Talk:Delta-v

As I already pointed out in the section above, I think the introduction should be re-written!

But also

"Producing Delta-V" seems rather pointless (little contents)

"Delta-v budgets" is (in my opinion!) not very good and is anyway misplaced in this article. And the statement "Additionally raising thrust levels (when close to a gravitating body) can sometimes improve delta-v" is nonsens, the Delta-V is the same but the effect of the Delta-V is larger (in terms of energy, hello Oberth!)

"Oberth effect" is (in my opinion!) not very good and anyway misplaced in this article. And this is a very obvious "effect" that hardly deserves a name (Sorry Oberth!).

"Porkchop plot" could possibly be made a separate article but this has little to do with the definition of "Delta-V" and is misplaced in this article.

I would propose to reduce the "Delta-V" article to:

+++++++++++++++++++++++++++++++++++

ΔV or Delta-V (literally "change in velocity"), is a term used in astrodynamics.

Orbit maneuvers are made by firing a thruster to produce a reaction force acting on the spacecraft. The size of this force will be

${\displaystyle f=V_{exh}\ \rho \,}$

1

where

${\displaystyle V_{exh}\,}$ is the velocity of the exhaust gas

${\displaystyle \rho \,}$ is the propellant flow rate to the combustion chamber

The acceleration ${\displaystyle {\dot {V}}\,}$ of the spacecraft caused by this force will be

${\displaystyle {\dot {V}}={\frac {f}{m}}=V_{exh}\ {\frac {\rho }{m}}\,}$

2

where ${\displaystyle m\,}$ is the mass of the spacecraft

During the burn the mass of the spacecraft will decrease due to use of fuel, the time derivative of the mass being

${\displaystyle {\dot {m}}=-\rho \,}$

3

If now the direction of the force, i.e. the direction of the nozzle, is fixed during the burn one gets the velocity increase from the thruster force of a burn starting at time ${\displaystyle t_{0}\,}$ and ending at ${\displaystyle t_{1}\,}$ as

${\displaystyle \Delta {V}=-\int _{t_{0}}^{t_{1}}{V_{exh}\ {\frac {\dot {m}}{m}}}\,dt}$

4

Changing the integration variable from time ${\displaystyle t\,}$ to the spacecraft mass ${\displaystyle m\,}$ one gets

${\displaystyle \Delta {V}=-\int _{m_{0}}^{m_{1}}{V_{exh}\ {\frac {dm}{m}}}\,}$

5

Assuming ${\displaystyle V_{exh}\,}$ to be a constant not depending on the amount of fuel left this relation is integrated to

${\displaystyle \Delta {V}=V_{exh}\ \ln({\frac {m_{0}}{m_{1}}})\,}$

6

which is the well known "rocket equation"

If for example 20% of the launch mass is fuel giving a constant ${\displaystyle V_{exh}\,}$ of 2100 m/s (typical value for a hydrazine thruster) the capacity of the reaction control system is

${\displaystyle \Delta {V}=2100\ \ln({\frac {1}{0.8}})\,}$ m/s = 469 m/s.

If ${\displaystyle V_{exh}\,}$ is a non-constant function of the amount of fuel left^[1]

${\displaystyle V_{exh}=V_{exh}(m)\,}$

the capacity of the reaction control system is computed by the integral (5)

The acceleration (2) caused by the thruster force is just an additional acceleration to be added to the other accelerations (force per unit mass) affecting the spacecraft and the orbit can easily be propagated with a numerical algorithm including also this thruster force. But for many purposes, typically for studies or for maneuver optimization, they are approximated by impulsive maneuvers having the ${\displaystyle \Delta {V}\,}$ given by (4) as illustrated in figure 1

This approximation with impulsive maneuvers is in most cases very accurate, at least when chemical propulsion is used. For low thrust systems, typically electrical propulsion systems, this approximation is less accurate. But even for geostationary spacecraft using electrical propulsion for out-of-plane control with thruster burn periods extending over several hours around the nodes this approximation is fair.

+++++++++++++

This covers the old "Introduction" and "Producing Delta-V" leaving out the misplaced text

Any objections?

Stamcose (talk) 16:17, 31 July 2011 (UTC)

I now update the article! If somebody feels that the "Porkchop plot" is useful and should be a separate article, please go ahead! I personally do not think so!

