TWR? Delta V? WTH?

[RocketScientistsSon]

So I have been doing some research on how to build a proper rocket and I understand the two things you need to get a rocket in to orbit are TWR and delta V. I have a pretty good grasp on TWR as it is just a ratio you want between 1.5 and 2.5. I can generally get my rockets built to inside that window. My problem is understanding delta V.

From what I have read delta V tell you if you have enough oomph (fuel? thrust? energy?) to get to where you want to go. For me, since I am in the early stages of the career game (havent gotten to orbit yet) I want to get to orbit. I have read that requires 4500 delta V. But I am not sure how to calculate that in the game from the information given.

So on to my questions:

1. is a TWR over 2.5 bad? or just wasteful?

2. Can someone explain a little more on delta V and how to figure it out in game?

3. Is the 4500 delta V just to break out of Kerbin's atmosphere or to get to a LEO?

thanks!

Edited December 9, 2014 by RocketScientistsSon

[Norpo]

Delta-V is, literally, "Change in velocity".

If you have a ship that's going 0 m/s relative to another object, a ship with a delta-v of 500 m/s after burning all it's fuel, will be going 500 m/s relative to that other object.

4500 Delta-V is the amount required to get into Kerbin orbit, a lot is lost to the inaccurate drag model, and to Kerbin's gravity as you go up. (It applies a LOT more drag than it should). The rest is used on the circuralization burn.

Delta-V is hard to figure out ingame, you can do it manually with a calculator and all that, but there's a mod that can do it for you. Kerbal engineer redux. I don't know anything about TWR (All I know is that it should be around that range, like you) or figuring delta-V out manually, i'm afraid.

[RainDreamer]

So, my amateur understanding is:

TWR is thrust to weight ratio. The higher it is, the faster you go up, because it indicate the amount of thrust you have to lift a given mass. So if you have a TWR of 1, you have the exact amount of thrust for...being in place. Because the amount of gravitational force from all that weight dragging down is exactly the amount of thrusting force that you can impart on your vehicle, and the two equal force counter each other out. When TWR is >1, you can achieve lift off. The more you have, the faster you go to space. I don't know the problems that may occur in stock for too high TWR, but in FAR/DRE, that is just asking for your rapid unplanned disassembly in mid air as you are essentially ramming super fast into the air.

An important note in the case of space planes or stuff with wings and launch horizontally - They have 2 things that allow them to take off with less than 1 TWR: Lift and horizontal vector of thurst. Lifting surfaces generate lift (obviously) and that assist you to fight against gravity. Horizontal vector of thurst means less of your thrust is used to counter gravitational forces. Since gravity drag you directly down, if you want to go directly up, you have to have more power than gravity to lift off. However, if you change that vector into one that is not directly opposite the direction of gravity, less of your thrust is being used to fight gravity and more being used for launching your craft. That said, don't try having less than 0.5 TWR unless you really know what you are doing. It is hard to fly those things and it is difficult to design properly. Also, they are slow.

[Taki117]

So I have been doing some research on how to build a proper rocket and I understand the two things you need to get a rocket in to orbit are TWR and delta V. I have a pretty good grasp on TWR as it is just a ratio you want between 1.5 and 2.5. I can generally get my rockets built to inside that window. My problem is understanding delta V.

From what I have read delta V tell you if you have enough oomph (fuel? thrust? energy?) to get to where you want to go. For me, since I am in the early stages of the career game (havent gotten to orbit yet) I want to get to orbit. I have read that requires 4500 delta V. But I am not sure how to calculate that in the game from the information given.

So on to my questions:

1. is a TWR over 2.5 bad? or just wasteful?

2. Can someone explain a little more on delta V and how to figure it out in game?

3. Is the 4500 delta V just to break out of Kerbin's atmosphere or to get to a LEO?

thanks!

1. The reason we chose TWR like we do is so that our rockets ascend at terminal velocity, this is the point during a freefall where air resistance equals the force of gravity, the same is true on ascent. If you ascent slower than terminal velocity you lose dV to gravity, if you ascend faster than terminal velocity you lose dV to drag. A good rule of thumb is that a TWR of 2.0 will give you exactly terminal velocity.

