Horus Lupercal
Primarch - Warmaster
Professor
Swingin' on a Star
Deja Vu
Biker Mice from Mars
ET phone home
Floater
Copycat
Registered
I can get 4,000m/s Dv for 57 tons.
If we go for whole tanks, I'll have 4050ishm/s, and it'll still be 5 tons lighter than yours.
And thats still not the most efficient method. 4100m/s for 57 tons, no ion engines.
The closest whole tank amount to 3651m/s for me is 3824m/s, which I can get down to 36 tons launch mass.
And yes, that is with a probe and landing gear.
Horus Lupercal
Primarch - Warmaster
Professor
Swingin' on a Star
Deja Vu
Biker Mice from Mars
ET phone home
Floater
Copycat
Registered
Your main problems are 2 fold.
Firstly, your addiction to Frontier engines. Broadsword is superior under any and all conditions. Under certain circumstances, any other engine is better than frontier (including Titan) because of its shocking TWR and I'd only use frontier if I was in space, had build space constraints and I couldn't fit the right number of Broadswords in.
Second, I get where you got your Dv budget numbers from, but you've read the map wrong. After you've spent your 821m/s allocation, you'll be in LLO, at 10km orbit. At this height you're doing 332 m/s, so your landing budget needs to be at least 10 times higher than what you've allocated to get that to zero for landing.
Firstly, your addiction to Frontier engines. Broadsword is superior under any and all conditions. Under certain circumstances, any other engine is better than frontier (including Titan) because of its shocking TWR and I'd only use frontier if I was in space, had build space constraints and I couldn't fit the right number of Broadswords in.
Second, I get where you got your Dv budget numbers from, but you've read the map wrong. After you've spent your 821m/s allocation, you'll be in LLO, at 10km orbit. At this height you're doing 332 m/s, so your landing budget needs to be at least 10 times higher than what you've allocated to get that to zero for landing.
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That low-mass design is really neat, and I'm sure you could shave off a little mass. Are we considering returning from the moon as well as landing?
I would also draw your attention to a form posted a while ago, diving into the math of SSTO spacecraft.
In this, I found that the best engine for SSTO spacecraft is the broadsword because of their twr. However, if you are designing stages meant to fly in space and which do not need a specific twr, the frontier engine should be considered for any spacecraft pushing a payload of more than 150 tonnes. That is, (payload of 150)||(stage fuel, ~225)(Frontier engine)<. (P)||(F)<. As such, broadsword engines are not always the best, but they are for most SSTO applications (Unless you're going for part count).
I'll try my hand at the math.
Using Altaïr's incredible DeltaV map, and this map, ground>LEO is 2800, LEO> LLO is (640+59+122) OR (650+140) (821 OR 790. Call it 800.) LLO>ground is 464 (call it 500). When returning, we need to go to LLO, then intersect earth, but use an aerobreak. Thus, we need dv=2800+800+500*2+(800-640)=4760 (with return) or
dv=2800+800+500=4100 (without return)
And after the math comes the engineering. After all, who needs landing legs, right? You can always try to cut corners, which works so well in aerospace.
Let's assume the design is 4 things- probe, parachute, fuel tanks, and engines.
Payload is then probe+parachute=2.5+.5=3.0 tonnes.
deltav=isp*g*ln(M0/Mfin) //Mass final over mass at t0
deltav/(isp*g)=ln(M0/Mfin)
M0/Mfin=e^(deltav/(isp*g))
Mfin=M0-Mfuel*.9 //Final mass is initial minus reaction mass. Mfuel is wet mass of fuel tanks
M0=Mp+Me+Mfuel // Mp= Payload = 3, Me=Engine mass
My=Mp+Me // Payload and engines can be lumped together. Y not?
Mfin=My+Mfuel*.1
R=e^(deltav/(isp*g)) // Just to simplify writing it, since this term will not be broken up
(My+Mfuel)/(My+Mfuel*.1)=R
(My+Mfuel)=R*(My+Mfuel*.1)
My+Mfuel=R*My+R*Mfuel*.1
Mfuel-R*Mfuel*.1=R*My-My
Mfuel*(1-R*.1)=My(R-1)
Mfuel=My(R-1)/(1-R*.1) // Basically this says that the fuel required is the mass of the payload and engine times a ratio
Ok, for one-way, using broadswords. I will assume broadswords are best for this case.
e^(4100/(9.8*281))=R=4.432.
