# Talk:Specific impulse

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## Second sentence

Article second sentence: It represents the impulse (change in momentum) per unit amount of propellant used. The reference says Specific impulse is the change in momentum per unit mass for rocket fuels, or rather how much more push accumulates as you use that fuel. This hardly sounds like a technical definition. In fact it is not even correct. For example, the rocket described in this article whose variable effective exhaust speed is equal to the rocket speed, giving the rocket constant momentum (because the momentum of the exhaust is zero, and the total is conserved). change in momentum per (anything) in that case would be zero, but the Isp is not zero because Veff is not zero (and the rocket is producing thrust). I have seen Isp definitions as thrust/flow. We frequently convert thrust to momentum by integrating because d(momentum)/dt = force. But this equation is not always true! d(mv)/dt = m dv/dt = ma = F IF the mass is constant, but for a rocket the mass is not constant.

It seems someone went to a lot of trouble to figure out a clean way to make kgf/kg go away, when all of the technical sources I have seen (not that many) have simply said "This is a kludge, but it is what people do and what they are used to, and maybe useful for ...". If someone has done this they have violated so many WP policies I do not know where to start. It does not matter what we think is clean, proper, or useful, or the way we think things should work. We may be right, but that is not the way WP works, trying to work that way leads to all sorts of silly conflicts, and in this case it has led to the second sentence of the article, the definition of the term, being incorrect. And no, it would not be useful for some clever person to argue that something has the appropriate change in momentum. My point is that the effort on this article is already too full of clever, and not enough of encyclopedia. Would someone please pull some references from textbooks, instead of trying to write one?

Get it together, people! 69.231.124.52 (talk) 09:40, 12 August 2011 (UTC)

I think you've identified some bad English in the reference article and perhaps some other should be used. But (other than need to delete this reference) can you find something that is actually wrong with the Wikipedia article? For example, I don't see any kgf/kg problems here. It's all in correct units (velocity OR time). SBHarris 00:05, 13 August 2011 (UTC)
I guess the source of errors in articles is not as important as correcting them. The second sentence uses impulse as (change in momentum), which follows from the unauthoritative source, but over 2 out of 3 physics sources I found in a quick search define impulse in terms of force and time, typically f times delta t or integral f dt. Then they show that, for constant mass, impulse is change in momentum, but change in momentum is not the actual definition in most cases.
The second sentence, which defines the term specific impulse, is incorrect. Instead of It represents the impulse (change in momentum) per unit amount of propellant used it should read It represents the impulse (force integrated over time) per unit amount of propellant used. A reference for this is "Rocket Propulsion Elements", George Paul Sutton and Oscar Biblarz, John Wiley and Sons, 2001, page 28. This reference says "per unit weight of propellant" instead of "amount", but hopefully you folks have another reference that says "amount".69.231.149.52 (talk) 00:08, 14 August 2011 (UTC)
The second sentence IS correct. Impulse is change in momentum for the general case, not just the constant-mass case. I = dp = Fdt. You must do the integral if your force is not constant, but only if you must write in terms of force. If you never do, you can keep change in impulse or momentum as dp and then specific impulse is Isp = dp/dm = change in momentum per change in mass. SBHarris 23:42, 14 August 2011 (UTC)
dp <> Fdt for something (such as a rocket) whose mass is not constant. There does seem to be a great deal of confusion here. For instance Newton's second law was equivalent to F = dp/dt, but modern sources point out that Newton was talking about the reaction of a specific body with a constant mass. In any case, this article might prefer sources that are relevant to specific impulse, such as those on rocket propulsion. One could define I = dp or I = Fdt but most authoritative sources define impulse in terms of force and time, e.g. dI = Fdt. Isp = F/wdot (assuming F constant) is a basic equation for specific impulse which works given dI = Fdt. These equations can be found in the source cited above. 69.231.117.240 (talk) 05:37, 15 August 2011 (UTC)