2.Delta-V (dV) is a measure of how much your rocket can change it's velocity. The term infact means Change in Velocity (Delta being Change and V being the symbol for velocity) To figure it out you use the Rocket Equation linked for your convenience, or you can get a mod such as Kerbal Engineer Redux or MechJeb that has a dV readout. That being said dV does not equal speed! Just because you expend 200 m/s of dV does not mean that your instantaneous speed changes by 200 m/s, in fact, if you do it right, it might not change at all, but your orbit sure will.

3. the 4500 m/s of dV is the average amount of dV required to get into a stable orbit.

[Red Iron Crown]

TWR and Delta-V are the two most important characteristics of a spacecraft, and they both relate to velocity change. Very simply:

TWR is the rate of velocity change. Delta-V is the amount of velocity change.

More TWR is generally good, up to a certain point. If your TWR is below 1 for the body you're landed on, the ship will not be able to accelerate upwards because its thrust is not enough to overcome gravity. A minimum of about 1.2 is required for an efficient ascent. Higher TWR will allow a more efficient ascent, especially on airless bodies, but too high a TWR means you're carrying excess mass in the form of engines. Once in orbit, TWR becomes much less important. Very low TWRs can be used there, the only real limit is your patience in doing long burns. A lower TWR in orbit means a more efficient ship, but a higher TWR in orbit means more efficient burns. I typically aim for around 0.5 (Kerbin-relative) for orbital craft.

Delta-V is a measure of how much a craft can change its velocity. This is critical because changing velocity is how we change the size and shape of our orbits to reach destinations. It is calculated using the Rocket Equation (if you only learn one bit of rocket science it should be how that equation works). It is a measure of a craft's capabilities that corrects for size and engine efficiency, a craft with a given amount of delta-V can reach the same destinations whether it's a 1000 ton chemical rocket monster or a 1 ton high efficiency ion probe. It is very useful to use a mod like Kerbal Engineer Redux to calculate it (it's tedious to do by hand) and learn to read a delta-V map to figure out requirements. Once you do this, you'll be able to design spacecraft much more efficiently.

[GoSlash27]

1) Just wasteful.

2) The delta v represents energy. It's the chemical potential energy in your rocket and also the kinetic energy of orbit+ energy lost as heat and noise during the ascent.

the formula is the Isp of your engine(s) x 9.81 x the natural log of ( the mass of your rocket fully fueled/ the mass of your rocket dry).

3) It's for a low orbit.

Best,

-Slashy

[Red Iron Crown]

The delta v represents energy.

I have to disagree with this. The units are wrong for energy (delta-V is in m/s, energy is in kg*m^2/s^2 or Joules), and a m/s of delta-V can change a craft's kinetic energy by varying amounts due to the Oberth effect. Delta-V represents potential velocity change, nothing more and nothing less.

[Troubletcat]

1. High TWR in the stock game are mostly just wasteful, but only inside an atmosphere. The problem is that as you go faster you experience more resistance from the atmosphere, so it's more efficient in terms of fuel consumption not to go too quickly when you're low inside atmospheres. As an aside, 2.5 is probably still considerably higher than you want for a launch TWR. Aim in the region of 1.2-1.8 for launch. Aside from being inefficient, it can make your craft more wobbly or harder to steer as you get to high speeds in the lower atmosphere. If you use FAR instead of the stock aerodynamics it can rip your craft apart completely.

2. Delta-V was covered pretty well by Norpo. It's hard (but possible) to figure out based on the information given to you in-game. If you want to know how, here's the wiki page that explains it: http://wiki.kerbalspaceprogram.com/wiki/Tutorial:Advanced_Rocket_Design . Otherwise use Kerbal Engineer.

3. 4,500m/s is enough for LKO in the stock game. As an aside, you can make it to orbit with considerably less than 4,500m/s delta-v using FAR or NEAR.

