galaxy at 200,000m/s is not relevant, as it applies to all three. You can therefore exclude them from the frame of reference. So why do I care right now, well acceleration changes relative velocities within the same frame of reference, or it can be said that it creates a new frame of reference. More on this later.

Kinetic and Potential Energy

Fortunately for us, Newton worked through F=ma into a kinetic energy equation:
Ek = 0.5mv²
This is a very important equation as it tells you how much energy is going to be in things that hit you once you know the relative velocity. The key thing to realise is the relativity of it. Its mirror is potential energy. You know if someone suspends a piano on a rope above your head and the rope is cut, the piano will definitely kill you. You know this because you know there is gravity pulling on it with a force, the force of gravity. You know that even though it is not moving there is potential energy there. The same applies if you throw a ball upwards. When you throw it, it has maximum kinetic energy but the force of gravity will decelerate it until a single instant where it’s motionless in the air. At the top of its flight it is still and has no kinetic energy but it has the equivalent potential energy to that of when you through it. As it accelerates down it will come back to your hand with the same kinetic energy as when you threw it... ignoring energy lost to friction with the air etc.

Thrust Energy and Momentum Energy Introduction

Be warned. This part is my own description of what is going on and so is not recognised as official physics. Do not use it in exams as it will not be understood. Also I haven’t been able to double check with a mathematician or physicist so it might just be plain wrong.

I propose subcategories of kinetic energy called momentum energy and thrust energy which deals with the shifting relativity of the velocity resulting from acceleration. It stems from something that I can best explain with the following thought experiment.

You are person A floating in space next to Space Station 0. With your space suit you mass 100kg. Your friend person B is about to come past at 2m/s so you fire up your thrusters for a second and match velocities. In that 1 second burst F=ma so you must have applied 200N of force. s=ut+0.5att so you must have travelled the distance of 1m. As E=Fs then you must have applied the energy of 200J. You can check this with the kinetic energy equation Ek=0.5mvv. Your fuel gauge confirms you’ve used 200J of energy (as everything is just perfect).

You have another friend, person C, passing Space Station 0 on your vector at a velocity of 4m/s. Relative to you and your friend, person B, they are doing 2m/s more. As they start to come past, you apply your thrust for 1 second again in the exact same way. You still have a mass of 100kg, you accelerate by 2m/s and so in doing so you travel 1m relative to person B. You’ve applied 200N of force and have a kinetic energy of 200J.

You do some mental arithmetic and work out that as A was 200J and B was 200J so A+B=400J. You check your fuel gauge and it says you’ve used 400J of energy. Wonderful. You realise that you are approaching a telescope called Telescope 0, and it is stationary relative to the Space Station 0, so you are approaching it at 4m/s. You decide to double check and confirm the kinetic energy. You accelerated your 100kg mass at 2m/s so the force remains 200N. The held thrust for a combined time of 2 seconds distance you travelled while under thrust was s=ut+0.5att which was 4m. Hang on this is twice as much. Ek=Fs so this would mean you have a kinetic energy of 800J! A+B=C, except it doesn’t!

Isaac Newton was wrong as is the kinetic energy equation Ek=0.5mv². This is so absurd it is completely laughable and I can assure you the equation is correct.
You’ve made energy from nothing. No experiment has ever shown this can happen. It can’t happen. The conservation of energy even underpins Newton’s work. It is the ultimate physical law and I cannot stress how important this rule is.
The workings were incorrect. Double check. You’ll find that everything is correct so it’s not this one.
Something else is going on here. This is true.

You probably already spotted that during the second thrust, you did travel 1m relative to person B and so gained 200J of kinetic energy relative to person B. However, during that time you were also already traveling at 2m/s so the distance you actually travelled in 1 second was 2m+1m=3m. So the total distance you travelled while under thrust relative to the space stations was the original 1m from the first thrust, plus the 3m from the second thrust. It still only cost you 400J to achieve that kinetic energy though so what is going on?

The answer is that after the first second, your acceleratory thrust shifted your relative base line so for the second acceleration you were operating on a plane of higher energy. (Technically speaking any acceleration is a shifting baseline but as infinitesimal moments of time are incomprehensible we as a people decided to quantify time by the second).

So I banged my head against a computer for 3 weeks working myself through my problem of having an engine with a defined wattage (energy per second). Reversing the equations to get the acceleration from the energy, my equations were telling me that after each second of sustained output, acceleration fell and after 10 seconds it was as good as nothing. The above example was how I eventually got my head around it and it explains why rocket engines don’t talk about chemical energy, just force/thrust or impulse. It leads me to the following conclusion. In order to represent the relativity shift brought on by acceleration, I’m breaking open kinetic energy into two components.

Thrust Energy per Second - þ aka Thorn

Thrust energy þ (Unicode 254) per second, I’ve labelled with the old English alphabet letter of lowercase thorn. For simplicity I’m now going to just call it thorn. It is the energy output of the engine in relation to the thrust over one second. It is therefore a measure of power. If acceleration is constant, the mass of the ship is constant, then thorn is constant. It goes back the source of energy calculation:
E=Fs
Energy = Force * distance
And
F=ma
Force = mass * acceleration

If you have an engine that is burning the same energy per second, it will be throwing out a constant force. As mass is constant so is acceleration. Knowing the distance travelled in a second from rest, we can calculate the energy. We can also crosscheck with the kinetic energy equation of
Ek = 0.5mv²
For example, if Engine X is going constantly at its top rate, it will be generating a constant acceleration of 2m/s². It has a constant mass of 1kg and so it’s generating a constant force of 2N. It will also be maintaining a constant þ of 2J/s or 2W. The engine is therefore power rated with a þ of 2 watts, although due to inefficiencies it actually burns 1.2þ (2.4J) of fuel each second. þ is constant relative to acceleration, that is the point of it.

With þ, all you need to know is the mass of the object the engine is pushing (including the engine) and you can work out the acceleration with the following equation:
acceleration = square root (2 * (thorn / mass)
a = √2(þ/m)

The HAAS European Class ships have a total mass of 500mkg (if the SI unit was grams and not kg then this would be 500bg) and 16 Merlin Engines, each with a þ of 16TW, totalling 256TW, then the acceleration is 32m/s/s.

Momentum Energy - Ð aka Eth

For a while I was tempted to call Momentum Energy by the name of Relativistic Energy as it relates to the total remaining kinetic energy not explained by þ. I have decided to call this Ð, which his uppercase Eth. Put simply
Ek = Ð + þt
so
Ð = Ek - þt

Ð represents the energy differential between two differing states of rest. Observationally and incidentally, the change in Ð, or the second order, seems to follow the relationship of
2þ*(t-1)
Time is very important to Ð.

In a Nutshell

In a nutshell, to convert engine power to acceleration just do all your engine calculations in a 1 second time frame and forget kinetic energy. Once known, just mark its thrust energy as þ. If you want, you can also note the force.

Notes

The origin of the word for a second of time is interesting. Time was ordered into 24 hours and sub-divided into 60 twice. The first division was into minutes and the second division was, well, the second division, or ‘the second’ for short.

Bonus Material

Ampersand & (Unicode 38)
Ash Æ (198) æ (230)
Eng Ŋ (330) ŋ (331)
Eth Ð (208) ð (240)
Ethel Œ (338) œ (339)
Ond (as in ‘and’ and ‘et’) ⁊ (8266)
That (Thorn with stroke) ꝥ (42853) Ꝧ (42854)
Thorn Þ (222); þ (254);
Wynn Ƿ (503) ƿ (447)
Yogh Ȝ (540) ȝ (541)