───────────────────────── I believe I see what you're driving at. I should remark that total impulse (I) already signals a change in momentum, so I (not ΔI) = ΔP. You must keep the same numbers of operators on each side of the equation unless dealing with something defined as a change in something else. You can write I = ΔP but you cannot write I = dp as I did. In differential form this actually is dI = dp, since d(ΔP) = d(P-P_o) = dP. Thus dI = dp = Fdt gives the differential impulse and differential change in momentum over the period of time dt, and is correct (since mass does not change during dt time, our changing mass is no problem). It is perfectly true that dp = Fdt, however Fdt cannot be integrated meaningfully over time if the mass changes and you're focused only on part of the mass. The problem is that the thrust force first pushes fuel molecules toward the exhaust that it had formerly accelerated in the other direction, when the same fuel was still in the rocket. For example when the rocket is traveling as fast as its exhaust, fuel leaves at net zero velocity in the starting frame, so what is the "impulse" on that fuel bit? It's been greatly accelerated, then stopped, so it has no net momentum change, and thus how can we calculate its "impulse"? It serves on one side of the momentum equation at the beginning of flight (as fueled rocket), and on the other side at the end (as reaction mass). The impulses in both directions cancel each other.

Thus, you cannot integrate Fdt to get total impulse in a changing-mass rocket, because the impulse has been applied to the rocket AND (in both directions) some of the fuel. However, do long as it stays in a differential form Fdt, you can divide it by dm to get F dt/dm = F/[dm/dt] = thrust/[fuel consumption] = I_sp. The reason for this is that neither differential time or differential mass are bothered by the changing mass-with-time problem. However, this works only if you DO NOT try to integrate Fdt. Any definition of p or I that involves integral-Fdt is wrong, if mass changes over time, even if F does not. Even Impulse = F*Δt is wrong, because some of that "impulse" has been doubly applied in two directions, and not to same object. So even if you figure it out, what are you doing to do with it?

However I_sp = F/[dm/dt] = dP/dm is fine. It will be a constant, if both F and dm/dt are constant, or their ratio is constant. This means the average exhaust velocity is indeed the differential change of momentum per change in rocket mass (or per unit of fuel consumption). So yes, the definition is right.

The idea that dp/dt can be used for F even in a changing mass rocket works only if the entire system is under consideration, so that total force from outside the system is zero, so dP/dt is 0. Then ΔP/Δt = 0 also, and from this you can derive the rocket equation. This equation is easy to integrate for constant thrust even if mass changes drastically, so obviously Isp can be defined in those circumstances (it's just the average exhaust velocity, which substitutes for it, when I_sp is in velocity units). The integration is more difficult if the thrust changes (as a function of time or mass), but the rocket equation is still valid. SBHarris 02:48, 17 August 2011 (UTC)