Stamcose (talk) 13:52, 1 August 2011 (UTC)

Articles are supposed to cover the whole subject, but you seem to have just removed things arbitrarily, so I've reintroduced them. You should really only take things out if they are off-topic or completely not neutral to include them, but clearly delta-v budgets (for example) are notable, and on topic.- Sheer Incompetence (talk) Now with added dubiosity! 15:53, 1 August 2011 (UTC)

I think "off-topic" is precisely the right word!

Here the very concept of "Delta-V" should be explained. Texts that for example discuss strategies to reach planets ("Porkchop plot") does not belong here although they necessarily refer to this concept "Delta-V"! And "Oberth effect" is already a seperate article!

There is for example an article Money in Wikipedia. But this article does not (and should not) contain any discussion about the state budget of US although such a text necessarily would contain many numbers with the dimension $! Nor should it contain any comparison of the cost of living in different cities just because dollars, euros, pound and yens are mentioned!

That should be separate articles!

Stamcose (talk) 06:34, 2 August 2011 (UTC)

Explicit reasons to delete sections:

Producing Delta-v

Delta-v is typically provided by the thrust of a rocket engine, but can be created by other reaction engines. The time-rate of change of delta-v is the magnitude of the acceleration caused by the engines, i.e., the thrust per total vehicle mass. The actual acceleration vector would be found by adding thrust per mass on to the gravity vector and the vectors representing any other forces acting on the object.

This is now clearly said in the introduction! This is now just a repetition!

The rocket equation shows that the required amount of propellant dramatically increases, with increasing delta-v. Therefore in modern spacecraft propulsion systems considerable study is put into reducing the total delta-v needed for a given spaceflight, as well as designing spacecraft that are capable of producing a large delta-v.

"reducing the total delta-v needed for a given spaceflight" is mission design, not propulsion system design and this has nothing to do with "modern", this has always been the target for mission design! "designing spacecraft that are capable of producing a large delta-v"? Should rather say "sufficent delta-v with a reaction control system as light as possible"

Additionally raising thrust levels (when close to a gravitating body) can sometimes improve delta-v.

That is obviously completely wrong! The delta-v is the same but its effect on the orbit is stronger!

"Delta-v budgets"

"When designing a trajectory, delta-v budget is used as a good indicator of how much propellant will be required. Propellant usage is an exponential function of delta-v in accordance with the rocket equation, it will also depend on the exhaust velocity"

A screwed-up statement! Yes, the "rocket equation" links delta-v with propellant use in terms of kg propellant

It is not possible to determine delta-v requirements from conservation of energy by considering only the total energy of the vehicle in the initial and final orbits since energy is carried away in the exhaust (see also below).

For example, most spacecraft are launched in an orbit with inclination fairly near to the latitude at the launch site, to take advantage of the Earth's rotational surface speed. If it is necessary, for mission-based reasons, to put the spacecraft in an orbit of different inclination, a substantial delta-v is required, though the specific kinetic and potential energies in the final orbit and the initial orbit are equal.

One should rather say what is true rather than to list all what not is true. But is there some special reason to think that the reader should belive that "conservation of energy" has anything to do with the matter?

When rocket thrust is applied in short bursts the other sources of acceleration may be negligible, and the magnitude of the velocity change of one burst may be simply approximated by the delta-v. The total delta-v to be applied can then simply be found by addition of each of the delta-vs needed at the discrete burns, even though between bursts the magnitude and direction of the velocity changes due to gravity, e.g. in an elliptic orbit.

Absolutely twisted text! Does the author wants to say that the "impulsive manoeuvre approximation" using one impulse can cover the effect of several burns if these burns are in rapid sequence?

Stamcose (talk) 09:53, 2 August 2011 (UTC)

OK, a few points:

• delta-v is a scalar, not a vector
• introductions in Wikipedia articles are supposed to be just a few paragraphs
• we nearly always only have one reference section in Wikipedia's articles, and it goes at the end (it is possible to have more than one, but not the way you've done it).
• this isn't an article about the rocket equation, because we have an article on that elsewhere, it's about delta-v, so replacing the article with a rederivation of that seems inappropriate
• I'm not entirely convinced you understand the topic well enough to be editing here if you don't know it's a scalar. Some of your comments above seem to be implicitly assuming it's a vector, but it makes no sense considered in that way.
• Wikipedia articles are supposed to cover all aspects of the topic as much as possible, removing other sections like delta-v budget in particular, that's wrong.
• The article can certainly be improved greatly, please just make sure that it's unarguable that that's what you're actually doing with your edits.

