I agree that item P (which I believe for Spacex is a 30% reduction in mass to orbit penalty for reuse) is appropriate to consider if what is being transported something like propellant, in other words if Spacex is being paid $x for each kg to orbit. However this calculation is not currently applicable to Spacex from a business perspective in my opinion because that is not how they are paid.

I also agree with Lar that K should likely be closer to .7 not .4, I think the cost of the 1st stage is closer to 70% of the launch costs, not 40% of the launch costs.

Musk believes that the most revolutionary aspect of the new Falcon 9 is the potential reuse of the first stage “which is almost three-quarters of the cost of the rocket.”

I also assume that while the .9 rate exponent may be appropriate for typical aerospace manufacturing, this number may not be the appropriate rate exponent for Spacex. Maybe someone could come up with another number for Spacex (considering the commonality of the upper stage, general manufacturing process and the fact Spacex might need the extra stages anyways for its small sat/mars projects) and then run all of these numbers and see how it looks.

Quote from: clongton on 04/24/2015 01:02 pmDo I correctly assume that scenario 1 is based on ULA's estimate of recovering the entire Vulcan booster stage? Based on his quotes above, I think it's pretty clear scenario 1 was meant to track SpaceX recovering its entire first stage.

Do I correctly assume that scenario 1 is based on ULA's estimate of recovering the entire Vulcan booster stage?

Lot's of interesting posts here but I worry about the total lack of credibility-analysis of the spreadsheet itself.Half of the parameters used are either estimates, guesstimates or assumptions.With that much uncertainty I personally would not give any credibility to any of the results of calculations based on this spreadsheet.

It's not valid to include fixed costs as part of k because the model assumes k is fixed even if the number of launches changes. Since the model is a $/kg to orbit, putting fixed costs in k implies you are comparing $/kg to orbit on the same number of flights, not the same number of kg to orbit.The $/kg model assumes something like propellant to a depot, where the number of kg is fixed. For example, if you lose 50% performance for re-use, you would have twice as many flights, and by having fixed costs in k that assumes your fixed costs double also.The model only makes sense if k includes only marginal costs, not fixed costs.The conclusions of the model are badly skewed by this error.

Lot's of interesting posts here but I worry about the total lack of credibility-analysis of the spreadsheet itself.Half of the parameters used are either estimates, guesstimates or assumptions.

I played with the numbers dr. Sowers provided and it seems to me that under slightly different assumptions Falcon 9 reuse becomes viable only after 5 flights. I changed two parameters:- rF - based on other posts in SpaceX treads, it is not always true that mass production reduces unit costs. This is why SpaceX is planning to recover payload covers - increasing their production rate would increase the cost instead of decreasing it.- RR_ratio - SpaceX goal is to make rockets as reusable as planes, so I assumed 0.01. I think it makes sense - if the booster costs $20M, refurbishment cost of $200K sounds reasonableTo make things even more interesting, if we assume rF being 1.1 (that would mean expanding production is expensive) makes F9 reuse viable only after 4 launches.

Quote from: georgesowers on 04/24/2015 12:59 amQuote from: Lar on 04/23/2015 10:18 pmDr. Sowers: I love the explanation of the spreadsheet (thanks for sharing the sheet and your exposition) but I was unclear as to the k values (k is "the fraction of the total cost of the expendable launch service represented by the production cost of the hardware to be reused." which is C(B)/ C(B) + C(~B) )You stated for the SpaceX scenario 1 case that you are using .4 for k ("based on internet chatter" ) and that ULA is <.3 but in the ULA scenario 2 case you state k at .18Could you clarify that? Are you saying that if ULA did a flyback booster you'd have a .3 or less but with the SMART proposal it's .18 or Also if I understand k, it's the fraction of the total cost of the vehicle that is recovered because that part of the vehicle is reused... internet chatter on NSF seems to use .7 as the fraction of an F9 cost based on Elon saying the first stage is 70% of the cost.Thanks for shedding light!The booster is <0.3 of our total cost. The engines, which we are recovering via SMART reuse are 0.18 of the total cost. I don't think 0.7 can possibly be relative to the total cost including fixed cost and overhead. It might be relative to the total production cost. If you use that with some reasonable assumption about other non-hardware costs, 0.4 is appropriate. I have a detailed understanding of our cost structure. I know how much is pads and factory and engineering and management and all those things other than pure hardware. Even though I don't have any insight into details of SpX, I know they have more as many pads as we do, a factory like we do, a test site we don't have, more employees, etc. All that has to be accounted for in the cost.It's not valid to include fixed costs as part of k because the model assumes k is fixed even if the number of launches changes. Since the model is a $/kg to orbit, putting fixed costs in k implies you are comparing $/kg to orbit on the same number of flights, not the same number of kg to orbit.The $/kg model assumes something like propellant to a depot, where the number of kg is fixed. For example, if you lose 50% performance for re-use, you would have twice as many flights, and by having fixed costs in k that assumes your fixed costs double also.The model only makes sense if k includes only marginal costs, not fixed costs.The conclusions of the model are badly skewed by this error.