[capi3101]

1. Wasteful. The game's default drag model is based on a number of factors: mass (the unrealistic bit), coefficient of drag, atmospheric density and velocity. The fast you go, the more drag you generate, and the formula uses the square of the velocity - the end result is that the increase in drag generated as your velocity goes up will more than counter the decrease in due to lowering atmospheric density. You start generating sufficient drag force to counter your rate of acceleration, which in turn means more energy is required to overcome the drag, which translates to delta-V lost. And speaking of...

2. Delta-V is a measure of your rocket's ability to make changes to its velocity (i.e. to change the speed and/or direction of its travel) under its own power. A rocket in space will travel forever, of course; this figure measures how much it can accelerate before it runs out of fuel. To determine the delta-V, you use the Tsiolkovsky Rocket Equation: dV = ln(M/Mo)*Isp*Go, where ln is the natural logarithm function (look for it on a calculator), M is the mass of the rocket when it's full of fuel, Mo is the mass of the rocket when it's out of fuel (i.e. its "dry mass"), Isp is the specific impulse of the engine (a measure of its efficiency, quantified in units of seconds) and Go is standard gravity (9.8 m/s^2 under ALL circumstances). The equation works for individual stages; for a multi-stage rocket, you determine the delta-V of each individual stage and then sum them all together to get the total delta-V of the rocket, treating any later stages along with the dry mass of the current stage.

That may have been confusing; hopefully not.

3. 4500 is generally what you're shooting for to get all the way into orbit. Bulk of that should be during the ascent; if done properly the final circularization can be done for somewhere between 50-100 m/s or so.

[GoSlash27]

I have to disagree with this. The units are wrong for energy (delta-V is in m/s, energy is in kg*m^2/s^2 or Joules), and a m/s of delta-V can change a craft's kinetic energy by varying amounts due to the Oberth effect. Delta-V represents potential velocity change, nothing more and nothing less.

You are, of course, 100% correct, so I'll revise the statement:

Delta V is *used as* an offhand representation of energy. It *should* only represent a change in velocity, but how it is applied in orbital mechanics (and around here) is as a measure of energy. It is incorrect to use it this way (which is why we have the Oberth effect and variations in the DV maps), but it's how it's done and how it should be thought of.

I didn't want to get this far in the weeds, and technically DV is only a change in velocity. Also technically there is no such thing as a "DV map" and rockets do not "have" DV.

When we speak of "DV budgets" to complete missions, we're talking about energy. When we talk about a burn to Jool, we're talking about kinetic energy. When we talk about building a rocket to "have" a certain DV, we're talking about potential energy.

We use DV as a shorthand description of Joules. It's not correct, it's just how it's done.

Best,

-Slashy

Edited December 9, 2014 by GoSlash27

[Red Iron Crown]

Why is there no such thing as a dV map? First I've heard that.

(I'd go into your other points but I think we are getting "into the weeds" as you say. )

[LethalDose]

So I have been doing some research on how to build a proper rocket and I understand the two things you need to get a rocket in to orbit are TWR and delta V. I have a pretty good grasp on TWR as it is just a ratio you want between 1.5 and 2.5. I can generally get my rockets built to inside that window. My problem is understanding delta V.

From what I have read delta V tell you if you have enough oomph (fuel? thrust? energy?) to get to where you want to go. For me, since I am in the early stages of the career game (havent gotten to orbit yet) I want to get to orbit. I have read that requires 4500 delta V. But I am not sure how to calculate that in the game from the information given.

So on to my questions:

1. is a TWR over 2.5 bad? or just wasteful?

2. Can someone explain a little more on delta V and how to figure it out in game?

3. Is the 4500 delta V just to break out of Kerbin's atmosphere or to get to a LEO?

thanks!

If you're asking if dV is thrust, I think you may need to go review the concept of TWR in addition to dV.

TWR is simply acceleration, but expressed in g's (9.8 m/s²) instead of raw SI units (m/s²). If you have a TWR of >1 (acceleration > 9.8m/s²) when you launch, you can accelerate straight up on Kerbin. The advantage of this should be obvious for rockets.