My=3+2.5*Ne // Ne is the number of engines. Assume 2
My=3+2.5*2 = 8 tonnes
Mfuel=8*(4.432-1)/(1-4.432*.1)=49.310 tonnes. Round to 50 tonnes
M0=58 tonnes initial mass.
TWR=Ne*T/M0=2*40/58=1.38 which is perfect
Assuming 4000 dv, that works out to 45.74 tonnes fuel. Make it 46 tonnes, overall mass is 54 tonnes, twr is similar. That's basically what Horus Lupercal found.
Using Frontier?
e^(4100/(9.8*285))=R=4.340.
My=3+8*Ne // Ne is the number of engines. Assume 1
My=3+8*1 = 11 tonnes
Mfuel=11*(4.340-1)/(1-4.340*.1)=64.90 tonnes. Round to 65 tonnes
M0=65+11=76 tonnes initial mass.
TWR=Ne*T/M0=1*112/76=1.47 which is in the same ballpark.
Result? The mass of a frontier makes a world of difference, in a negative manner.
Ok, what about for a return trip, 4800 delta-v, using broadsword engines.
10 engines, fuel is 308 tonnes.
M0 is 336 tonnes
TWR is 1.19, which is absolutely minimally marginal at best. A low TWR also means more deltav is wasted on takeoff.
I'm a bit rushed to finish this up, but as you can see, just going from 4100 to 4800 deltav is a massive increase in mass. This is why it's so important to stage, of course.
I hope this helps.
I would also draw your attention to a form posted a while ago, diving into the math of SSTO spacecraft.
In this, I found that the best engine for SSTO spacecraft is the broadsword because of their twr. However, if you are designing stages meant to fly in space and which do not need a specific twr, the frontier engine should be considered for any spacecraft pushing a payload of more than 150 tonnes. That is, (payload of 150)||(stage fuel, ~225)(Frontier engine)<. (P)||(F)<. As such, broadsword engines are not always the best, but they are for most SSTO applications (Unless you're going for part count).
I'll try my hand at the math.
Using Altaïr's incredible DeltaV map, and this map, ground>LEO is 2800, LEO> LLO is (640+59+122) OR (650+140) (821 OR 790. Call it 800.) LLO>ground is 464 (call it 500). When returning, we need to go to LLO, then intersect earth, but use an aerobreak. Thus, we need dv=2800+800+500*2+(800-640)=4760 (with return) or
dv=2800+800+500=4100 (without return)
And after the math comes the engineering. After all, who needs landing legs, right? You can always try to cut corners, which works so well in aerospace.
Let's assume the design is 4 things- probe, parachute, fuel tanks, and engines.
Payload is then probe+parachute=2.5+.5=3.0 tonnes.
deltav=isp*g*ln(M0/Mfin) //Mass final over mass at t0
deltav/(isp*g)=ln(M0/Mfin)
M0/Mfin=e^(deltav/(isp*g))
Mfin=M0-Mfuel*.9 //Final mass is initial minus reaction mass. Mfuel is wet mass of fuel tanks
M0=Mp+Me+Mfuel // Mp= Payload = 3, Me=Engine mass
My=Mp+Me // Payload and engines can be lumped together. Y not?
Mfin=My+Mfuel*.1
R=e^(deltav/(isp*g)) // Just to simplify writing it, since this term will not be broken up
(My+Mfuel)/(My+Mfuel*.1)=R
(My+Mfuel)=R*(My+Mfuel*.1)
My+Mfuel=R*My+R*Mfuel*.1
Mfuel-R*Mfuel*.1=R*My-My
Mfuel*(1-R*.1)=My(R-1)
Mfuel=My(R-1)/(1-R*.1) // Basically this says that the fuel required is the mass of the payload and engine times a ratio
Ok, for one-way, using broadswords. I will assume broadswords are best for this case.
e^(4100/(9.8*281))=R=4.432.
