Math formulas for sub-orbital flight

In summary: The reason you would need a rocket that can shut off and fire again is because the airframe may be at high temperature during reentry. The "brute thrust" you are using is not enough to get the vehicle over the hump. You will need a heavier engine to do the job.
  • #1
jgb
12
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I would like some "simple" (yeah, right!) math formulas that I can create an EXCEL spreadsheet to calculate flight profiles for a sub-orbital space plane with a 2,000 to 12,000 mile range. Please note, my math skills are 45 years old. If, as I suspect it will, calculus is involved, I would need some way to convert those formulas to EXCEL multi-cell formulas.

I found some flat Earth formulas, but they are only valid for a few dozen miles.

The base vehicle would be a 2 stage flying wing air breathing subsonic booster to about 50,000 ft launching the sub-orbiter.

The SO would be rocket powered on accent, coast sub orbital, then rocket braked to slow below M1.5 - M2 above the denser air, with a short jet powered conventional flight to destination. Rocket braking avoids high temperature aerodynamic braking, and reduces the G forces. This makes for a much lighter and less exotic vehicle.

Yes, I fully realize this is not the most efficient way to do sub-orbital flight, but the technology is far simpler and I can use brute thrust to get over the hump. Plus I've developed a proprietary way to make these vehicles about 60% the weight of conventional construction.

I need the formulas that can account for weight to thrust, specific maximum G forces on acceleration and deceleration (nominally max 2G) so I can calculate fuel volumes that will control the airframe design.

If there is honest interest I can define more details here, but for the really seriously interested I will provide far more via PM, except for some proprietary stuff that is being patented.

Anybody interested?

jgb
 
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  • #2
I don't have any ready, and I don't really have the time to work them out, but I wanted to comment on your braking method.

I don't think a rocket brake is the way to go. Do you plan to do a roll during your coast to get the thrust vector right? If so, you are adding a complex step to every flight which is always risky. Then, you have to consider that you will require a rocket that can shut off and fire again on descent; not all rockets can do this. Also, consider that you will be burning fuel and have fuel filled tanks during reentry with all the abuse and heat that goes with it, that too is risky.

How fast will your vehicle be going? Does it have control surfaces? Why can't it slow down conventionally. If you are sub orbital, you won't be going at space shuttle speeds. A long descent at shallow angles will have a lower G rating than a rocket brake...
 
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  • #3
I doubt Excel is the right tool for that job. I'm not entirely sure what you're looking to do (trajectory optimization? Latitude/longitude/altitude vs. Time?), but it looks to me like you're going to want a crash course in MATLAB programming.
 
  • #4
Fo Sho. Matlab is what you will want to use. I am pretty sure you will be able to model relatively simple trajectories using really complex excel calc's, but Matlab will get you there quicker (that is assuming you have access to it).
 
  • #5
I know Matlab is the way to go, but my math skills are not up to even its basic understanding, and I don't have the formulas to begin with. I know EXCEL very well, and I'm not after precision calcs. I need "close enough".

All I am trying to do is some basic calcs at various ranges, accelerations, launch altitudes and fuel SFC's to get a good idea of what fuel mass I need to get the airframe sized and thrust levels ball-parked. Nowhere near optimizing anything. That's for real rocket scientists to figure out, assuming feasibility in the first place.

It's an idea I've had for over 5 years now, and some recent developments of my method of very lightweight manufacturing the 2 airframes lead me to believe it could be done.

As for using reverse thrust rather than aerobraking; I plan to pivot the rocket engines back about 120+ degrees. That way the craft stays in a forward attitude, no pitch maneuvers.

I am using linear spike technology, pairing a dual set in an over/under configuration. LS types are far easier to pivot along 1 axis than bell types, and have vacuum restart ability.

Yes, I need to carry additional fuel for lofting and rocket braking, plus jet engine operation for a conventional landing with go-around or alternate site capability.

But if I slow the craft down to low Mach before I get into resistive air I solve several issues, far lower thermal stress on the airframe, lower loads stress on the airframe, lower and consistent G's decelerating, and a more benign flight path without much of a sonic boom. The real trade-off is more fuel for less complexity of design and manufacturing.

I am assuming max speeds for max range to be in the order of 10,000 to 13,000 mph, but again, I need those formulas to get a better approximation.

I hope that helps somewhat.

jgb
 
  • #6
  • #7
Thanks Enigma. Looks good, but and a BIG BUT here, it is way beyond my math comprehension. And even at that, how does it translate to EXCEL level math?

I know there are long involved equations that reduce calculus to algebra, but I think these have to be individually developed. I don't think there are "standard" algebraic translations for given calculus functions. In EXCEL I can breakdown a complex equation into a series of cell references, and let the spreadsheet do the grunt work.

So, while that reference goes a long way to answer my question, I cannot make use of it. I need algebraic formulas, not calculus.

And if I can get these rocket equations into a workable EXCEL spreadsheet, I will openly publish that sheet for anyone to use.
 
  • #8
If you're willing to just push the "I believe" button, ideal rocket equation does not use calculus. You won't understand where its derived, but all you need to use is a logarithm.