Hope this helps.- Sheer Incompetence (talk) Now with added dubiosity! 14:41, 2 August 2011 (UTC)

Ok, I close this matter now. I am not going to continue the argumentation with Mr Incompetence and als not wage any "editing war". What should be said about the article as it stands in this moment has already been said! The Wikipedia system has in it advantages and its disadvantages! Some articles get good but the system does not always work! Actually, it is wonder that it works well that often considering that absolutly everybody is allowed to edit everything!

"if you don't know it's a scalar"

I guess this introduction was written by Mr Incompetence himself as it so much stresses "scalar" which seems to be his favourite thesis!

Anyway, somebody interested in the topics will also read this "Discussion"! That is what I always do myself to see if an article is disputed or not!

Stamcose (talk) 16:44, 2 August 2011 (UTC)

${\displaystyle \Delta {v}=\int _{t_{0}}^{t_{1}}{\frac {|T|}{m}}\,dt}$

change to

${\displaystyle \Delta {v}=\int _{t_{0}}^{t_{1}}{\frac {T}{m}}\,dt}$

where

${\displaystyle T}$ is the instantaneous thrust

${\displaystyle m}$ is the instantaneous mass

+++++ stop here ++++++

If there are no other external forces than gravity, this is the integral of the magnitude of the g-force.

Pointless to mention "g-force"

In the absence of external forces, and when thrust is applied in a constant direction this simplifies to:

${\displaystyle =\int _{t_{0}}^{t_{1}}{|a|}\,dt=|{v}_{1}-{v}_{0}|}$

which is simply the magnitude of the change in velocity. However, this relation does not hold in the general case: If, for instance, a constant, unidirectional acceleration is reversed after ${\displaystyle (t_{1}-t_{0})/2}$ then the velocity difference is ${\displaystyle {v}_{1}-{v}_{0}=0}$, but delta-v is the same as for the non-reversed thrust.

Superflous (and somewhat ridiculos!) discussion, skip

For rockets the 'absence of external forces' usually is taken to mean the absence of atmospheric drag as well as the absence of aerostatic back pressure on the nozzle and hence the vacuum Isp is used for calculating the vehicle's delta-v capacity via the rocket equation, and the costs for the atmospheric losses are rolled into the delta-v budget when dealing with launches from a planetary surface.^{[citation needed]}

Strange and superflous discussion, skip

Stamcose (talk) 20:35, 9 August 2011 (UTC)

What you guys think of deleting this part of the article: "Delta-v is produced by the use of propellant by reaction engines to produce a thrust that accelerates the vehicle." It is only a orbital manouver, it can be by an explosion or something like that, not only by propellant. Gusta (talk) 03:12, 24 March 2013 (UTC)

I don't think that's a good idea, delta-v is something to work out how much propellant you need.Teapeat (talk) 04:18, 24 March 2013 (UTC)

How about ""Delta-v is typically produced in spacecraft by the use of propellant in reaction engines to produce a thrust that accelerates the vehicle."? --agr (talk) 16:01, 24 March 2013 (UTC)

Fine for me. Gusta (talk) 17:43, 24 March 2013 (UTC)

I don't think it's a good idea for the same reason I already gave.Teapeat (talk) 18:30, 24 March 2013 (UTC)

Delta-v doesn't always involve propellant, there is aerobreaking, tank venting, etc. I think "typically" covers the situations you are concerned with.--agr (talk) 23:07, 24 March 2013 (UTC)

Many things are roughly equivalent to delta-v, but real delta-v follows the rocket equation (tank venting forms a reaction engine, but many types of aerobraking don't follow the rocket equation and can have infinite Isp.)Teapeat (talk) 02:46, 25 March 2013 (UTC)

Please could someone write a sentence or two explaining how delta-v is different from acceleration. "For a typical debris avoidance maneuver, the [ISS] station will be subjected to delta V of between 0.5 and 1 meter per second." -- is this not an acceleration? (from http://arstechnica.com/science/2013/07/how-nasa-steers-the-international-space-station-around-space-junk)

It is unclear why delta-v is used at all instead of acceleration.

