dV calculate and dV actual are way off - where am I being stupid?

[mjl1966]

OK, I'm just trying to sort out how to use the in game parms to correctly plug into the rocket equation. Please help me understand where I went wrong.

TEST VEHICLE:

MKI

FL-T400

LV-T30

MASS (dry, I think) Using the reported component mass sans fuel mass.

M(mki) 0.84

M(flt400) 2.25*

M(lvt30)1.25

So, Mo = 4.34. (i box reports 4.3, so far so good)

FUEL MASS

Prop 0.9

Ox 1.1

So, Mf = 2.0

Using the *BIG FAT SCIENTIFIC ASSUMPTION that the reported mass of the fuel tank (2.25) is its dry mass, we get

MT=6.34

Isp = 320.

Plugging into tyranny equation:

dV=320*9.82*ln(6.34/4.34) = 1190.

Emperical vertical launch at 100% and 50% throttle only gets me to v=570

So, questions:

a.) Am I calculating this incorrectly?

b.) Does drag really suck up that much dV?

c.) all of the above?

*Of note: assuming reported fuel tank mass includes fuel mass, the calculated dV is 1966, which is even more way off.

Thanks!

Edited December 18, 2014 by mjl1966

[OhioBob]

The calculated delta-v of a vehicle is what it would produce in the absence of gravity and drag. When you launch from Kerbin you are experiencing gravity and drag losses, so the actual velocity attained by your vehicle will be considerably lower than your calculated delta-v. For instance, if you are accelerating off the launch pad with a thrust-to-weight ratio of 1.5, then 2/3rd of you thrust is simply cancelling out gravity. You'll lift off the pad with an apparent acceleration of only 0.5 g.

Does this sound like your problem?

(edited to add)

The solution is to over design you delta-v to account for the expected losses. For example, it is generally accepted that in the stock game it takes about 4500 m/s delta-v to reach low Kerbin orbit. However, actual orbital velocity is a little less than 2300 m/s. The difference is the gravity and drag losses.

Edited December 16, 2014 by OhioBob

[Norpo]

Did you launch in a vaccum far away from anything that produces gravity, or just from the launch pad at Kerbin?

If the latter, then you're losing velocity gradually to gravity and drag. The delta-V is the change in velocity when ABSENT of gravity and drag.

[WanderingKid]

Dv with MK 1 command (0.84t), FL-T400 tank(2.25t wet, .25t dry), and TV-30 Engine (1.25t)

Wet mass: 4.34

Dry Mass: 2.34

dV = 320 * 9.6 * ln( 4.34/2.34)

dV = 3072 * ln( 1.86)

dV = 3072 * .62461...

dV = 1918.808 m/s

That's not really enough to get anywhere.

However, TWR = Thrust / (m * g). You start with a TWR of 215 / ( 4.34 * 9.8) = 5.05. That's WAAAAAAYYY over what you need.

My guess is you need to seriously throttle back on that machine. You're probably dumping absolute tons of dV into drag against the Terminal Velocities. Keep in mind at 5k you only want to be doing ~165 m/s, and at 10k ~265 m/s. That's where you're most likely losing a lot of your expected value.

[Starman4308]

Using the *BIG FAT SCIENTIFIC ASSUMPTION that the reported mass of the fuel tank (2.25) is its dry mass

Nope. That's the mass with fuel. The dry mass of the FL-T400 is 0.25t. That's true of all fuel tanks: the listed mass is when filled with fuel.

You're also, as mentioned, losing giant quantities of velocity on launch. When launching straight up from KSC, you lose 9.8 m/s each second to gravity. You are also losing velocity to atmosphere, at a rate proportional to atmospheric density and the square of velocity.

As it works out, the most efficient way to do things is to ascend at terminal velocity: when gravity exactly matches atmosphere. This means that, for the first part of the ascent, the most efficient TWR is slightly greater than 2.0*: 1.0 to counter gravity, 1.0 to counter atmospheric drag, and a little bit left over to accelerate as terminal velocity increases (due to thinner atmosphere). As such, you generally design for 1.6-2.0 launch TWR: the advantage of going a bit less than 2.0 TWR is that you will burn fuel on the way up, and if you need to throttle down to stay at terminal velocity, it means you're carrying wasted engine mass.

*If you use traditional staging, the most efficient launch TWR seems to be about 1.65: you'll burn fuel and increase your TWR on your way up. It'll be different for different staging schemes: something like asparagus, onion, or twisted candle can remain very close to optimal TWR by frequently staging off excess engines.

Please note that much of this advice gets thrown out the window if you install FAR or NEAR, which change aerodynamics significantly.

Edited December 16, 2014 by Starman4308

[Rusty6899]

Firstly, the reported mass of a fuel tank is its dry mass + mass of fuel, so the dry mass in this case is 0.25

Secondly, Delta-V, isn't the speed you can obtain from Kerbin's surface. It's a property of a fueled rocket that, in general, tells you how far you can go.

From the surface of Kerbin it takes around 4550m/s to get into orbit, but that is a measured, rather than calculated value, and is only applicable for an efficient ascent. Your orbital velocity at a low orbit will be around 2200m/s.

