Talk:Delta-v

message: sorry I did the last edit while not logged in (the need for delta-v to change inclination). Carrionluggage 05:04, 23 January 2006 (UTC)
thread_title: Untitled

message: could a list of approximate Delta-v's between various locations (planetary surfaces, important planetary orbits (low orbit, rotationally synchronous orbits, etc.), Lagrange points for various planets, moons of the gas giants, etc.) be given? Or is there already one somewhere? Linguofreak 03:25, 19 March 2006 (UTC)
thread_title:  Delta V chart 

message: The "Delta-V" is not really a "scalar" but rather a "vector quantity" with the dimension "velocity" defined
:\bar{\Delta v} = \int_{t_0}^{t_1} {\frac {\bar{T}} {m}}\, dt
where \bar{T}\, is the force caused by the thrust (vector)
The integral
:A=\int_{t_0}^{t_1} {\frac {|T|} {m}}\, dt
is of interest as a measure of the fuel usage only, from the rocket equation
:M_1=M_0\ e^{-\frac{A}{V}}\, dt
where
M_1\, is the mass at time t_1\,
M_0\, is the mass at time t_0\,
V\, is the exhaust velocity
If the direction of thrust is constant this is more or less the same thing but for a rotating spacecraft operating radial thrusters in pulsed mode it is not.
And that is it! Why discuss "g-force" as a concept here?  Pointless!
Stamcose (talk) 15:53, 27 July 2011 (UTC)
replies: text: Doesn't the bar notation either side of the T mean that the magnitude of the force is taken, and hence wouldn't the magnitude of the force being generated be the same in a rotating spacecraft at all times?- Sheer Incompetence (talk) Now with added dubiosity! 15:56, 1 August 2011 (UTC)
thread_title:  Critics of introduction 

message: :\Delta{v} = \int_{t_0}^{t_1} {\frac {|T|} {m}}\, dt
::change to
:::\Delta{v} = \int_{t_0}^{t_1} {\frac {T} {m}}\, dt
where
:T is the instantaneous thrust
: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:
: =  \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 (t_1 - t_0)/2 then the velocity difference is  {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.
::Strange and superflous discussion, skip
Stamcose (talk) 20:35, 9 August 2011 (UTC)
thread_title:  Critics of section "Definition" 

message: 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)
replies: text: 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)
replies: text: 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)
replies: text: Fine for me. Gusta (talk) 17:43, 24 March 2013 (UTC)
replies: text: I don't think it's a good idea for the same reason I already gave.Teapeat (talk) 18:30, 24 March 2013 (UTC)
replies: text: 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)
replies: text: 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)
thread_title:  Simplification 

message: 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".  </br>
LP-mn (talk) 15:22, 29 July 2014 (UTC)
thread_title:  Pronunciation 

message: 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)
replies: text: 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)
thread_title:  KSP 

message: 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)
thread_title:  Scope of this article is very poorly defined 

message: 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)
replies: text: 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)
thread_title:  Section for stupid people 

message: 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:
:\Delta v = \int^{t1}_{t0} \frac{|T|}{{m_0}-\Delta{m} {t}} ~ dt
where m_0 is the initial (wet) mass and \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,
:\int F ~ dt = J
The Integral is found to be:
: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,
: \frac{J}{\Delta m} = \frac{F}{p} = V_{exh}
you can equate your integral to
:\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)
thread_title:  A simpler derivation of the Rocket Equation from Delta V 

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thread_title:  External links modified 

message: 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)
thread_title:  LEO to Sun calculation looks like original research