Which velocity formula is more accurate?

[Arugela]

I'm running something with 600 seconds of burn time at 7.3392 acceleration from 0 starting velocity.

Normal forumula:

600^2x7.3392x0.5= 1321056

My formula:

(total units /2 +.5) * total units (times variable of basic unit(IE: velocity)

((600/2)+0.5) *600 *7.3392 = 1323257.76

The second formula is for finding total iterations over a time period or number of iteration.

IE:

1+2+3+4+5+6+7+8+9 = 45

((9/2)+0.5)*9 = 45

Difference:

1323257.76-1321056 = 2201.76

 

difference of difference:

 

1323257.76 / 2201.76 = 601

1321056 / 2201.76 = 600

 

So, one is the 601st value and one is the 600th value.

What does this matter to real world calculations?

 

Easy solution is to add or subtract 1/600th or 1/601th dividend from the appropriate number to convert to whichever is more appropriate. Actually I need to make sure 600 isn't the result of the 600 seconds. If so then change 600 to time or time+1. If it's a seperate value then keep it the same.

 

I'm assuming one takes into account the 0 second and one doesn't in some manner. But my brain is too tired to think it out.

 

 

Edited September 5, 2023 by Arugela

[darthgently]

3 hours ago, Arugela said:

I'm running something with 600 seconds of burn time at 7.3392 acceleration from 0 starting velocity.

Normal forumula:

600^2x7.3392x0.5= 1321056

My formula:

(total units /2 +.5) * total units (times variable of basic unit(IE: velocity)

((600/2)+0.5) *600 *7.3392 = 1323257.76

The second formula is for finding total iterations over a time period or number of iteration.

IE:

1+2+3+4+5+6+7+8+9 = 45

((9/2)+0.5)*9 = 45

Difference:

1323257.76-1321056 = 2201.76

 

difference of difference:

 

1323257.76 / 2201.76 = 601

1321056 / 2201.76 = 600

 

So, one is the 601st value and one is the 600th value.

What does this matter to real world calculations?

 

Easy solution is to add or subtract 1/600th or 1/601th dividend from the appropriate number to convert to whichever is more appropriate. Actually I need to make sure 600 isn't the result of the 600 seconds. If so then change 600 to time or time+1. If it's a seperate value then keep it the same.

 

I'm assuming one takes into account the 0 second and one doesn't in some manner. But my brain is too tired to think it out.

 

 

This would be what is called a fencepost problem in programming.  Are you counting fenceposts or the spans of fence between posts?  There will be one less span than posts. You probably want the velocity at the end of each span with the first post as post number zero and the last post as number of spans + 1. 

So if you want the final velocity, I think, given what you wrote, you want the 601st velocity as that "post" is at the end of the last "span".  The reason it is 601 instead of 600 is because there are 600 "spans" and 601 "posts" (or seconds) starting at t=zero

All that said, I'm not certain I am reading your situation correctly so my answer may be junk, but maybe it will at least give another way to visualize the situation

Edited September 6, 2023 by darthgently

[Superfluous J]

The engineer in me wants to say, "What's the difference? The final velocity is the same in both cases: 1.32 million units."

[Shpaget]

Formula is

v=a*t, so 7,3392 *600 = 4 403,52

[RCgothic]

You appear to be calculating s=0.5*a*t^2, which is the formula for distance traveled under constant acceleration starting from rest over a time interval.

@Shpaget has the correct formula for velocity under constant acceleration starting from rest over a time interval.

Edited September 6, 2023 by RCgothic

[sevenperforce]

3 hours ago, RCgothic said:

You appear to be calculating s=0.5*a*t^2, which is the formula for distance traveled under constant acceleration starting from rest over a time interval.

@Shpaget has the correct formula for velocity under constant acceleration starting from rest over a time interval.

As usual @RCgothic is entirely correct but I will also point out, just in case OP missed it, that the constant acceleration equation will not work properly for a reaction engine. If you are shoving exhaust out of a nozzle, then you also have to account for the decrease in your total vehicle mass across the period of the burn. You'd need to use the Tsiokolvsky rocket equation for that.

15 hours ago, Arugela said:

Actually I need to make sure 600 isn't the result of the 600 seconds. If so then change 600 to time or time+1. If it's a seperate value then keep it the same.

The 600 is most definitely the result of the 600 seconds. You're just doing iteratively what s=0.5*a*t^2 does non-iteratively.

