SpaceX progress towards a 100 launch year

[Barley]

The actual rate (+100%, +50%, +20%...) clearly isn't exponential, so the cumulative graph can't be either.

That's not actual rate data.  That's estimated and projected rate data. 

[Robotbeat]

That's an interesting detail actually, which some of us touched upon a while back - but since there was such a strong reaction it was dropped in the interest of more fundamental points.

The historic launch graph is visibly non linear when studied since 2010. It can be fitted by an exponential with a certain goodness of fit value, but because of the arguments meekGee mentions, it can probably also be fitted to other functional forms with similar success as trend model - such as a quadratic equation.

This is actually motivated by the fact that the launch rate is seen to increase linearly over certain time period, rather than also exponentially (as in the above remarks: the derivative of a quadratic function is a linear one [ d(ax^2+b)/dx = 2ax ] but the derivative of an exponential is another exponential [ d(e^ax)/dx = a·e^ax ]). On the other hand, this rate has stronger deviations from an ideal function over the last decade than the cumulative launch tally, so it may have limited value as a predictive model.

The point about all this is which model can give a greater predictive power with the fewest assumptions, leading to lower errors - not whether there's an true, underlying absolutely analytical model, which is clearly not the case.

Moreover, it's apparent a linear approximation is pertinent when analyzing smaller (up to yearly) periods of time, because the overall function -be it exponential or quadratic- is too flat to generate deviations over the noise in such a short period of time.

Quadratic has more free variables than exponential does. It also behaves poorly when extrapolated backwards, like linear does.

[eeergo]

Quadratic has more free variables than exponential does. It also behaves poorly when extrapolated backwards, like linear does.

ax²+b has one parameter less than a•e^bx+c .

All functions extrapolate badly when applied piecewise on yearly periods because of the noisy dataset - including piecewise exponential as you demonstrated. Full-dataset fitting hasn't been tried with quadratic and exponential side-by-side yet (remember, the exponential function fitting all data predicted 85 launches this year when the linear extrapolations were doing just that), but the linear rates meekGee points out do suggest it's not a terrible choice. And its potential error due to mismatches between the model and the analytical approximation is less extreme due to the more slowly-varying function.

[Robotbeat]

Quadratic is ax^2+bx+c (has to be to do a fit, otherwise you can’t move the x axis offset). Exponential has just Ae^(bx) or equivalently e^(b*(x-a)). No constant “c” is needed.

[steveleach]

I tried fitting various other types of trend, again using Google sheets, but everything else I tried produced weird results. Linear works in the short term, exponential works in the long term and the short term (but with somewhat different exponents), and everything else goes mad.

Maybe someone with better analytics skills than me can make something fit, but as others have pointed out, there's a lot of noise in the signal so more parameters doesn't make a better prediction.

[meekGee]

The actual rate (+100%, +50%, +20%...) clearly isn't exponential, so the cumulative graph can't be either.

That's not actual rate data.  That's estimated and projected rate data.

Yes.  But that's what's being discussed.  "Actual" in that sense, as opposed to the other quantities that got graphed and mixed in.

The estimate is ours, the projection is by SpaceX. Missing any of them by a bit doesn't change anything.  So suppose his year is 97 or 103..   it's still a far cry from +100% of last year.  And suppose next year  is a few flights off SpaceX's projection - it's still not going to be anywhere near +50%.

It was a parallel discussion.  The graphs were tracking one thing, but the labels were applied to another.

[Barley]

Quadratic is ax^2+bx+c (has to be to do a fit). Exponential has just Ae^(bx). No constant “c” is needed.

You can hide the extra parameter in a quadratic by sneaky choice of origin, but the extra parameter still exists.

a (t-t₀)² + c = at² - (2at₀)t + (c + t₀²)

Solve for b' and c'.


In contrast for an exponential a change in time origin rolls into A

A e^B(t-t₀) = (A e^-Bt₀ ) e^Bt

Solve for A' and B'

[steveleach]

For what it's worth, and at risk of further derailing things, another model that kinda works is a long term exponential trend with 2019, 2021 and 2022 all significantly below trend due to external factors such as covid, and a return to trend in 2022 & 2023.