Thank you for your long reply. I do not have time right now to reply to it in justice, but I am still not sure we're really agreeing here. I dont think you can say dp = Fdt because the mass doesn't change (very much) during differential time dt. Even though m doesn't change very much during dt, dp and dt are also small values so it is not right to consider it constant. dp = Fdt is equivalent to dp/dt = F, and if p=mv, then dp/dt = d(mv)/dt <> mdv/dt (because m is not constant) = ma = F, so dp/dt <> F. It is somewhat confusing, since we are considering F in this case to be "on the rocket", but the rocket is not a particular fixed thing (set of particles). However, the math works out pretty cleanly as long as you don't assume mass is constant, or that dp/dt = F (or equivalently, delta p = F delta t), or worry about the particles in the rocket. Once you do that, if you want to use consistent terms, you have to make a choice does I = integral F dt or I = delta p, and most sources seem to go with I = integral F dt, with the corollary that I = delta p if the mass is constant. However, one could go the route of not really defining F on a rocket, because as you said it gets confusing determine exactly what particles at what time constitute the rocket. That is what I meant by my initial comment (which I think I struck out) that I hoped this conversation did not get into a complicated discussion of what particles the momentum exactly was being taken of. As I said the math works just taking a rocket as a variable mass object, and letting dp <> Fdt. More importantly, I don't really think I am qualified for this discussion, I am not competent to write a textbook on rocket propulsion. However, the cite I gave, Rocket Propulsion Elements, Suttan and Biblarz, page 28, does this entire derivation. It uses I = integral F dt, and does not assume F = dp/dt or that the mass is constant, and pretty simply derives all the math. Moreover, many sources I have looked at define I in terms of F and t, and then say I is change in momentum but not as the definition, and usually with the caveat that the mass must be constant. I really think we should stick with the sources and their definitions. The reason I am going to all the trouble of this discussion is that when I was a little kid some teacher of mine defined impulse as change in momentum, then I tried to put that into the definition of specific impulse and became a very confused little kid because the math does not work out very easily if you define momentum that way. I knew about how change in mass effects the relation between momentum and force, so if I had known the specific impulse was in terms of F delta t and not delta(m v), I would not have been nearly so confused. Poor me... In any case, show me some sources that say dp/dt = F for non-constant mass, or ask for my sources that it doesn't, and I guess we can get on from there. I hope I understood your reply and this is understable because of my hurry 69.231.117.240 (talk) 09:53, 17 August 2011 (UTC)
I think you are right, I changed it.--Patrick (talk) 12:19, 17 August 2011 (UTC)
No. I'm quite sure that your text never defines impulse except in the case of a relatively constant-mass object/rocket, because it is an undefined quantity otherwise, by the definition of "impulse." Will you now seek to change the impulse article also? Please read it. The reason impulse (and momentum change also) are undefined is that if your mass changes, you have no "object" to apply "impulse" (or momentum-change either) to, so what is it that undergoes this "impulse?" This is not a complicated discussion of some arcane detail-- it's essential to the meaning of impulse. I = integral Fdt of course assumes mass is constant (or it changes inconsequentially over the time period Δt integrated), since (again) otherwise I(Impulse) is something you don't know the meaning of. The number you get from FΔt hasn't only been applied to the rocket, but in both directions on some of the fuel, and these cancel, so the number doesn't tell you anything if fuel-mass is significant. In that case, a force has been applied to rocket + fuel, then the force applied to the SAME fuel in the OTHER direction. This does not result in any total momentum change you can add up.

Your argument that dp <> Fdt because dp/dt = d(mv)/dt <> mdv/dt is a bad one. There is no logical chain, there. Yes, indeed d(mv)/dt <> mdv/dt when mass is not constant, but this does not imply that the previous equality is therefore wrong. It means you simply can't go beyond dp/dt = d(mv)/dt in the constant mass case, is all, because d(mv) <> mdv. Rather, of course, dp/dt = mdv/dt + vdm/dt if mass is changing. However, F = dp/dt = d(mv)/dt (before you expand the differential) is still fine. Expanding the differential improperly is where you got off track, but only at that point. IOW, the equation F = ma is only correct if mass is constant, but that doesn't mean that F = dp/dt requires mass must be constant.

On the same track, I fail to see what you think you are gaining by defining Isp = d(integral Fdt)/dm when the fundamental theorem of calculus (so long as the integral remains indefinite and unperformed) simply makes this equation Isp = Fdt/dm = F/[dm/dt]. You can (and should) use this simpler form, even if you don't accept that dp = Fdt is always true. Although, as it happens it IS always true, and thus by substitution Isp = dp/dm. As a little kid you may have tried to put change in momentum into this equation, but that must have been trying to figure Isp as Δp/Δm (wrong for changing mass except in the limit of dp/dm). Thus, if you tried to integrate Fdt to get FΔt, or tried to integrate Fdt to get Δp, you got the wrong answer if the mass wasn't constant. If mass is not constant, then FΔt <> integral Fdt <> Δp = I. None of this means or implies that it is incorrect that dp = d(ΔI) = Fdt. These are all correct.

As for the second sentence, we having a problem with English. The written out phrase "Change in momentum" divided by "change in mass" is correct if it refers to dp/dm, but not correct if it refers to Δp/Δm (which you seem to assume it does). The last implies that mass changes significantly, which means there is no way to calculate a meaningful Δp, since there is nothing for Δp to apply to.