In orbit, TWR is basically a measure of how fast you can spend your dV. High dV in orbit means you can make large burns precisely because they take less time.

dV is how much total acceleration you have in a vessel or stage. In orbital mechanics, it effectively measures the range you can travel, e.g. 1500 m/s of dV is enough to get from LKO to Minmus and back, but not enough to go to Duna and back. In practice, it's similar to knowing how far a the fuel in your car's tank will take you.

~ 4500 is the dV cost for surface to LKO, given a reasonably efficient ascent. This changes with mods, like FAR.

These answers are pretty simplistic. I may write a larger blog entry on this.

[GoSlash27]

Why is there no such thing as a dV map? First I've heard that.

(I'd go into your other points but I think we are getting "into the weeds" as you say. )

It's a case of the two of us saying the exact same thing from opposite ends.

On the "No DV map" thing, You've already answered that. If somebody tells you it takes precisely a certain change in velocity to get to Duna, then you'd point out (correctly) that it's not so. What it *takes* is a precise change in kinetic energy. Nevertheless, we have these DV maps and treat DV as if it's energy.

That's my point...

Best,

-Slashy

[Red Iron Crown]

On the "No DV map" thing, You've already answered that. If somebody tells you it takes precisely a certain change in velocity to get to Duna, then you'd point out (correctly) that it's not so. What it *takes* is a precise change in kinetic energy. Nevertheless, we have these DV maps and treat DV as if it's energy.

That's my point...

Every dV map specifies the orbit at each origin/destination node. The Oberth effect and other factors are calculated from there, so the map is valid as long as we understand its assumptions. Though I definitely agree that we do tend to treat dV as if it is energy, that's part of why the Oberth effect is so confusing.

[LethalDose]

The delta v represents energy. It's the chemical potential energy in your rocket [...]

I have to disagree with this. The units are wrong for energy (delta-V is in m/s, energy is in kg*m^2/s^2 or Joules), and a m/s of delta-V can change a craft's kinetic energy by varying amounts due to the Oberth effect. Delta-V represents potential velocity change, nothing more and nothing less.

Slashy's technically correct, but it's not... (REALLY thinks hard about the right word here) It's not the most useful way of explaining dV. The rocket equation translates the potential chemical energy into dV, and Tsiolkovsky's rocket equation can be rearranged to show that, for any given in KSP (where ISP and dry mass are effectively constants per vessel), dV is simply a function of fuel mass.

dV is simply a convenient way to represent the potential chemical energy held in a tank.

Edited December 9, 2014 by LethalDose

[justidutch]

There are also a number of tools out there to help you calculate Delta-V outside the game. I built my very own when wanting to discover more about it myself. You can find it here:

http://jscript.ca/ksp.aspx

[LethalDose]

Oh, and as has been mentioned before, calculating dV by hand is a real PITA. Get VOID or KER.

TWR is easy. Acceleration is just total thrust divided by total mass, and then TWR is that divided by 9.8 (dividing by 10 is easier and will still give you a very accurate answer)

[Red Iron Crown]

Slashy's technically correct, but it's not... (REALLY thinks hard about the right word here) It's not the most useful way of explaining dV. The rocket equation translates the potential chemical energy into dV, and Tsiolkovsky's rocket equation can be rearranged to show that, for any given in KSP (where ISP and dry mass are effectively constants per vessel), dV is simply a function of fuel mass.

dV is simply a convenient way to represent the potential chemical energy held in a tank.

*Thinks hard about this.* I think I see what you're saying here (and what slashy said, too). It's not useful to represent the chemical energy directly (i.e. give the kJ of chemical energy), so delta-V can be thought of as a way to express the application of that energy that takes into account the changing kinetic energy stored in the fuel as well. That's a different way of thinking about it that I hadn't encountered before (or at least I hadn't put the pieces of it together). Thanks to both of you for educating me a bit today.

[GoSlash27]

Slashy's technically correct, but it's not... (REALLY thinks hard about the right word here) It's not the most useful way of explaining dV. The rocket equation translates the potential chemical energy into dV, and Tsiolkovsky's rocket equation can be rearranged to show that, for any given in KSP (where ISP and dry mass are effectively constants per vessel), dV is simply a function of fuel mass.

dV is simply a convenient way to represent the potential chemical energy held in a tank.