Quote from: Lar on 04/23/2015 10:18 pmDr. Sowers: I love the explanation of the spreadsheet (thanks for sharing the sheet and your exposition) but I was unclear as to the k values (k is "the fraction of the total cost of the expendable launch service represented by the production cost of the hardware to be reused." which is C(B)/ C(B) + C(~B) )You stated for the SpaceX scenario 1 case that you are using .4 for k ("based on internet chatter" ) and that ULA is <.3 but in the ULA scenario 2 case you state k at .18Could you clarify that? Are you saying that if ULA did a flyback booster you'd have a .3 or less but with the SMART proposal it's .18 or Also if I understand k, it's the fraction of the total cost of the vehicle that is recovered because that part of the vehicle is reused... internet chatter on NSF seems to use .7 as the fraction of an F9 cost based on Elon saying the first stage is 70% of the cost.Thanks for shedding light!The booster is <0.3 of our total cost. The engines, which we are recovering via SMART reuse are 0.18 of the total cost. I don't think 0.7 can possibly be relative to the total cost including fixed cost and overhead. It might be relative to the total production cost. If you use that with some reasonable assumption about other non-hardware costs, 0.4 is appropriate. I have a detailed understanding of our cost structure. I know how much is pads and factory and engineering and management and all those things other than pure hardware. Even though I don't have any insight into details of SpX, I know they have more as many pads as we do, a factory like we do, a test site we don't have, more employees, etc. All that has to be accounted for in the cost.

Dr. Sowers: I love the explanation of the spreadsheet (thanks for sharing the sheet and your exposition) but I was unclear as to the k values (k is "the fraction of the total cost of the expendable launch service represented by the production cost of the hardware to be reused." which is C(B)/ C(B) + C(~B) )You stated for the SpaceX scenario 1 case that you are using .4 for k ("based on internet chatter" ) and that ULA is <.3 but in the ULA scenario 2 case you state k at .18Could you clarify that? Are you saying that if ULA did a flyback booster you'd have a .3 or less but with the SMART proposal it's .18 or Also if I understand k, it's the fraction of the total cost of the vehicle that is recovered because that part of the vehicle is reused... internet chatter on NSF seems to use .7 as the fraction of an F9 cost based on Elon saying the first stage is 70% of the cost.Thanks for shedding light!