Surely you can see that this is true, as Isp is perfectly well-defined for rockets with drastically changing mass-fractions, whereas Δp is NOT definable, and thus not calculatable. If you disagree, tell me how to find change in momentum Δp for a rocket that burns 9/10ths of its mass, as fuel? You can't do it. Momentum change (at least for the rocket) is not FΔt = thrust*time, in this case, even if thrust is constant. Δp <> FΔt. However, it is easy to define Isp in this situation, and aeronautics routinely does it, Isp = dp/dm = Fdt/dm = F/[dm/dt].

BTW, there is a definition of total impulse that is useful in finding specific impulse, but this is total impulse expended upon the fuel in the rocket's frame, which means it is the momentum change of the fuel as seen by the rocket. It is Δp in the rocket frame where the velocity of the exhaust gas is constant, so Δp = Δ(mv) = Δm * Ve. This means that Δp/Δm = Ve = Isp. It is easy to see here, why Isp ends up being effective Ve. And yes, this is F (dt/dm) where (dt/dm) is appoximated by (Δt/Δm). This expression for Isp can also be written as Isp = Δp/Δm = FΔt/Δm, and FΔt can be thought of as the "total impulse," so long as it's understood to be the total impulse on the fuel from the rocket's frame of reference. You can even do Isp = (integral Fdt)/Δm, so long as you remember that it's still [momentum change of fuel as seen by rocket]/mass change = [momentum change of fuel as seen by rocket]/fuel mass = effective exhaust velocity of fuel. But if you use the definite integral form, then "Δm" here has to be over the limits of the integration (the total fuel burned), and cannot simply read "per unit fuel" as in the Wiki article, as it was recently changed.

The differential form of Isp = (integral Fdt)/Δm is still Isp = dp/dm or Fdt/dm. I don't mind if it's changed to fix this, but it's confusing in the lede. The impulse, here, since it is on the fuel not the rocket, can't be used to easily calculate the rocket's final velocity from the start. That's the job of the rocket equation, since the rocket mass changes, so the acceleration with a fixed thrust changes throughout the burn. SBHarris 19:51, 17 August 2011 (UTC)

Your claim, that F=dp/dt is wrong because the mass changes is simply incorrect.
In each moment, the mass changes from m to m-dm, but the momentum dp provided by the burn is evenly divided up over the mass. The amount that ends up in lost mass in the exhaust is negligible in any instant of time, because it's dp*(dm/m), since dm/m is the fraction of the mass that is in the exhaust. But that is a second order correction, you have an infinitesimal multiplied by another infinitesimal, but for infinitesimal quantities second order corrections tend to precisely zero as you take the derivative to zero.Teapeat (talk) 19:28, 17 August 2011 (UTC)
Meanwhile the amount of momentum change that ends up in the body of the rocket is dp (m-dm)/m which tends to dp as the derivative is taken to zero.Teapeat (talk) 19:28, 17 August 2011 (UTC)
So f=dp/dt is a perfectly valid equation here.Teapeat (talk) 19:28, 17 August 2011 (UTC)
Yes. Our IP user .240 (why don't you pick out a username?) wants to use total impulse FΔt but doesn't like using Δp in its place, since this momentum "change" doesn't look physical. He even goes too far, and resists dp = Fdt, which is true even here-- I think because he doesn't want to integrate both sides to get Δp = FΔt. Well, the total impulse isn't very physical either, when seen in unaccelerated frames like the "rocket start frame". However, the total impulse is just the momentum change the rocket sees imposed upon the exhaust, by means of thrust alone (it is not the total momentum change on the rocket, since the rocket can be mostly fuel, which is due to diappear). The reason this momentum change doesn't apply in the rocket in the start frame is that the rocket is that its mass is changing. Worse still, in the rocket rest-frame its velocity is (obviously) not changing, so it has a momentum change of zero; momentum is not conserved in this open system, and there are fictitious forces opposing the thrust, keeping the rocket still, in its own rest-frame. These problems apply to the fuel also, but in the rocket's accelerated rest-frame, even the fuel that has been earlier ejected, seems to keep accelerating "away," and thus acquiring new momentum all the time, by operation of the fictitious forces at work at all places in an accelerated frame. So the momentum change imparted by the rocket on the fuel is the least of this momentum, as seen in the wacky rocket frame.