[edit] meant to say "2019, 2020 and 2021 all significantly below trend"

[Nomadd]

 I don't know why I keep hoping for more, but this whole thread looks more and more like a numerology convention every time I see it.

[steveleach]

I don't know why I keep hoping for more, but this whole thread looks more and more like a numerology convention every time I look.

The increase in numerology convention-ness is definitely exponential though.

[eeergo]

Quadratic is ax^2+bx+c (has to be to do a fit, otherwise you can’t move the x axis offset). Exponential has just Ae^(bx) or equivalently e^(b*(x-a)). No constant “c” is needed.

Sure, if you don't need to shift the exponential in the Y axis you don't need the "c". As is evident, because there never was a negative launch rate, and there was a time the launch rate was 0, this is such a particular case. But the same applies in the quadratic case: the "c" constant has the same role.

The shift in the X axis is what needs the linear term "bx" in the general quadratic equation you wrote, because b/2a is what regulates the parabola's origin in X. But needed only by convention, because you can set a certain year as the origin without ill effects. That's the "t0" parameter in Barley's formula.

So yeah, in this particular case you can do away with the "c" in both formulas, freeing yourself of a parameter. You can also do away with the "t0" by convention (t0=0), effectively eliminating "b" in the quadratic and equaling e^(b·t0) = 1. But you still strictly need both A and B modulators for the exponential, while you only strictly need the "a" in the quadratic case.

[Robotbeat]

Quadratic is ax^2+bx+c (has to be to do a fit, otherwise you can’t move the x axis offset). Exponential has just Ae^(bx) or equivalently e^(b*(x-a)). No constant “c” is needed.

Sure, if you don't need to shift the exponential in the Y axis you don't need the "c".

I would say shifting it in y would be improper.

Plotting the y axis as log scale would allow you to treat exponential the same as linear but with the advantage (over pure linear) that you wouldn’t have negative launch rates when extrapolated to the past.

[eeergo]

Quadratic is ax^2+bx+c (has to be to do a fit, otherwise you can’t move the x axis offset). Exponential has just Ae^(bx) or equivalently e^(b*(x-a)). No constant “c” is needed.

Sure, if you don't need to shift the exponential in the Y axis you don't need the "c".

I would say shifting it in y would be improper.

Plotting the y axis as log scale would allow you to treat exponential the same as linear but with the advantage (over pure linear) that you wouldn’t have negative launch rates when extrapolated to the past.

Semilogarithmic plots are just different visualizations, the underlying data doesn't change. If the function has negative values, they will just be "spaghettized" towards negative infinity in the logarithmic axis.

Regardless, the point was about the free parameters, hope it's clear now.

[Barley]

But you still strictly need both A and B modulators for the exponential, while you only strictly need the "a" in the quadratic case.

No.  The arbitrary choice of origin remains a degree of freedom.  If you believe it isn't, let me choose it and see how your model works.

You could make the other constants "go away" by some carefully choosing a time unit instead of months and some chosen measure of launch mass instead of a simple count.   Neither of these gets rid of a parameter, although it's more commonly done to ease manual calculations rather than as straight mendacity.

[eeergo]

But you still strictly need both A and B modulators for the exponential, while you only strictly need the "a" in the quadratic case.

No.  The arbitrary choice of origin remains a degree of freedom.  If you believe it isn't, let me choose it and see how your model works.

You could make the other constants "go away" by some carefully choosing a time unit instead of months and some chosen measure of launch mass instead of a simple count.   Neither of these gets rid of a parameter, although it's more commonly done to ease manual calculations rather than as straight mendacity.

If it remains so, the results will be unphysical (negative or non-vanishing launch rates, if you keep the linear term in the general quadratic), or you're adding an unnecessary parameter that should also be added in the exponential case (t0).

Anyway, this is all an academic discussion. Even if you could create an exponential and a quadratic model with the same number of degrees of freedom for this particular case, this would just mean they're equivalent in complexity - while the exponential case always has a more extreme behavior that will in principle increase errors in extrapolations far from the fitted datapoints. And then there's the non-exponential launch rates...

