Calculating maximum vertical height with DV

[JP_Magoo]

This is a bit of a technical question, so there might not be a simple answer. (Or an answer that KSP players know) Does anyone know how to calculate your end apoapsis based on your total DV. For example, if I had 500m/s of DV, and I burnt straight up. How high would my apoapsis be (Ignoring air resistance.) I'm sure there's some equation for this, (Based on DV, gravity, and the Oberth effect) but I'd rather not have to figure it out if someone else knows it, or if there's a simpler answer.

 

[OhioBob]

If a object is positioned at a radial distance of R₁ from the center of the planet, and we want to give it an upward velocity that will propel it to a radial distance of R₂, the require initial velocity is,

V = SQRT[ 2 * μ * (1 / R₁ - 1 / R₂) ]

where μ is the gravitational parameter.

So in your example, we know V and R₁, so we must solve for R₂.

Rearranging the above equation, we get

R₂ = 1 / ( 1 / R₁ - ( V² / (2 * μ) )

Of course this is not exactly what you're asking for.  V is the initial velocity that you must give the object, which is not the same as the ΔV of a rocket.

I think I derived an equation for this like about 20 years ago.  I might still actually have it somewhere.  If I can find it, I'll post it.
 

[OhioBob]

Wow, I actually found my equations.  It's not real simple, however.

There are two equations, and they only apply to the time that the engine is burning.  The first computes the velocity of the rocket at burnout,

v = V_e * LN[ M / (M - q * t) ] - g * t

and the second computes the distance traveled by the rocket during the burn,

d = V_e * { t + t * LN[ M / (M - q * t) ] + M * LN[ (M - q * t) / M ] / q } - g * t² / 2

where V_e is the exhaust gas velocity, t is the burn time, M is the initial mass of the rocket, q is the propellant mass flow rate, and g is the acceleration of gravity.

The equations assume the rocket is traveling straight upward, there is no atmospheric drag, and g is constant.

Once the engine burns out, the rocket is in free flight and you can use the equation from my previous post to compute how much higher the rocket will go.  Just use v from the equation above for the initial velocity.  And if we say that the radius at the launch pad is R₀, then we have R₁ =  R₀ + d.  The total height reach is R₂ - R₀.
 

Edited September 8, 2018 by OhioBob

[Mahnarch]

You can use my, much easier, equation;

 

F5 + Space Bar + F9 = X

[JP_Magoo]

I found a less accurate but simpler equation: hg = 2v/2. I think it's a bit more at-my-level. xD

Edited September 14, 2018 by JP_Magoo

[OhioBob]

2 hours ago, JP_Magoo said:

I found a less accurate but simpler equation: hg = 2v^2. I think it's a bit more at-my-level. xD

It should be, hg = v^2/2.

[JP_Magoo]

Yeah, i found that out the hard way. 

[Leszek]

The problem is that even ignoring atmosphere the final altitude of your vehicle doesn't depend only on DV.  The initial TWR matters.

Looking over the complicated formulas provided earlier, I think they account for this.  But I personally don't want to do a page of math every launch.  So you have two options, make a spreadsheet to do the math for you, or you can think of it by rules of thumb.

For the same DV, the rocket with the higher TWR will go higher.  For the same TWR, the rocket with the higher DV will go higher.  For best height, maximize both.

In atmospheres, maximize both until .9 Mach.  Then throttle down to 1.25 or so TWR until 1.25 Mach.  Then give'r.  I know there isn't any math in this post but it once you distill down to HG = v^2/2 I think you are loosing enough accuracy to not really matter.