Quote from: ChrisWilson68 on 04/25/2015 09:16 amQuote from: georgesowers on 04/24/2015 12:59 amQuote from: Lar on 04/23/2015 10:18 pmDr. Sowers: I love the explanation of the spreadsheet (thanks for sharing the sheet and your exposition) but I was unclear as to the k values (k is "the fraction of the total cost of the expendable launch service represented by the production cost of the hardware to be reused." which is C(B)/ C(B) + C(~B) )You stated for the SpaceX scenario 1 case that you are using .4 for k ("based on internet chatter" :) ) and that ULA is <.3 but in the ULA scenario 2 case you state k at .18Could you clarify that? Are you saying that if ULA did a flyback booster you'd have a .3 or less but with the SMART proposal it's .18 or ???Also if I understand k, it's the fraction of the total cost of the vehicle that is recovered because that part of the vehicle is reused... internet chatter on NSF seems to use .7 as the fraction of an F9 cost based on Elon saying the first stage is 70% of the cost.Thanks for shedding light!The booster is <0.3 of our total cost. The engines, which we are recovering via SMART reuse are 0.18 of the total cost. I don't think 0.7 can possibly be relative to the total cost including fixed cost and overhead. It might be relative to the total production cost. If you use that with some reasonable assumption about other non-hardware costs, 0.4 is appropriate. I have a detailed understanding of our cost structure. I know how much is pads and factory and engineering and management and all those things other than pure hardware. Even though I don't have any insight into details of SpX, I know they have more as many pads as we do, a factory like we do, a test site we don't have, more employees, etc. All that has to be accounted for in the cost.It's not valid to include fixed costs as part of k because the model assumes k is fixed even if the number of launches changes. Since the model is a $/kg to orbit, putting fixed costs in k implies you are comparing $/kg to orbit on the same number of flights, not the same number of kg to orbit.The $/kg model assumes something like propellant to a depot, where the number of kg is fixed. For example, if you lose 50% performance for re-use, you would have twice as many flights, and by having fixed costs in k that assumes your fixed costs double also.The model only makes sense if k includes only marginal costs, not fixed costs.The conclusions of the model are badly skewed by this error.This would only be the case if there was a massive increase in launch rates, at a rate of 10 launches per year, which is with in the current capabilities and launch demands of both of these companies, assuming the fixed costs don't change from their current ratios is fairly reasonable. However, if launch rates increase dramatically, then we would see a shift in K, becoming larger as the fixed costs are spread out more.

Quote from: georgesowers on 04/24/2015 12:59 amQuote from: Lar on 04/23/2015 10:18 pmDr. Sowers: I love the explanation of the spreadsheet (thanks for sharing the sheet and your exposition) but I was unclear as to the k values (k is "the fraction of the total cost of the expendable launch service represented by the production cost of the hardware to be reused." which is C(B)/ C(B) + C(~B) )You stated for the SpaceX scenario 1 case that you are using .4 for k ("based on internet chatter" :) ) and that ULA is <.3 but in the ULA scenario 2 case you state k at .18Could you clarify that? Are you saying that if ULA did a flyback booster you'd have a .3 or less but with the SMART proposal it's .18 or ???Also if I understand k, it's the fraction of the total cost of the vehicle that is recovered because that part of the vehicle is reused... internet chatter on NSF seems to use .7 as the fraction of an F9 cost based on Elon saying the first stage is 70% of the cost.Thanks for shedding light!The booster is <0.3 of our total cost. The engines, which we are recovering via SMART reuse are 0.18 of the total cost. I don't think 0.7 can possibly be relative to the total cost including fixed cost and overhead. It might be relative to the total production cost. If you use that with some reasonable assumption about other non-hardware costs, 0.4 is appropriate. I have a detailed understanding of our cost structure. I know how much is pads and factory and engineering and management and all those things other than pure hardware. Even though I don't have any insight into details of SpX, I know they have more as many pads as we do, a factory like we do, a test site we don't have, more employees, etc. All that has to be accounted for in the cost.It's not valid to include fixed costs as part of k because the model assumes k is fixed even if the number of launches changes. Since the model is a $/kg to orbit, putting fixed costs in k implies you are comparing $/kg to orbit on the same number of flights, not the same number of kg to orbit.The $/kg model assumes something like propellant to a depot, where the number of kg is fixed. For example, if you lose 50% performance for re-use, you would have twice as many flights, and by having fixed costs in k that assumes your fixed costs double also.The model only makes sense if k includes only marginal costs, not fixed costs.The conclusions of the model are badly skewed by this error.

Quote from: Lar on 04/23/2015 10:18 pmDr. Sowers: I love the explanation of the spreadsheet (thanks for sharing the sheet and your exposition) but I was unclear as to the k values (k is "the fraction of the total cost of the expendable launch service represented by the production cost of the hardware to be reused." which is C(B)/ C(B) + C(~B) )You stated for the SpaceX scenario 1 case that you are using .4 for k ("based on internet chatter" :) ) and that ULA is <.3 but in the ULA scenario 2 case you state k at .18Could you clarify that? Are you saying that if ULA did a flyback booster you'd have a .3 or less but with the SMART proposal it's .18 or ???Also if I understand k, it's the fraction of the total cost of the vehicle that is recovered because that part of the vehicle is reused... internet chatter on NSF seems to use .7 as the fraction of an F9 cost based on Elon saying the first stage is 70% of the cost.Thanks for shedding light!The booster is <0.3 of our total cost. The engines, which we are recovering via SMART reuse are 0.18 of the total cost. I don't think 0.7 can possibly be relative to the total cost including fixed cost and overhead. It might be relative to the total production cost. If you use that with some reasonable assumption about other non-hardware costs, 0.4 is appropriate. I have a detailed understanding of our cost structure. I know how much is pads and factory and engineering and management and all those things other than pure hardware. Even though I don't have any insight into details of SpX, I know they have more as many pads as we do, a factory like we do, a test site we don't have, more employees, etc. All that has to be accounted for in the cost.