[Arugela]

I'll assume that has to do with the relative inefficiency of rockets? I assume because of the excessive thrust needed.

2 hours ago, sevenperforce said:

As usual @RCgothic is entirely correct but I will also point out, just in case OP missed it, that the constant acceleration equation will not work properly for a reaction engine. If you are shoving exhaust out of a nozzle, then you also have to account for the decrease in your total vehicle mass across the period of the burn. You'd need to use the Tsiokolvsky rocket equation for that.

The 600 is most definitely the result of the 600 seconds. You're just doing iteratively what s=0.5*a*t^2 does non-iteratively.

You are always technically throwing fuel out something... Do batteries loose weight also? Not sure how electric works.

Yea, I was trying to figure out distance over delta V. I was having a hard time finding how to figure that out. If constant fuel use is a thing can't you just add to the acceleration from the fuel mass loss? I couldn't think how to change that from delta v to actual distance produced simply. I'll assume that is the formula for rho etx and air density mixed in.

Stats:

Starting mass: 86400 tons

Fuel mass: 64800 tons

Fuel use: 108 tons per second

Maximum burn time: 10 minutes (600 seconds)

We really need standard formulas for acceleration in imperial as well as metric.. It gets confusing.

Edited September 6, 2023 by Arugela

[sevenperforce]

29 minutes ago, Arugela said:

I'll assume that has to do with the relative inefficiency of rockets? I assume because of the excessive thrust needed.

No, that doesn't have anything to do with inefficiencies; it's a fundamental aspect of rocket engines. A rocket engine has to push something out of the back end in order to produce forward thrust.

29 minutes ago, Arugela said:

You are always technically throwing fuel out something... Do batteries loose weight also? Not sure how electric works.

Well, some batteries lose weight during discharge and others gain weight during discharge, but most remain about the same. However, that's not quite the right comparison. You can use up a battery in any number of ways on board a spacecraft, but unless you're yeeting some sort of exhaust out of the back of your spacecraft, your spacecraft isn't going to move. Electric ion thrusters operate by using electrical energy to accelerate tiny amounts of ionized gas at high velocity, but even they are subject to changes in acceleration as the gas is used up and the vehicle becomes lighter. The Dawn spacecraft, for example, carried a total of 425 kilograms of xenon gas, about 35% of its total weight at launch.

29 minutes ago, Arugela said:

Yea, I was trying to figure out distance over delta V. I was having a hard time finding how to figure that out. If constant fuel use is a thing can't you just add to the acceleration from the fuel mass loss?

If fuel use is constant then yes, you have a constant change in acceleration and you can determine the distance covered accordingly. You just have to go to a third-order kinematic equation instead of the more familiar second-order kinematic equations.

Velocity is the rate at which position changes; an object moving at 10 m/s is changing its position by 10 meters every second. Acceleration is the rate at which velocity changes; an object accelerating at 10 (m/s)/s or 10 m/s² is changing its velocity by 10 m/s every second. The rate of change in acceleration is called "jerk" (or sometimes "jolt") and it has units of ((m/s)/s)/s or m/s³. 

The third-order kinematic equation for distance can be readily derived by integration, but I'll skip that step and just give it to you:

x = x₀ + v₀t + 0.5*a₀t² + 1/6jt³

In order to solve this equation, you'll need to know each of the starting values (x₀ , v₀, and a₀) as well as your value for jerk (j). In your case, the starting position x₀  seems like it is going to be zero, and the same appears to be true for the starting velocity v₀ as well. So you need to define a₀ and j.

If your initial acceleration is 7.3392 m/s², then that's your a₀. How do we calculate j? Well, with constant propellant use, j will be a constant, so we can use this equation (should be self-evident; let me know if it's not):

j = (a_f - a₀)/t

In order to accelerate a starting mass of 86,400 tonnes at 7.3392 m/s², you'll need an engine producing 634 meganewtons. By the end of your burn, your vehicle mass has dropped to 21,600 tonnes but you're still thrusting at 634 meganewtons, giving you a final acceleration of 29.36 m/s². Using our equation above, we get jerk equal to 29.36 m/s² minus 7.3392 m/s² divided by 600 seconds, or 0.0367 m/s³.