[woods170]

@Robotbeat, Barley, Eeergo, et al.,

Would you kindly move the math discussion over to mathforums.com or the math stackexchange, or something similar? It's OT for this thread.

Thank you.

[Robotbeat]

But you still strictly need both A and B modulators for the exponential, while you only strictly need the "a" in the quadratic case.

No.  The arbitrary choice of origin remains a degree of freedom.  If you believe it isn't, let me choose it and see how your model works.

You could make the other constants "go away" by some carefully choosing a time unit instead of months and some chosen measure of launch mass instead of a simple count.   Neither of these gets rid of a parameter, although it's more commonly done to ease manual calculations rather than as straight mendacity.

If it remains so, the results will be unphysical (negative or non-vanishing launch rates, if you keep the linear term in the general quadratic), or you're adding an unnecessary parameter that should also be added in the exponential case (t0).

Anyway, this is all an academic discussion. Even if you could create an exponential and a quadratic model with the same number of degrees of freedom for this particular case, this would just mean they're equivalent in complexity - while the exponential case always has a more extreme behavior that will in principle increase errors in extrapolations far from the fitted datapoints. And then there's the non-exponential launch rates...

Literally the opposite is true. The exponential/compounding model has less extreme behavior than linear (implies negative launches in the past) or quadratic (implies high launch rate in the past). That’s why it’s a better overall model. And we’ll soon find out by the end of the year.

I think people get wrapped up in the word “exponential” as if it’s exotic, but it’s literally just the same thing used when people say a company grows at 20% per year, or a loan has an 8% interest rate, or population is growing at 3% or inflation is 4%… these are all exponential models. And it’s totally normal and would be weird if people tried to shoe-horn linear or quadratic models into those sorts of conversations. When something, like launch rate, is growing, it’s due to companies applying learning continuously to improve on past capability. Companies don’t flip on a switch On January 1st and start operating at a constant unchanging rate after that, they improve continuously on past capability, with actual results being noisy as expected.

Anyway, we’ll see soon enough which model worked best. I think it’s pretty likely for SpaceX to achieve 100 launches by the end of the year or close. And they’re making the continuous progress necessary to reach that goal.

[alugobi]

Or close?

Where's that certainty you started with in this thread? 

[Robotbeat]

Or close?

Where's that certainty you started with in this thread? 

Reality is never certain. I think 100 launches is still likely. Whether 99 or 101, I’m not sure. When did I ever claim 100 exactly?

Mostly I was responding to those CERTAIN they wouldn’t get to 100 launches… And the linear estimate at the beginning of the year was about 80-85 launches.

The point is simply SpaceX gets better at launching rockets over time.

I thought they’d do 37 launches last year, but they did 61. They’re currently on pace for like 80-85 launches per year, but as they are improving over time, that’s likely an underestimate. Linear extrapolations from the beginning part of an exponential curve will undercount.

Like my GPA in college, if you fall behind the required average early on it gets harder to make up. …

This doesn’t make sense when applied to an exponential curve (say, an assumption they’ll grow capacity at 60% per year). Your previous year is always going to be less and, if the function is smooth, early months in the year will always have fewer launches than later months.

For an exponential function like that, an annualized rate of 85 launches per year for the first month and a half is actually ahead of schedule.

EDIT: I just did a curve fit to an exponential function with assumption they’d get 100 launches this year and got 61 launches last year. It has been 52 days into 2023, and they should’ve gotten 11.4 launches so far to keep on pace. They’ve gotten 12.

So they’re right on track if you assume a steady progression of capability. Not behind one iota. If the next launch occurs on time on February 23rd, they’ll be ahead by about one launch.

[alugobi]

I don't recall anyone stating outright that they'd not make 100, but I'm not going back through all the pages to look.  I expressed skepticism because there are so many variables out of SX's control, but I still said that they'd get in the ballpark.  I don't particularly care; they're doing very well for themselves and continuously improving. 

I also don't care about methods to predict it.  At this stage, I'm not sure that it can be with accuracy to say that one or the other particular formula is "the one".