I'd like to calculate the approximate time left to my stage-1 velocity target.

A simple way to do this would be to divide my Delta V by my acceleration. Trouble is... my acceleration changes drastically from 1g at launch to 3g before engine cutoff... so this calculation would give an unrealistically large number early in the launch.

Is there a better way?

I know I could divide my fuel mass by my mass flow rate, but since my cutoff time is set by the delta V between my current velocity and my target velocity, this (m/mdot) ratio would only tell me how much time I have left before I run out of fuel---which would hopefully be many seconds after my engine cutoff...

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    $\begingroup$ "my acceleration changes drastically from 1g at launch to 3g before engine cutoff" sounds like a short reminder about intergrals would solve your problem? $\endgroup$ – CallMeTom Jan 14 at 11:00
  • $\begingroup$ @CallMeTom Bingo! But I do wonder how the OP is measuring/estimating his function a(t) . $\endgroup$ – Carl Witthoft Jan 14 at 13:29
  • $\begingroup$ a(t) is being calculated at each time step. The problem is that early in flight acceleration is low and dv is high, so the fraction dv/a gives unrealistically large numbers. These converge on the actual time-to-go as you approach MECO, but early in flight the fraction is useless, unless I account for how acceleration will change from now through MECO. That change is unknown because it depends on my pitch angle profile, among other things, which belongs to my list of unknowns to solve for... $\endgroup$ – user36480 Jan 14 at 18:59
  • $\begingroup$ So I guess I'm looking for an approximate analytical function a(t) that I can integrate into the future to get my projected dv(t) and solve that for the time it will take to hit my velocity target... I think I've seen a function of that sort in a paper on Powered Explicit Guidance, but it was expressed in terms of unknown coefficients... $\endgroup$ – user36480 Jan 14 at 19:16
  • $\begingroup$ You've mentioned a couple of times now that acceleration depends on your pitch. Why is that? $\endgroup$ – Organic Marble Jan 14 at 23:08

Your first stage cutoff should be based on propellant level, not velocity. In real rockets the cutoff is based on either fuel or oxidizer levels reaching a threshold, because you don’t want to risk an uncontrolled engine shutdown or ox-rich situation. Burning as long as possible provides some insurance against underperforming upper stages.

In simulation, assuming you don’t model uneven fuel consumption, cutoff by propellant level is equivalent to cutoff by run time, so estimating time to go via flow rate is straightforward.

  • $\begingroup$ @RusselBorogove: That was my first assumption too: that fuel mass determines the cutoff point for MECO. But the space shuttle manual made me rethink, because it says: "Guidance software commands the minimum throttle setting, calculates the cutoff velocity, and sets the MECO time. The primary task performed by guidance immediately before and during fine count is computing the appropriate MECO time, based on the desired cutoff velocity." Fine count here is the period of time leading to engine cutoff. So at least in one case it seems a prespecified velocity target determines the cutoff point. $\endgroup$ – user36480 Jan 14 at 19:05
  • $\begingroup$ Plus, stage 1 of Falcon 9 routinely shuts down with significant amounts of fuel left. And that stage has to return to a landing pad by following a trajectory sensitive to its initial conditions---altitude, downrange distance, and velocity. It wouldn't seem unreasonable to control for at least one of those variables, maybe using fuel threshold---15%, say---as a hard cutoff point in case you still haven't reached your target velocity when you hit your critical fuel threshold. $\endgroup$ – user36480 Jan 14 at 19:11
  • $\begingroup$ @Alex that shuttle guidance is for orbital targeting, not the first stage. The shuttle's "first stage" - the SRBs - separated when they ran out of fuel. Maybe you're using the term 'stage-1' in a way we are misinterpreting. For orbital targeting, at least for shuttle, it was highly undesirable - read, there would be lots of meetings about it - if it ran out of prop (it happened a few times). $\endgroup$ – Organic Marble Jan 14 at 23:05
  • $\begingroup$ Ah! Thanks for clarifying, OrganicMarble. That makes sense. The question remains, though, just in a different context: how would you calculate the time to go (e.g. to hit your velocity target for orbital insertion)? Even if you start after stage separation, your acceleration changes wildly up to your cutoff point. How do you get an estimate that isn't too far off early on? $\endgroup$ – user36480 Jan 15 at 1:55
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    $\begingroup$ Time-to-go for insertion has to be done differently, yes -- I answered as I did specifically because you said first stage! On insertion, assuming constant throttle, your acceleration is inversely proportional to your mass, and your mass is linearly decreasing with your propellant, which makes it a relatively straightforward calculus problem, but one I don't have the patience to work. You could precalculate some tables iteratively like some calculus-impaired chump with CPU cycles to burn, as well. (I say this as a calculus-impaired chump.) $\endgroup$ – Russell Borogove Jan 15 at 2:00

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