This is actually precisely what I'm saying, although I must stress that Red Iron Crown is absolutely correct in stating that it's incorrect to use it this way.

In addition to describing potential chemical energy in a rocket, we also use it to describe changes in kinetic energy and losses due to gravity and drag.

This is not the proper way to use the term either... it's just how it's used.

Best,

-Slashy

Edited December 9, 2014 by GoSlash27

[LethalDose]

dV is simply a convenient way to represent the potential chemical energy held in a tank.

This is actually precisely what I'm saying, although I must stress that Red Iron Crown is absolutely correct in stating that it's incorrect to use it this way.

No, it's correct, it's just unit conversion. All of the follow are correct:

• I can represent the chemical energy in a rocket using dV (in the reference of a given vessel or stage, conditions I made explicit in my initial explanation)
  
• I can represent the mass of an object in terms of the force it exerts on the surface below it (in the reference of a a given planet and altitude)
  
• I can represent the money I will have in a savings account in 20 years using the money I have in it now (in the reference of a certain set interest rate)
  
• I can represent the amount of British pounds I have in terms of of American dollars (in reference of a certain exchange rate)
  

Those are all correct because the term "represent" is not synonymous with the "equal", at least in how the terms are used in mathematics and science. So long as there is a valid function that f(X) = Y (which denotes that there are no variables other than x that define f(x)), X can be used to represent Y.

Another way of saying this is that X may represent Y, even if X does not equal Y. In fact, this is done all the freaking time in real science. For example, we use weight to measure mass, and in RTq-PCR, we represent the number of viral genomes present as cycle count.

And it's correct.

I get that this was the point you were making (as you highlighted), but it's wrong to try to say it's incorrect. It would be incorrect to simply substitute one value for the other, e.g. 500 m/s of dV means 500 KJ of potential energy, but no one is doing anything like that here.

In addition to describing potential chemical energy in a rocket, we also use it to describe changes in kinetic energy and losses due to gravity and drag.

This is not the proper way to use the term either... it's just how it's used.

I think you're reading way to far into what other posters have written. I don't know that I've seen anyone on the forums who actually understood what they're talking about make such a simplistic equivalency. There are some posters asking questions about it, but they're the minority, and IMO it's incredibly inaccurate to draw some generalization about "it's just how it's used" based on what I've read here.

Edited December 9, 2014 by LethalDose

[GoSlash27]

I get that this was the point you were making (as you highlighted), but it's wrong to try to say it's incorrect. It would be incorrect to simply substitute one value for the other, e.g. 500 m/s of dV means 500 KJ of potential energy, but no one is doing anything like that here.

To the contrary; *everybody* does that here. The fact that it's incorrect is why we have the Oberth effect. When we refer to DV maps, we are treating m/sec as if they were Joules. When we apply the rocket equation, we are treating m/sec as if they're Joules. When we say "it takes 4,500 m/sec to get to LKO" we are talking about m/sec as if they're Joules.

It's incorrect in the same sense that expressing speed in units of distance is incorrect.

We all do it even though it's incorrect because it works. Or at least well enough to get by.

Best,

-Slashy

[Empiro]

This is actually precisely what I'm saying, although I must stress that Red Iron Crown is absolutely correct in stating that it's incorrect to use it this way.

In addition to describing potential chemical energy in a rocket, we also use it to describe changes in kinetic energy and losses due to gravity and drag.

This is not the proper way to use the term either... it's just how it's used.

Best,

-Slashy

The thing is, while you can relate energy to dV, this relationship is extremely non-linear, so it's rarely ever helpful in expressing things in terms of energy.

For example, doubling the amount of fuel doubles the amount of chemical energy in the fuel, but in general, it doesn't double the amount of dV. Likewise, two maneuvers that necessitate the same change in kinetic energy might require vastly differing amounts of dV because of the Oberth effect. This is the reason why we use dV maps instead of delta-Joule(per kg) maps -- in the former case, you don't need to break out the calculators to see exactly how much fuel you'll actually wind up burning.