Dr. Sowers: I love the explanation of the spreadsheet (thanks for sharing the sheet and your exposition) but I was unclear as to the k values (k is "the fraction of the total cost of the expendable launch service represented by the production cost of the hardware to be reused." which is C(B)/ C(B) + C(~B) )You stated for the SpaceX scenario 1 case that you are using .4 for k ("based on internet chatter" :) ) and that ULA is <.3 but in the ULA scenario 2 case you state k at .18Could you clarify that? Are you saying that if ULA did a flyback booster you'd have a .3 or less but with the SMART proposal it's .18 or ???Also if I understand k, it's the fraction of the total cost of the vehicle that is recovered because that part of the vehicle is reused... internet chatter on NSF seems to use .7 as the fraction of an F9 cost based on Elon saying the first stage is 70% of the cost.Thanks for shedding light!

Here's an example to illustrate what I mean. Suppose your annual fixed costs are $1 billion. Suppose your marginal costs are $50 million per launch for hardware and $50 million per launch for non-hardware. Suppose you are doing 10 launches a year and your vehicle has no re-use. Then your fixed costs are amortized over 10 launches and it comes to $100 million in fixed costs per launch. So total cost per launch is $200 million including fixed costs. k is .25 ($50 million out of $200 million) if you consider fixed costs as part of your k.Now suppose re-use costs you a 50% payload hit. Then in the re-use case you need to do 20 launches. Now, your $1 billion in fixed costs is $50 million per flight. Now your k becomes .33 ($50 million out of $150 million) for the re-use case if you consider fixed costs as part of your k.But the model assumes one k is applicable for both the re-use and non-reuse case. That's not true here.The fundamental problem is that fixed costs are divided by a different number of flights in the re-use and non-reuse case.If you include fixed costs in your k, you're essentially treating them as marginal costs, because the model uses k in calculating the costs for the extra flights you have to fly to make up for the lower payload per flight because of re-use. In this example, k can only be the same if fixed costs are $2 billion for the re-use case and $1 billion for the non-reuse case.These problems go away if you remove the fixed costs from k.

You need to be comparing $/kg to GTO not actual launch cost for fair comparison. F9R will be approx $11,000/kg ie <4t at $45m.F9E will be $11,000/kg ie 6t at $65m Vulcan ACES version will be approx $11,000/kg ie 8t at $90m (Dimitry's estimate for core without SRB).

I see what you're saying there and you're right. Is there an easy way to modify the spreadsheet to do that?

and would it matter anyway?

Quote from: Rocket Surgeon on 04/28/2015 06:15 amI see what you're saying there and you're right. Is there an easy way to modify the spreadsheet to do that?The spreadsheet is fine as long as k doesn't include fixed costs. Leave out all fixed costs and it works fine.Quote from: Rocket Surgeon on 04/28/2015 06:15 amand would it matter anyway?Yes, it matters. It's the difference between a 0.4 and a 0.7 for k, and we saw in the discussion above what a drastic difference there is between a 0.4 and 0.7 k for the SpaceX case plugged into the model.The conclusion that partial reuse as proposed for Vulcan makes more economic sense than full first-stage re-use changes if you exclude the fixed costs from k.

The question this is used to ask is whether or not the Vulcan should boost back or use the modular recovery and the conclusion is that the Vulcan should use modular recovery as it pays back quicker, .... problem is in reality, the Falcon 9 is already a very different beast. Unless they have a much more in depth analysis than this spreadsheet (I sure hope they do) then the mistake is comparing apples-to-apples, we really have apples-to-oranges.