My=3+2.5*Ne // Ne is the number of engines. Assume 2
My=3+2.5*2 = 8 tonnes
Mfuel=8*(4.432-1)/(1-4.432*.1)=49.310 tonnes. Round to 50 tonnes
M0=58 tonnes initial mass.
TWR=Ne*T/M0=2*40/58=1.38 which is perfect
Assuming 4000 dv, that works out to 45.74 tonnes fuel. Make it 46 tonnes, overall mass is 54 tonnes, twr is similar. That's basically what Horus Lupercal found.
Using Frontier?
e^(4100/(9.8*285))=R=4.340.
My=3+8*Ne // Ne is the number of engines. Assume 1
My=3+8*1 = 11 tonnes
Mfuel=11*(4.340-1)/(1-4.340*.1)=64.90 tonnes. Round to 65 tonnes
M0=65+11=76 tonnes initial mass.
TWR=Ne*T/M0=1*112/76=1.47 which is in the same ballpark.
Result? The mass of a frontier makes a world of difference, in a negative manner.
Ok, what about for a return trip, 4800 delta-v, using broadsword engines.
10 engines, fuel is 308 tonnes.
M0 is 336 tonnes
TWR is 1.19, which is absolutely minimally marginal at best. A low TWR also means more deltav is wasted on takeoff.
I'm a bit rushed to finish this up, but as you can see, just going from 4100 to 4800 deltav is a massive increase in mass. This is why it's so important to stage, of course.
I hope this helps.
T
Right... Mercury.
... that took about 4 seconds to liftoff...
That's a huge budget.
First Mercury assist.
Second assist.
And the situation after that. Not bad. But, 3% fuel...
Fourth assist and then the capture... scary. Considering I have no intention of using my frontiers.
That's a HUGE gap to close for a pair of Ions at 49%...
Nervous yet?
You know, I honestly don't think I'll make it...
Shit this is killing me... Just maybe...
But, no...
Ridiculously close though!!!
... that took about 4 seconds to liftoff...
That's a huge budget.
First Mercury assist.
Second assist.
And the situation after that. Not bad. But, 3% fuel...
Fourth assist and then the capture... scary. Considering I have no intention of using my frontiers.
That's a HUGE gap to close for a pair of Ions at 49%...
Nervous yet?
You know, I honestly don't think I'll make it...
Shit this is killing me... Just maybe...
But, no...
Ridiculously close though!!!
This is incredible! I'm sure you'll get it soon. It looks like this could get you into orbit, at least, but with those margins, I don't know about landing.
Just some thoughts- 2 ion engines at 50% thrust is the same as 1 at 100%. Also, since the game has no shadows, batteries are actually unneeded. You could, if you so choose, remove all the batteries, which would make sense if you reduce it to 1 ion engine at full power. Of course, two ion engines and batteries allows for higher thrust (for a time), which is also important. It comes down to how you want to do it.
Also, using broadsword engines might be a little better, but might be too bulky. You want to avoid using a twr that's too low- A rule of thumb I picked up and follow (possibly from ksp) is between 1.7 and 1.2, with 1.4 being ideal. For SSTO, I target around 1.2. What is definitely undesirable is anything around 1. Consider the case of a spacecraft with twr at liftoff of .98. It must burn fuel until the twr is 1.00 before it can clear an inch of the pad- and though the fuel is gone, the empty part of the fuel tanks still weighs the craft down. It's always better to just take the mass off to start with. The other reason for a higher twr is the gravitational drag. The lower the twr, the longer you fight gravity and the more wasted potential.
Thus, adding more engines could actually get you more fuel into orbit.
I redesigned the craft to use broadsword engines, as well as one ion engine and no batteries.
Behold, the Monstrosity (1)! (Monstrosity was already taken in my save files)
This craft actually has the same amount of fuel as the one you built- 315 tonnes- but now with a twr of 1.18.
With 10 broadsword engines, the engine mass is 10*2.5=25 tonnes. With 3 Frontiers, 3*8=24 tonnes. The engine mass is very similar, though the thrust is very different (112*3=336 vs 40*10=400).