However, things are clearer once you realize that the expressions Δp = I = FΔt are all fine provided they are defined narrowly in the rocket rest-frame, and as regards the rocket's impulse upon the fuel it ejects. They don't apply to the rocket itself (with whatever fuel it carries), except in differential form: F = dp/dt or dp = Fdt. The reason that total-impulse does not equal Δ momentum for the rocket, except at each instant, is due to this wacky accelerated frame property, which causes momentum to accrue from fictitious forces in some places (like the early exhaust plume). Or, these fictitious forces can cause momentum to FAIL to accrue. This is why our rocket doesn't gain momentum when seen in its own frame-- the total force on it must be zero, which is why it never moves, even as the thrust force continues, its mass falls, and its acceleration increases when seen from its starting-frame. SBHarris 20:46, 17 August 2011 (UTC)

I'm confused. Of course I read the impulse (physics) article, it completely agrees with what I am saying. Of the many sources I mentioned, some are sources to that article. In classical mechanics, an impulse (abbreviated I or J) is defined as the integral of a force with respect to time. It also states that impulse is equal to the change in momentum, but does not give this as the definition. Lower down in the article, and in the authoritative sources, it mentions that having impulse equal to the change in momentum requires the mass to be constant.
The text I cited is for rocket propulsion, and it defines impulse as one of the terms relevant to its derivations for specific impulse. As this article is on specific impulse, I would think we would want to use derivations from sources that are authoritative on specific impulse, including the definitions that those sources use, instead of picking and choosing definitions based on how we interpret the physics. We are not writing a textbook here, but an encyclopedia, so putting together things taken from difference sources according to our understanding does not seem like the right approach to me. However, in this case the majority of the definitions of impulse I could find, even those not specific to rockets, define impulse in terms of force and time, including the one in Wikipedia that was linked to above.
Thinking too much can make this very complicated, especially for people that are obviously intelligent such as the above posters. Rockets do exist, they are objects with characteristics such as velocity. Rockets have a mass, which might be a little ambiguous because it is not clear exactly where the rocket itself ends and where the exhaust starts, but if you draw a line in or at the entry to the combusion chamber you can define it, and the amount of ambiguous mass is pretty small. Given a mass and a velocity, rockets have a momentum, which is their product. This is not complicated stuff if you don't overthink it. Rocket engines produce a force, which can be measured on a test stand or by having force measuring sensors built into the rocket itself at the junction of the engine and the rest of the rocket. A rocket can have its engine fired over a specified amount of time at a constant force. This force, multiplied by this time, is not equal to the change in the momentum. F delta t <> delta p. Not even to first order. Sometimes F and delta p change in completely different ways, depending on the way the mass is expended. Dividing by delta t and taking the limit as delta t goes to zero, F <> dp/dt. One could argue that the momentum I am using here is not somehow a valid momentum, because it does not consist of the momentum of a fixed set of particles. However, momentum is defined as the product of the mass and the velocity of an object, and the argument that the momentum is not valid then comes down to saying that a rocket is not an object. This is way overthinking things. This is what I keep getting back to. We should be talking about sources, and clear ways of expressing things, and not getting into exactly what particles are considered in the momentum of something. That is textbook writing, philosophizing, and something we could argue about all day. The sources on rocket propulsion define momentum of a rocket. They give a valid definition of impulse in terms of force and time, not momentum, a definition that does not assume the mass is constant. This definition of impulse is then used in following derivations, including the rocket equation.
I'm sorry I tried to abbreviate my mathematical explanation showing dp/dt <> F if the mass is not constant, and it was not clear. What I should have written was dp/dt = d(mv)/dt and also mdv/dt = ma = F. Since d(mv)/dt <> mdv/dt if the mass is not constant, then dp/dt <> F if the mass is not constant, by substitution of both sides of the inequality. This is a simple piece of math, but I tried to abbreviate it too much. If anyone shows me an authoritative source which says that dp/dt = F for non-constant mass, I will be glad to eat my words. Also, if anyone can show me a single authoritative source that says F <> ma, even for non-constant mass, I would really have to eat all my words, plus every physics textbook I've ever read. (in Newtonian mechanics etc., I'm not up to eating a bunch of textbooks because of some SR or quantum issues).
The reference I have been citing does not define Isp = d(I)/dm, or Isp = d(integral Fdt)/dm, it defines Isp = I/(weight of propellant) over some burn of propellant, where I = integral Fdt. I did not make the change to this article where it says derivative of the impulse with respect to amount of propellant used and that definition is not consistent with the source I have been using. What I proposed was It represents the impulse (force integrated over time) per unit amount of propellant used. My proposal now seems incorrect, perhaps it should be per amount of propellant, in line with the source, instead of per unit amount of propellant, but this issue was already in the article and I try to make minimal changes at one time. The source uses this definition to derive the basic Isp = F/[dm/dt], but nowhere does it require using the momentum of the rocket for this derivation, as impulse is not defined in terms of momentum. It does use momentum for other things. The point of my little kid story is that putting momentum in the definition leads to problems, even in deriving this basic equation, where momentum is not normally in the definition of impulse at all in most sources. Also, regarding integral Fdt, one certainly can integrate Fdt to get F delta t if the force is constant irregardless of the mass, that is just basic math.