Also please include a formula connecting delta-v and acceleration. Thanks. Lehasa (talk) 19:06, 5 July 2013 (UTC)

It's not an acceleration. A metre per second has units of velocity not acceleration.

And as the article already points out delta-v is the integral of the magnitude of the acceleration over time.GliderMaven (talk) 19:34, 5 July 2013 (UTC)

I recently heard of an audiobook reader who pronounced it "Delta-Five". I'm going to insert a comment in the leading paragraph about this. If you feel it needs to be moved, so be it; but please don't discard or delete it as being too "obvious".
LP-mn (talk) 15:22, 29 July 2014 (UTC)

Is the section "Simulation" actually needed? I feel that it's simply a plug for the game instead of giving any actual information. — Preceding unsigned comment added by SpinkZeroZero (talk • contribs) 19:50, 1 September 2014 (UTC)

I agree that the section does not belong in this article at all. I'll be removing the section and request temporary protection as the article is currently on the front page of the KSP subreddit. Andyops (talk) 01:22, 2 September 2014 (UTC)

I haven't been able to find a cleanup template that precisely says this, but this article suffers greatly from a lack of direction. The term "delta-v" is in fact rather generic; it is a fundamental parameter in kinematics, and means simply a change in velocity of any object or particle. It happens to be applied to astrodynamics of space vehicles, and the introduction rather arbitrarily usurps that single definition. You don't have to be talking about rockets or spacecraft necessarily, to use the term delta-v. JustinTime55 (talk) 18:42, 15 December 2014 (UTC)

Articles are not on terms, they are on topics. The topic here is the scalar delta-v that is used in rocketry. You can see from the equation it's a scalar, it's the integral of a scalar, the magnitude operator turns the vector into a scalar.GliderMaven (talk) 16:17, 17 February 2015 (UTC)

Your response to my recent edit doesn't have anything to do with this particular thread the above thread on "Scope of this article". And your objection to the word vector is no excuse for performing an unthinking reversion of all the copyedit changes I made to the lead, the most significant of which is removing the lame phrase "amount of effort". You could have very easily just changed "vector" back to "scalar" without reverting. And a scalar quantity (which never had any direction asociated with it) is not really the same as the magnitude of a vector. (And if you really think delta-v is not a vector, I certainly wouldn't want you to navigate any spaceflight I made!) JustinTime55 (talk) 18:57, 18 February 2015 (UTC)

There's another article on the vector quantity, and nothing links to it and that article has no references either, basically because it's unimportant. Whereas the topic this article is on, is critical to astrodynamics; high priority.GliderMaven (talk) 20:05, 18 February 2015 (UTC)

Formally, this isn't a vector quantity, vectors components add and subtract linearly. This doesn't do that; you can only add delta-v.GliderMaven (talk) 20:12, 18 February 2015 (UTC)

(I refactored this, because the topic is completely different. You apparently have trouble grasping that my recent edits have absolutely nothing whatever to do with the fact I thought the topic scope wasn't general enough back in December. I accept the fact that the scope of this article is limited to rocket calculations. The existence of other articles is completely irrelevant to this.

You are quite wrong about spacecraft delta-v not being a vector; spacecraft do not fly in one dimension. If a craft is traveling in one direction, and its desired velocity is at, say a 60 degree angle from this, then you can't simply subtract the new velocity from the old one, and the rocket thrust is not pointed in either the new nor the old direction; you must do the vector calculation to get the proper magnitude of the delta-v. Then it becomes a scalar quantity, because the thrust and the delta-v are both in the same direction. Read what I have written; I went to great pains to explain this.

I repeat what I said about not wanting you to navigate a spacecraft on which I would fly; you would surely get us lost. JustinTime55 (talk) 20:34, 18 February 2015 (UTC)

Ignoring your personal attack, that's not ever referred to in rocketry as delta-v, nor is the delta-v they use in rocketry the integral of the change in velocity as you wrote, nor have your changes been based on any references.GliderMaven (talk) 20:41, 18 February 2015 (UTC)

