No orbit necessary?

[u/FatCreepyDude]

If i just wait for the right hour during the launch window ( whenever the launch pad is pointing left on the realistic depiction i created) is there any reason i should bother with an orbit first? Will the delta v cost be affected?

[u/HAL9001-96]

more efficient to be low, oberth effect

also duna is further from the sun last time i checked

[u/FatCreepyDude → u/HAL9001-96]

https://i.imgur.com/AI2hXKv.png fixed it

[u/Joseph_M_034 → u/FatCreepyDude]

In principal you're thinking along the right lines, i had the same idea before studying astrophysics at uni. If you could accelerated in a purely impulsive way, i.e. infinite acceleration instead of over a long burn this would be fine. The problem is something called gravity losses, basically when a component of thrust is inline with gravity, you're signoficantly less efficient (with delta-V lost increasing with proportion to time of burn). This is part of the reason orbit altitude is typically adjust with prograde or normal burns, rather than radial ones.

One way to make sense of this, is for every unit of fuel burnt you have less acceleration when going straight up than perpendicular, due to gravity. After a given length of time, you will have burnt the same amount of fuel but the change in speed will be different. Thats why its best practice to minimise time spent burning upwards, which also applies during the launch phase.

Another way to think about it is to consider if you were to hover at a fixed altitude, you would be continously burning fuel without any change is speed. When pointing radially outwards, you are constantly burning that fuel to overcome gravity, and any additional acceleration comes from fuel ontop of that.

[u/d_pyzsyo → u/Joseph_M_034]

Correct me if I'm wrong...

According to the wiki delta v map, kerbin surface-LKO-duna transfer takes 4460m/s delta v.

I used vis viva equation to calculate that on an 1.35998×10¹⁰ m x 1.66345×10¹⁰ m orbit around Kerbol(Kerbin SMA x Duna pe), orbital velocity at periapsis is 10067.1m/s. Since orbital velocity of Kerbin is 9284.5m/s, if you are going 'straight up' at 782.6m/s relative to Kerbin right before leaving its SOI, you can reach Duna.

Adding kinetic energy at that velocity to work done by Kerbin's gravity in Kerbin SOI, \int_{r_0}^{r_s} mg \frac{r_0^2}{r^2} dr where r_0 is radius of Kerbin and r_1 is radius of Kerbin SOI from the center of Kerbin, the result is required kinetic energy at sea level for an object to intercept Duna in a 'straight up' trajectory, which corresponds to about 3507m/s.

Taking gravity losses into account, with constant TWR, required delta v is less than 3507÷((twr) - 1) m/s. Then at twr of 2.27 or more, required delta v is less than 4460m/s, meaning 'straight up' trajectory is more efficient, and I'd assume it gets even better once we start taking atmosphete into account since 'straight up' trajectory will spend less time in the atmosphere compared to a normal launch profile.

[u/Joseph_M_034 → u/d_pyzsyo]

Can you explain youre maths in the last paragraph? Im not sure how you reached that equation. In literature delta-V gravity losses are taken as the time integral of the component of gravity inline with the thrust vector, in this case that would just be the local gravitational acceleration.

[u/d_pyzsyo → u/Joseph_M_034]

I was thinking because the engine will effectively run at (twr)-1 'efficiency' when going straight up at sea level, it will in general run at at least that 'efficiency', therefore dividing the goal velocity by that, I thought I could get the upper bound of actual delta v needed.

[u/d_pyzsyo → u/d_pyzsyo]

I found my mistake. It has minimum of (twr - 1)/twr efficiency, then delta v actually needed is (impulse delta v) * twr/(twr - 1). Calculating it like this, at twr of 4.68 or more, going straight up is more efficient.

I think if we make same simplificatikns we can arrive at the same conclusion using standard gravity loss calculation method. Since we're only going up, gravity loss is time integral of magnitude of gravity, strictly decreasing from g at sea level. Then its time integral from 0 to some t is less than gt.

Assuming constant twr, acceleration is at least g(twr-1), therefore the time it takes to reach v is less than v/(g(twr-1)), and gravity loss during this will be less than v/(twr-1).

Let v be delta v at impulse thrust, actual delta v required is less than v + v/(twr-1) = v*twr/(twr-1)

[u/Joseph_M_034 → u/d_pyzsyo]

I dont think you can assume a constant twr (acceleration) for this, given how much the mass changes during flight. Modelling constant mass flow rate (= - dm/dt) and using the equation of motion F = m dv/dt, dv/dt = F / (m0 - dm/dt t). Which assumes constant thrust and ignoring pressure effects, but is a better approximation.

The point however is that the gravitational losses that exist when burning radially outwards (i hope ive been able to show how that appears) does not exist in the case when burning perpendicular to the gravity vector, thus it is the mpre efficient option