Regarding whether "change in momentum" divided by "change in mass" refers to dp/dm or delta p/delta m, the source I have been referring to uses a definition for Isp over an interval, i.e. integral F dt/delta m. But for the issue of this discussion, it doesn't matter. F <> dp/dt if mass is not constant. Therefore F/(dm/dt) <> (dp/dt)/(dm/dt), assuming dm/dt is not zero. But (dp/dt)/(dm/dt) = dp/dm. Therefore F/(dm/dt) <> dp/dm. So using Isp = dp/dm is also not valid since it causes Isp to violate the basic derivation that Isp = F/(dm/dt). Defining impulse in terms of momentum doesn't work either way.
Regarding Isp = (integral Fdt)/Δm meaning that the impulse, here, is on the fuel is not what is meant. The F is the force of the rocket engine on the rocket, so the impulse is on the rocket. Perhaps phrasing it as impulse(engine thrust integrated over time) would be more clear than my original phrasing of impulse(force integrated over time). The force on the combusted fuel is of course equal and opposite to the force on the rocket, but the reference is referring to the force on the rocket and uses that for the derivations.
The argument that "F=dp/dt is wrong is simply incorrect because second order differentials can be ignored" is not right. The actual momentum gain (or loss) in dt is d(mv) = v*dm + m*dv. From the specific impulse equation, F = -veff*dm/dt since m is decreasing but positive veff is in the negative direction, but F = m*a = m*dv/dt so m*dv = -veff*dm. Therefore dp = v*dm - veff*dm = (v - veff)*dm. In this case, the v comes from the change in mass of the rocket, and the veff comes from the engine thrust. This is not the same order as dp*(dm/m) and none of it is second order. I do not understand the derivation of dp*(dm/m) well enough to see why it is not giving the same momentum loss. If you stand by your derivation here, I will try to understand it better to see why it does not come out the same as my simple, more or less algebriac, one.
dp = Fdt is a linear relation involving differentials that is equivalent (given a few assumptions such as dt <> 0, i.e. p and t are not on a vertical segment of a curve) to dp/dt = F, at a particular value for p and t. For a rocket with a changing mass, dp/dt <> F, even at this one point without regard to any interval around the point. There is another term in dp/dt, vdm/dt, which skews the value from F. dp = Fdt is equivalent to a particular linear relationship between dp and dt, and it just isn't true if the mass changes. Mass changing happens at point t also, at the rate dm/dt, and effects the linear relationship between dp and dt even at just that point. For this argument it does not really matter whether one uses differentials or changes over intervals. There is a defined momentum for an object (mass times velocity), it can be interesting in various ways even if the mass of the object changes, but it doesn't follow the same rules as in the constant mass case.
Of course one could start picking out, at any given time, exactly the particles that were in the rocket initially but are now in the exhaust stream of the rocket. This allows working with a constant set of particles and a constant mass. But it is not necessary, it is a complicated explanation, and it is not what the sources I have seen do.
Also, one could try to change reference frames, such as using the rocket's frame itself. In this case, the rocket has zero momentum, and it also has zero velocity so it is clearly not what we have been using in any parts of this discussion, although it can be a pretty interesting place to be, especially near c. I hope I have done justice to the well expressed statements of the previous posters.69.231.116.245 (talk) 05:02, 18 August 2011 (UTC)
So I don't have much experience with editing Wikipedia- since there has been no response to my previous post for almost 24 hours, is there now a consensus that impulse is in terms of force and time? And that delta momentum <> force * delta time for a rocket (which in the limit implies dp <> Fdt), so we can get past the physics, and on to the article? Or can anyone suggest a source that I can read to find the errors of my ways? (Sorry, it was rude of me to write so much yesterday.)
Should we work to a consensus on whether specific impulse is defined as "impulse / amount of fuel" versus "(dI/dt)/(damount/dt)" versus "dI/damount"? I don't see that it matters much as long as thrust and flow rate are constant, but the latter definitions do conflict with my main source and are currently being used in the article. Looking at the article history, I see a number of edits regarding this.
Also I see that The higher the specific impulse, the less propellant is needed to gain a given amount of momentum was the last sentence of the first paragraph before it was deleted. This seems very useful for a casual reader. How about something like The higher the specific impulse, the lower the propellant flow rate required for a particular thrust, and in the case of a rocket the less propellant is needed for a particular change in velocity per the Tsiolkovsky rocket equation. 69.231.117.173 (talk) 03:17, 19 August 2011 (UTC)
I would rather say delta-v instead of "change in velocity", there can be a huge difference when moving slowly against gravity (imagine a launch with hardly enough thrust) and/or make clear that this does not apply for a launch (also a kind of "change in velocity" operation, but one which is not even possible for rockets with a very high specific impulse because of the low thrust).--Patrick (talk) 08:33, 19 August 2011 (UTC)