As a very (very, very) rough estimate, an efficient ascent to orbit will have a TWR of around 1.5-2. It would normally take 2000m/s to get to 10,000m, at which point you'll be travelling at around 260m/s and will want to pitch 45 degrees East, another 1500 to get your apoapsis to above 75km and another 1000, when you get to your apoapsis, to burn east parrallel to Kerbin's surface to reach orbit. This means that a LV-909 engine fully throttled would probably be a better choice for your rocket than a LV-T30 as it would have an initial TWR of 1.4, would use a more fuel efficient engine and would be cheaper. It would have between 2500 m/s and 3000 m/s delta-V.

Delta-V as a concept makes more sense when you are in orbit. If you are travelling at 2300m/s and have 500m/s delta-V, then if you burn all of your fuel prograde, you will be travelling at 2800m/s and delta-V requirements can be calculated rather than estimated.

[mjl1966]

Thanks to you all! You answered my question + a whole lot more.

After reading these responses, I did more experiments and found that a 5 tank single stack will give me 4375, which is juuust enough to establish an orbit at 70K peri. Actual net dV was ~2300 so it does seem that a whole herd of vees are being rounded up by Drag and Gravity and other members of the Great Rocket Tyranny.

Great stuff fellas. This really helped.

[Th3F3aR]

You could "simply" calculate your Drag by v0/APg where v0 is the m/s on your desired orbit ~2300m/s at 70km and APg is your a/g | a is calculated by g*(TWR-1)

then you'll see at least how much dV you'll lose on fighting gravity.

Remember it's a simple calculation not a 100% correct one

This is another Drag equation : Sqrt(GM*(1/r2-1/r1)) where r1 is the 600km on Kerbin (sea level) and r2 your desired orbit

Please don't ask why this works, for a 70km orbit this basically gives 4555m/s(but actually you can't combine them :-). I am a little confused by those 2 equations cause they should both display gravitationalDrag according to the parameters but they spit out different results.

But i think the main reason why they spit out different results is cause the 1st displays gDrag according to your ascent speed and the 2nd does it by the planets gravitational field not regarding how fast you ascend.

Nevertheless, as the others have said, drag costs you a lot of dV, that is why some genius said that "if you get to Orbit you are half way to everywhere".

[mjl1966]

How do I change this ti "ANSWERED"?

[Red Iron Crown]

How do I change this ti "ANSWERED"?

Edit the original post, go advanced, the option to change the prefix should be there.

Another thing to note is that dV spent does not always end up as speed. If you imagine a circular orbit in which you perform a prograde burn, at the far end of the new, elliptical orbit your vessel will actually have less speed. It is better to think of dV buying an energy change in the orbit, that 4500m/s of dV spent getting to LKO can be thought of as adding about 2300m/s of actual speed as well as adding the potential energy of another 70km or so of altitude (minus a small bit for gravity and drag losses).

This can be seen more clearly once in orbit. Do a prograde burn from LKO to raise Ap to 1000km and another to circularize when you get there, and you'll have far less speed than when you started even though nothing has been lost to gravity or aero drag. Instead, that dV expenditure has resulted in the vessel having much more potential energy in the form of additional altitude.

[Kesa]

You could "simply" calculate your Drag by v0/APg where v0 is the m/s on your desired orbit ~2300m/s at 70km and APg is your a/g | a is calculated by g*(TWR-1)

then you'll see at least how much dV you'll lose on fighting gravity.

Remember it's a simple calculation not a 100% correct one

This is another Drag equation : Sqrt(GM*(1/r2-1/r1)) where r1 is the 600km on Kerbin (sea level) and r2 your desired orbit

Please don't ask why this works, for a 70km orbit this basically gives 4555m/s(but actually you can't combine them :-). I am a little confused by those 2 equations cause they should both display gravitationalDrag according to the parameters but they spit out different results.

But i think the main reason why they spit out different results is cause the 1st displays gDrag according to your ascent speed and the 2nd does it by the planets gravitational field not regarding how fast you ascend.

The first equation give the exact dv requirement for achieving a speed of v0 for an upward acceleration in abscence of air, that's why it is a rough approximation for orbit. With TWR = ~1.6 and v0 ~2300m/s, it gives 4555m/s, which is, astonishely, the actual typical cost of orbiting with a typical TWR, because errors compensate them selves :

(1) the atmospheric drag is not taken on account

(2) the TWR often raise up to 2 or more in the 10 000 first meters, as the fuel of the first stage deplete (should reduce drag)

(3) Though TWR of the mid and last stages are often significantly smaller than 1.6, accelerating horizontally saves alot compared to accelerating vertically

So it's not drag, it's drag + desired v, but should be used with caution. I don't expect it to be true on any other body than Kerbin, and it is an over estimation for achieving orbit on an airless bodies, and a quite dramatic one if TWR is between 1 and 6.

Since there is no TWR in it, I think the second is related to transfer cost, so is not related to ascent. This two equations does not describe the same gravity drag : the first one describe loss due to vertical trusting, where the second describes... Well, I don't know, I'd be glad to know where it comes from. During transfer, you burn horizontally (prograde and retrograde), so no gravity loss like described by 1st equation or by some of it's refinements. However, it cost more to go from one circular orbit to an other one and then to athird one than to go from the first circular orbit directly to the third. That may be the kind of gravity loss described by the second equation.

Edited December 18, 2014 by Kesa