• I never wrote "delta-v is the integral of the change in velocity"; where exactly do you see that? Do you have trouble interpreting what you read objectively? (And that is not a "personal attack", but a good-faith attempt to understand your reactions.) You have demonstrated a tendency to "shoot from the hip" and have made several incorrect assumptions about what I wrote and why I wrote it.
• The whole reason they call it "delta-v" is because it's based on the velocity change required (including direction, which is often not in the same direction as current or final velocity), which they must know at the start. This is determined by methods outside of the scope of this article; you can't start talking about rocket calculations until you know the direction in which the rocket fires. Once you know that, then it becomes a "scalar" to calculate how much fuel is required, and this is becasuse the thrust is in the same direction as the delta v.
• A spacecraft hovering against gravity doesn't change this; as the text said, simplifying assumptions are made (no gravity). If it is hovering against gravity, then that term (negative) must be added, and the delta-v balances against gravity (for a net delta-v (physics) of 0.)
• Your wholesale reversion, while easy, also took out other improvements I made, such as calling it the measure of "effort", and the confusing change of symbol from T to f for thrust.
• Pot calling the kettle black: what I wrote is as much based on references, as what you reverted it to (which is, not at all). Where are the reliable sources which support (your?) version? JustinTime55 (talk) 21:10, 18 February 2015 (UTC)

The existing definition of delta-v in the article does not ignore gravity. Your changes give the wrong answers. A rocket hovering against gravity uses enormous delta-v if maintained for any length of time, whereas what you wrote defines it as zero.GliderMaven (talk) 21:39, 18 February 2015 (UTC)

Could we get like a section that contains a laymans summary for people who don't have a degree in physics or mathematics? People who don't know what the difference is between a "scalar" and a "vector" is, and don't really care, but are interested to know what "delta-v" is , and how it compares to other measurements like "pounds of thrust", or "kilonewtons". There are some articles that pretty much require an assumption that the person reading it will have a working knowledge of the subject, but I think "delta-V" is basic enough that the concept ought to be within the grasp of mere humans, if it was written right. I note above in the comments that someone once advocated re-writing the entire article basically as a series of formulas interspersed with lines of text leading from one formula to the next. While I'm sure this is ideal for people who know what they are looking at, to many people, complex formulas, with their strange symbols and terms, are about as clear as Greek, or a fraternity cipher. And it would seem unfair to write an entire article in such a fashion that it is inaccessible to those who haven't passed the initiation and joined the club. There should be at least a nod towards those who haven't got more than a high school education, but are curious to learn what they can. Even if you find such things vulgar..45Colt 00:15, 6 September 2015 (UTC)

Well, this topic quite literally is rocket science, so many people will find it confusing, at least initially. I'd recommend playing with Kerbal Space Program for a while and it may start to come into focus for you. Once you understand it, feel free to make suggestions on how it may be improved.GliderMaven (talk) 11:29, 6 September 2015 (UTC)

Although the page's is more detailed and makes perfect sense, There is a much simpler derivation of the rocket equation, which goes as follows. By representing the Delta-v equation as the following:

${\displaystyle \Delta v=\int _{t0}^{t1}{\frac {|T|}{{m_{0}}-\Delta {m}{t}}}~dt}$

where ${\displaystyle m_{0}}$ is the initial (wet) mass and ${\displaystyle \Delta m}$ is the initial mass minus the final (dry) mass,

and realising that the Integral of a resultant force over time is Total Impulse, assuming thrust is the only force involved,

${\displaystyle \int F~dt=J}$

The Integral is found to be:

${\displaystyle J~{\frac {Ln({m_{0}})-Ln({m_{1}})}{\Delta m}}}$

Realising that Impulse over the change in mass is equivalent to Force over propellant mass flow rate (p), which is itself equivalent to exhaust velocity,

${\displaystyle {\frac {J}{\Delta m}}={\frac {F}{p}}=V_{exh}}$

you can equate your integral to

${\displaystyle \Delta v=V_{exh}~ln({\frac {m_{0}}{m_{1}}})}$

which is the Tsiolkovsky rocket equation.

I added this to the Rocket equation page instead. — Preceding unsigned comment added by 86.171.45.62 (talk) 22:14, 3 February 2016 (UTC)

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Text says "the figure of 30 for LEO to the sun is also too high.[7]" but the reference [7] looks like an OR calculation and seems wrong. Perhaps that comment should go on the image talk page so it can be discussed ? (perhaps it would be better if the line went from the Sun to "earth C3=0" ?) - Rod57 (talk) 13:51, 29 May 2018 (UTC)