2013/8/25 "It represents the force with respect to the amount of propellant used per unit time.[1]" To many people, "per" means "divide by", and the units are expected to be inverse seconds. Could the sentence be "It represents the impulse (change in momentum) divided by the amount of propellant used."? — Preceding unsigned comment added by Kinzele (talkcontribs) 15:43, 25 August 2013 (UTC)

## Isp v. effective exhaust velocity

Article should explain why is specific impulse more commonly used in rocket science than effective exhaust velocity. When using specific impulse, force of thrust is:

${\displaystyle F=I_{sp}\cdot {\dot {m}}\cdot g_{\oplus }}$

and it is defined as:

${\displaystyle I_{sp}={\frac {v_{ee}}{g_{\oplus }}},}$

where ${\displaystyle v_{ee}}$ is effective exhaust velocity. Both the definition and usage contain arbitrary and seemingly unnecessary factor ${\displaystyle g_{\oplus }}$ (surface gravity of Earth). On the other hand, if one uses effective exhaust velocity, force of thrust is simply:

${\displaystyle F=v_{ee}\cdot {\dot {m}}}$

and there is no need for an additional definition. Also, effective exhaust velocity has clearer and more intuitive meaning than the time measured by specific impulse. Therefore it should be explained, using some reliable sources, why is ${\displaystyle I_{sp}}$ preferred over ${\displaystyle v_{ee}}$ in RS, because for non-RS people it would seem that it shouldn't be. --94.253.206.109 (talk) 10:20, 7 December 2013 (UTC)

This is apparently because of the imperial-metric disparity. People from Europe (like Wernher von Braun) would use m/s for sp and the Americans would of course try using ft/s. Because cooperation of the two was important, they chose to divide by a known constant of appropriate unit to give a unit of sp that is the same in metric and imperial.Misiu mp (talk) 10:15, 25 June 2016 (UTC)