Edited December 9, 2014 by Empiro

[LethalDose]

To the contrary; *everybody* does that here. The fact that it's incorrect is why we have the Oberth effect. When we refer to DV maps, we are treating m/sec as if they were Joules. When we apply the rocket equation, we are treating m/sec as if they're Joules. When we say "it takes 4,500 m/sec to get to LKO" we are talking about m/sec as if they're Joules.

It's incorrect in the same sense that expressing speed in units of distance is incorrect.

We all do it even though it's incorrect because it works. Or at least well enough to get by.

Best,

-Slashy

Again, you are reading way too much into this.

To the point where you are incredibly wrong.

First off, I can state with a fair amount of certainty that I have never drawn such an idiotic equivalency. Go ahead and show me where I changed units from m/s to Joules, or claimed a dV map was showing required transfer energy.

Second, I'd like to see a posted example of someone who has a clue drawing this equivalency that you claim everybody makes.

Third, the rocket equation certainly does not treat the m/s as Joules. The units track there directly and simply. The way in which the dV is "spent" to change KE or orbital specific energy is purely dependent on the direction relative to travel in which the dV is spent.

Finally, there is no way someone could interpret the dV for Joules (or any other measure of energy) on the maps because the dV provided is equivalent across separate vessels where kinetic energy is not. In fact, the dV values being independent of mass is the reason maps are written using dV units instead of kinetic energy. Orbital energy itself is often actually written as energy per [vessel] mass to get around this issue, but I'm guessing talking about J/kg instead of m/s of transfer energy would make half the community's heads explode...

For example, it would X m/s dV for ship to transfer to the Mun from a 75 km parking orbit around Kerbin if it weighed 2 t or 20 t (given an equivalent set transfer orbit), but it would Y Joules of KE for the 2 t ship to transfer and 10Y Joules for the 20 t ship to transfer.

Is it that you don't realize that there are very specific assumptions about transfers and orbits included in every one of those dV maps?

Actually, you can ignore the first two points and that last question. I'm done with this discussion. You either don't understand the concepts you're arguing about or you can't express them in a way to be understood. In either case, further discussion is a waste of time.

Peace.

[GoSlash27]

The thing is, while you can relate energy to dV, this relationship is extremely non-linear, so it's rarely ever helpful in expressing things in terms of energy.

For example, doubling the amount of fuel doubles the amount of chemical energy in the fuel, but in general, it doesn't double the amount of dV. Likewise, two maneuvers that necessitate the same change in kinetic energy might require vastly differing amounts of dV because of the Oberth effect. This is the reason why we use dV maps instead of delta-Joule(per kg) maps -- in the former case, you don't need to break out the calculators to see exactly how much fuel you'll actually wind up burning.

Also agreed here.

Long way around to the point that we all agree on: When we refer to DV, we are actually referring to energy although we do not *express* it as such. It is far more convenient for us to work in terms of DV, much like a pilot will prefer to calculate fuel economy in terms of minutes.

Best,

-Slashy

[DrD]

Wow you guys are making this way too complicated for a new guy, although the conversaton is very interesting.

Delta V is change in velocity which is speed and direction. It's easiest to understand when you make a maneuver node, you will have to point in a certain direction, and change your speed in that direction by some amount. This amount will be shown as a bar to the right of the nav ball. You need enough fuel to bring the bar all the way down to zero.

Calculating delta V is very tricky and it's best to get KER. Or you can just over-build your rockets and play around until you get a feel for what you need in a rocket to get to orbit. You will fail but that's half the fun. The first time I got into orbit I didn't have any fuel left to get back so I had to get out and push (EVA and use jetpack to slow down spacecraft enough to get into atmosphere.)

As an aside, one can easily understand that velocity is direction plus speed when you realize that in one complete (circular) orbit your velocity changes quite a lot while speed is constant because your direction changes. In fact you velocity changes by the acceleration provided by gravity multiplied by the time in orbit. It's just that all the velocity changes occur in direction not speed (assuming a circular orbit, of course in eccentric orbits you speed does change as well.)