You'll need to know the starting mass, final mass, and specific impulse of your motors (which would be a look up or estimate). From there, its plug and chug.
 
  • #9
enigma said:
If you're willing to just push the "I believe" button, ideal rocket equation does not use calculus. You won't understand where its derived, but all you need to use is a logarithm.

You'll need to know the starting mass, final mass, and specific impulse of your motors (which would be a look up or estimate). From there, its plug and chug.

The "I believe button":confused: Are you being sarcastic? I know what I am asking for is a stretch and I'm not looking for the "ideal equation", just something that can ballpark the fuel loads and trajectories.

If I can calculate the max fuel load and thrust, just to loft the fuel load, I can initially size the vehicle, which gives me its mass. That, in turn, gives me the thrust requirements which refines the fuel mass, and iteratively will zero in on the final design. Using EXCEL I can graph the iterations to get to the final values quickly.

I tried to interpolate/extrapolate existing rockets but I can't find range/mass/fuel loads for ballistic missiles, only a few orbital rockets. I am not designing an orbiter. Complicating that were different fuels used by various rockets, so I had insufficient data to get to 1st base, other than an intuitive feel for the size of the ballpark.
 
  • #10
Sorry - I must have hit you with an inside joke that I assumed translated well to text.

Back in college, if one of my study group couldn't follow the derivation of an equation we "hit the 'I believe' button" (imagine frustrated student pantomiming pushing a button) which meant that we were lost in the derivation but we'd use the equation anyway without understanding why it worked.

The ideal rocket equation isn't 'ideal' because it's the best. It's 'ideal' because it takes into account no real-world inefficiencies - dealing instead with the "ideal rocket".

You don't need to know thrust. You don't need to simulate the burning of fuel. The ideal rocket equation states that knowing the "Specific Impulse" of the rocket (it'll be somewhere in the mid 200s for a solid motor and somewhere in the mid 300s to very low 400s for a hydrogen/oxygen motor - this can be looked up and approximated), the beginning mass (with fuel), and the ending mass (after the fuel is burned), you will know how much velocity your craft will have gained at the end.

If you hit ~9-10km/sec (it's only ~7.75km/sec to "be" in orbit, but you'll need to deal with the real world) you can approxomate orbit. You'll need significantly less for sub-orbit.

You can also re-arrange the equation if you know how much velocity you want to get how much fuel you need relative to your ship's dry mass.
 
  • #11
OK, and that equation is ... ?
 
  • #13
Thank you, I now have some studying to do.

But it looks like the equation relates to orbital insertion, or attaining about 17,000+mph. I need sub orbital, or around (wild guess at this point) 10,000mph, or am I missing something?

The one big part of my problem, which I need to do first, is calculate what velocity I need to obtain various ranges between 2,000 and 12,000 miles. Although the airframe/fuel capacity design point will be 12,000 miles with a defined payload, the intermediate ranges will determine either the reduction in fuel, or the increase in payload.

I still want to maintain a max ~2G in any case.

Also I guess the retro rocket fuel consumption would be the inverse of a similar acceleration, only a bit more complex.

jgb
 
  • #14
Same equation. You're just calculating changes in velocity as a function of fuel burned.
 
  • #15
enigma said:
Same equation. You're just calculating changes in velocity as a function of fuel burned.

I am confused. How do I derive the range for a given velocity or vice versa? My quick glance at the equation does not reveal a factor for range.

There is no "range" for an orbit, only an "altitude" or orbital radius/diameter. Do I then have to intersect that orbital diameter with the Earth's diameter to derive range?
 
  • #16
Ok. First, you have to decide on some things. There are not magic numbers that will give you a flight plan. You have to work from both sides. How sub-orbital are we talking here? Virgin Galactic's ships only go to around Mach 3 and they head into "space" for a few minutes on their parabolic trajectories. Depending on how high you want to go, you can go faster, but just keep in mind that as you increase your desired speed, you are increasing the propellent you need, which increases the size and weight of the ship, which increases the amount of propellent you need, which...and on and on.

Anyway. Flight trajectories can be visualized by basic kinematics. You throw a tennis ball at an angle. That ball has both horizontal and vertical components of velocity. The vertical component works against gravity and will tell you how high you will go. The horizontal component will work against drag and tell you how far you will go.

With air vehicles, you have a radial velocity and a tangential velocity which correspond to your horizontal and vertical components, respectively. You have an angle of attack and a thrust angle. These are not necessarily the same, but for your calculations, you might want to take them to be.
(Just to confuse you more, imagine that a rocket is flying at a 45 degree angle. It ignites a booster and pitches upward thrusting toward the sky at a 65 degree angle. It doesn't start straight up at 65 degrees, since it already has radial velocity, so while it is thrusting at 65 degrees to the horizontal, the actual angle it is moving at, is somewhere in between and changing until it reaches a sort of steady state.)

The easiest way to do this calc (and I can't provide the equations now, sorry) is to take the most simplified model and create your equations. Imagine a flat plane with gravity, zero drag, and a single force vector (your rocket it always pointed in one direction).