How is propellant quantified by a force where the force is something other than the weight of the propellant [[1]] (the Specific impulse in seconds / General definition section says the propellant is quantified by it weight on Earth))? Thanks --catslash (talk) 20:35, 8 August 2014 (UTC)

Specific impulse is said at the start to have units of velocity, that is, meters/second, but later it has units of time, seconds. Please clarify this major discrepancy. Thanks. -- pjt111

## Use of "specific impulse" for airbreathing engines

I think there are a couple of major things wrong with the article as currently written:

• Misunderstanding of the term "propellant": we define specific impulse as: the total impulse (or change in momentum) delivered per unit of propellant consumed. The propellant of an airbreathing (jet) engine consists of the air flow through the engine, plus the fuel consumed (the mass flow of which is much smaller than the air).
• The article claims that specific impulse is used as a performance parameter for jet engines as well as rockets, and attempts to compare the "specific impulse" of several jet engines with rocket engines. I happen to work in the US jet engine industry, and in my forty-year career, I have never, ever heard or seen anyone talk about the "specific impulse" of a jet engine. We always use the term thrust specific fuel consumption, which is defined as the fuel only (no air) divided by thrust. — Preceding unsigned comment added by JustinTime55 (talkcontribs) 21:00, 15 July 2016 (GMT)
Do you have a reference for your use of the term propellant in that context? I'm very sure that's not how it's strictly defined. Propellants are the chemicals that are carried in a vehicle to generate the motion. In an airbreathing jet engine, or a car, it's just the fuel, the fuel is the propellant (the air is not propellant, because it's not carried). In a rocket the fuel and oxidiser are both propellants because you need both of them for obvious reasons.GliderMaven (talk) 00:52, 16 July 2016 (UTC)
How did I know you would be the one to give me blowback on this? As a matter of fact, I do have a reference. If you go to reference 1, the Qualitative Reasoning Group website already cited, and click on What is a propellant?], you find it's defined as "a material that spews out of the back of the spacecraft giving it thrust". In the context of specific impulse, it has nothing to do with fuel (even though the word is often ambiguously used for that). It has nothing to do with what goes into the vehicle's tanks, but rather the stuff that comes out of the nozzle. And it's completely pointless to talk about cars, which do not use jet propulsion at all; in that case specific impulse is completely meaningless.
Consider the bottle rocket, where the propellant is pressurized water (no fuel or combustion at all), or in the solid state nuclear rocket where it's unburned hydrogen heated by the nuclear fuel (the consumption rate of which in no way at all relates to specific impulse; think, if you tried to use the mass decay rate of the fuel, you would get a ridiculously huge number.)
It is absolutely ridiculous to list "specific impulse" of jet engines in the thousands, and put those numbers next to those of rocket engines. You will notice there are no sources cited for the numbers in the two Examples tables, and the citation given in the sentence An air-breathing jet engine typically has a much larger specific impulse than a rocket; for example a turbofan jet engine may have a specific impulse of 6,000 seconds or more at sea level whereas a rocket would be around 200–400 seconds is a verification failure; the numbers 6,000 and 200-400 do not appear anywhere on the cited page [2]. The calculation of those numbers is original research, as is the chart in the General definition subsection.
It is equally meaningless to compare "specific fuel consumption" of rockets to jet engines, if it is arbitrarily calculated as mass of fuel in pounds per hour (without oxidizer) divided by thrust. It is only meaningful to compare specific impulse of jets to that of rockets if it is calculated properly: thrust divided by (airflow + fuel flow rate). JustinTime55 (talk) 19:41, 18 July 2016 (UTC)
"The propellant is a material that spews out of the back of the spacecraft" - they're solely talking about rockets and spacecraft, air breathing jet engines are neither. The term propellant is, in all reputable sources, used for material that is carried inside a vehicle; and it's useful because it means you have to make provisions for propellants, build tanks for it, and the engines have to be big enough to accelerate the tanks and propellant etc. And the term 'propellant' is used in jet fuels as well; for example 'JP-8' stands for "jet propellant 8".GliderMaven (talk) 20:59, 18 July 2016 (UTC)

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