Now that we have that, we plug all the rest in. 

x = x₀ + v₀t + 0.5*a₀t² + 1/6jt³

 

x = 0 + 0*600 sec + 0.5*7.3392 m/s² * (600 sec)² + 1/6*(0.0367 m/s³)*(600 sec)³

x = 0 + 0*600 sec + 3.6696 m/s² * 360,000 s² + 0.00612 m/s³ * 216,000,000 s³

x = 1,321,056 m/s² * s² + 1,321,920 m/s³ * s³

x = 2,642,976 meters

So there's your answer -- which you will note is about double the answer you would have gotten without accounting for change in acceleration.

29 minutes ago, Arugela said:

I'll assume that is the formula for rho etx and air density mixed in.

Nope, it is not. Kinematic equations assume no air resistance. If you're talking about something moving through an atmosphere then we're gonna have a whole other set of issues.

45 minutes ago, Arugela said:

We really need standard formulas for acceleration in imperial as well as metric.. It gets confusing.

The standard kinematic formulae are unit-agnostic.

[Arugela]

Here's a bad joke or a real world measurement:

86400tons*2000=172800000*47.880172=8273693721.6/2205=3752242.05061224489795918367

What measurement unit is it?

Or is that the wrong math?

Should it be 8273693721.6/2000=4136846.8608*2205= 9121747328.064

Are any of these an an existing unit?

Edited September 6, 2023 by Arugela

[Kerbart]

Your calculation doesn't take into account the diminishing mass during the burn -- check the rocket equation for that.

[Arugela]

Does that mean that the two formulas added together is simply the complete distance traveled?

And what measurement unit is that then? My brain goes screwy still when going between lb and metric.

Edited September 6, 2023 by Arugela

[sevenperforce]

16 hours ago, Arugela said:

86400tons*2000=172800000*47.880172=8273693721.6/2205=3752242.05061224489795918367

What measurement unit is it?

Or is that the wrong math?

Should it be 8273693721.6/2000=4136846.8608*2205= 9121747328.064

Are any of these an an existing unit?

This is the wrong math. You're not providing any units at all (except for "tons" up at the top, and even then I don't know whether you're talking about metric tonnes (1000 kg) or Imperial tons  (2000 lbs).

Include all your units for the entire equation. Units are algebraic; treat them like variables and multiply/add/divide as necessary. Without including units, your numbers are gibberish.

Also, can you contextualize the system a little bit for us? I have no idea what you're trying to figure out, which makes this all very challenging to characterize. Is this a launch vehicle? Are you burning in a straight line or along a curved trajectory? Are you already in orbit when the burn begins? All of this is important.

16 hours ago, Arugela said:

Does that mean that the two formulas added together is simply the complete distance traveled?

No. Use one formula at a time. If you need to combine formulae, use unit substitution (e.g. if you know that a = b + c², but c = b-3, then you can substitute b-3 in place of c to get a = b + (b-3)² so as to solve for a in the equation).

16 hours ago, Arugela said:

And what measurement unit is that then? My brain goes screwy still when going between lb and metric.

Best way to avoid your brain going screwy over units is to actually use units consistently. 

Metric units, unlike Imperial units, have the auspicious property of automatically cancelling if you use the correct base units. For example, if I know the propellant consumption (kilograms/second) and effective exhaust velocity (meters/second) of a jet airliner traveling at cruising speed, I know I can multiply them together to get cruising thrust, because (kilograms * meters) / (second * second) equals kg*m/s² which is equal to Newtons. And since Newtons of thrust are equal to Newtons of drag at constant speed, I can divide those Newtons by ρ*C_D*(velocity)²/2 (where ρ is air density and C_D is the drag coefficient) to get the cross-sectional area of the aircraft, because Newtons are kg*m/s², and kg*m/s² divided by density (kg/m³) and velocity squared (m²/s²) gives me units of m² which is area.

In those situations I may not even bother writing out my units because I know they are going to cancel out properly. But since you're not familiar with the correct base units I strongly suggest keeping track of the units consistently and doing the cancellations manually, or you'll miss something.

Obligatory funny:

Spoiler

 

[Arugela]

That is why I showed the dividing in the formula. The answer was the newtonne...

[sevenperforce]

10 minutes ago, Arugela said:

That is why I showed the dividing in the formula. The answer was the newtonne...

A division sign doesn't tell anyone anything about units. And do you mean newtons or tonnes? Good grief.

Pick your units and stick to them.

[Arugela]

It was a joke... If I divided by 2000 then multiplied by 2205 that was the hint. That was the unit.

Edited September 7, 2023 by Arugela