The total deltaV is as follows (Using this design vs this design with Frontiers. All hardware except engines is unchanged)
281*9.8*ln((321.6+25)/(321.6+25-315*.9))=4690.96 deltav with my design // These deltav values don't consider the ion engines
285*9.8*ln((321.6+24)/(321.6+24-315*.9))=4794.28 deltav with frontiers // It's all about getting max fuel to LEO
Change in deltav is 104 deltav, but that is more than made up by the increased twr.
Removing one broadsword engine gives 4782.35 deltav at 1.07 twr,
compared to 3 frontier engines and their 4794.28 deltav at 1.00 twr.
NOTE: all twr values are generated in the building menu and include the thrust from the ion engine.
Just some thoughts- 2 ion engines at 50% thrust is the same as 1 at 100%. Also, since the game has no shadows, batteries are actually unneeded. You could, if you so choose, remove all the batteries, which would make sense if you reduce it to 1 ion engine at full power. Of course, two ion engines and batteries allows for higher thrust (for a time), which is also important. It comes down to how you want to do it.
Also, using broadsword engines might be a little better, but might be too bulky. You want to avoid using a twr that's too low- A rule of thumb I picked up and follow (possibly from ksp) is between 1.7 and 1.2, with 1.4 being ideal. For SSTO, I target around 1.2. What is definitely undesirable is anything around 1. Consider the case of a spacecraft with twr at liftoff of .98. It must burn fuel until the twr is 1.00 before it can clear an inch of the pad- and though the fuel is gone, the empty part of the fuel tanks still weighs the craft down. It's always better to just take the mass off to start with. The other reason for a higher twr is the gravitational drag. The lower the twr, the longer you fight gravity and the more wasted potential.
Thus, adding more engines could actually get you more fuel into orbit.
I redesigned the craft to use broadsword engines, as well as one ion engine and no batteries.
Behold, the Monstrosity (1)! (Monstrosity was already taken in my save files)
This craft actually has the same amount of fuel as the one you built- 315 tonnes- but now with a twr of 1.18.
With 10 broadsword engines, the engine mass is 10*2.5=25 tonnes. With 3 Frontiers, 3*8=24 tonnes. The engine mass is very similar, though the thrust is very different (112*3=336 vs 40*10=400).
The total deltaV is as follows (Using this design vs this design with Frontiers. All hardware except engines is unchanged)
281*9.8*ln((321.6+25)/(321.6+25-315*.9))=4690.96 deltav with my design // These deltav values don't consider the ion engines
285*9.8*ln((321.6+24)/(321.6+24-315*.9))=4794.28 deltav with frontiers // It's all about getting max fuel to LEO
Change in deltav is 104 deltav, but that is more than made up by the increased twr.
Removing one broadsword engine gives 4782.35 deltav at 1.07 twr,
compared to 3 frontier engines and their 4794.28 deltav at 1.00 twr.
NOTE: all twr values are generated in the building menu and include the thrust from the ion engine.
T
I don't care. Lol.
... Actually I do. Let me explain: Aero was my first consideration. So a long narrow rocket with as few as possible solar panels was quite high on my priority list. Then, sufficient power for reasonably long burns. Then, enough thrust for liftoff. In truth, I only failed because I switched from frontiers to ions about 3 seconds too late. The difference it made... the rocket is perfectly capable of landing on Mercury.
... Actually I do. Let me explain: Aero was my first consideration. So a long narrow rocket with as few as possible solar panels was quite high on my priority list. Then, sufficient power for reasonably long burns. Then, enough thrust for liftoff. In truth, I only failed because I switched from frontiers to ions about 3 seconds too late. The difference it made... the rocket is perfectly capable of landing on Mercury.
Hey, I did it! Single Stage to Mercury with drag on!
I made a first attempt with drag off and I got to orbit with 40 percent of fuel, So I said "nah, too easy".
Launch again, exit, then resume...
I used as many Gravity Assistances as I could...
Fuel ended during descent, but luckily, speed was Low enough to ensure a soft landing...
Behold, the Mercury Obelisk!
I made a first attempt with drag off and I got to orbit with 40 percent of fuel, So I said "nah, too easy".
Launch again, exit, then resume...
I used as many Gravity Assistances as I could...
Fuel ended during descent, but luckily, speed was Low enough to ensure a soft landing...
Behold, the Mercury Obelisk!