From there, you can add drag. However, this becomes tricky as the density and temperature of the atmosphere changes significantly throughout the flight of trans-atmospheric vehicles. So you will have to make some sort of approximated model of the atmosphere to estimate the drag forces your vehicle will experience.

Then, you will have to add the round Earth (non rotating) and understand all the new angles that you have created. This will be a tough one as you will have to translate all of your "X" components into r*theta comonents, among other things, and that can get tricky.

From there you can do all sorts of things. Do you want to account for Coriolis? Do you want to account for thrust vectoring? Do you want to account for a rotating Earth (obviously, you will want to take off flying in the direction of the spin of the Earth as you already have that angular velocity and don't want to fight against it)?
You can account for the fact that you will be burning fuel. Then you can account for the fact that burning fuel will change your center of mass, which will change your pitch during thrusting, which will change your stability, angle of attack, and other things.

They don't call it rocket science because it is easy! haha. But you can make simplified models that work decently as estimates.
 
  • #17
Travis, I really appreciate your response. There were a few things I didn't figure on, but no show-stoppers.

I think, in fairness to you and Enigma, and the others that responded, I need to explain what exactly I am trying to define. With the exception of exactly how I will make these vehicles (subject of a patent development) that are significantly lighter than by today's manufacturing, I don't think there is much proprietary in my mission model, but the business model is.

We all know of the various projects to get a few very well heeled passengers to the edge of space and back. But all these are basically just an up/down flight providing at best a few minutes of micro-g with a significant price tag.

My mission is a sub-orbital commercial passenger transport for 30 to 50 well heeled passengers traveling between major cities of the world, on a daily scheduled service. ie: New York to London in about 90 minutes with about 20 min of micro-g. Or New York to Singapore in under 2 hours with about 40 min of micro-g. Pricing will be in the order of 2 to 5 times today's first class fare. There would be significant business traffic; there and back in 1 day, as well as "zero-G" junkies. NASA/universities could do 80 minutes of Earth bound micro-g experiments in a day, every day for less than the cost of a rocket with payload recovery today.

It is a 2 vehicle affair. The 2 vehicles will look similar, flying wings, but the booster is about 65% bigger.

The booster, (of which 1 or 2 resides at each major airport) will carry the sub-orbiter to about 50,000 to 65,000 feet under conventional jet power. Maybe a short rocket boost near the separation point. After separation, the booster flies back to the departure airport to be quickly readied for another flight with another sub-orbiter to a different destination.

The booster is simply there to get the sub-orbiter off the ground and is way over powered by itself. I plan 6 (maybe 8) GE90-115b class engines each generating 115,000 lbsT. That can lift a total mass of around 2,400,000 lbs. It carries only enough jet fuel for the 450 mile loft and return trip, and is essentially an empty airframe. The return after lofting will be on 2 engines. Its empty weight will be around 275,000 lbs. most of that engines, and landing gear.

The sub-orbiter is also a flying wing, with a delta planform. Other than the passengers and crew, it is a flying fuel tank. The mission profile is to be launched from the booster at about 50,000 ft. in the general direction of the destination. It too has 2 GE90-115b jet engines for takeoff assist and airport approach and landing at destination. With now 8 engines for boost, that equates to a total vehicle weight of 3,200,000 lbs. The rockets ignite at separation and the jets shut down. On a (nominal) 45 degree angle the acceleration will be limited to 2G, as we are not transporting trained astronauts here.

While accelerating the sub-orbiter will roll 180 deg to roof towards earth. As it is a flying wing there is no provision for conventional windows. The large windows are on top, sun roof style. I envision the passengers to free float in micro-g and congregate around the windows peering at the earth. Or maybe not.

MECO is determined by range required. Then a micro-g coast for 20 to 40 minutes or so, and the rockets swivel into reverse thrust position, thereby eliminating a 180 deg pitch maneuver. I want to slow the craft down at a max -2G to about M2 - 2.5 before reentering. Like I said before this significantly reduces airframe heating and stresses, and makes for a simpler design and manufacturing. Passenger seats will rotate to allow a normal recline into the flight vector. Again, they ain't astronauts.

After MECO2 there will be a short glide down to about 35,000 feet when the 2 jet engines are air started. The short 150 - 200 mile flight to destination is like any plane today. Having jet power at destination allows for a go-around or short hop to an alternate.

At destination, it is checked and refueled in about 2 hours, and readied for another flight.
In the event of flying to an alternate, the sub-orbiter is very capable of taking off unassisted with its own jets, having just jet fuel, and no rocket fuel on board. The empty weight will be around 250,000 lbs. 2 GE90-115b engines lift a Boeing 777 weighing over 700,000 lbs.

As for fuels, the rockets will burn jet fuel and an oxidizer. Even though LH2 is way more efficient, it means far more complexity of design and for ground support. The oxidizer may be LOX, because I want to investigate H2O2, even though I will need more than LOX. I am shying away from LOX for several reasons. H2O2 is easier to make and store than LOX, and considering we are at commercial airports, that is critical. Also cryo-fuels are far harder to tank and control on the sub-orbiter.

I fully understand this is not a very efficient and esoteric design. I am carrying extra fuel to compensate, but the business case proves out the concept.

In addition, there is of course a military model. It can perform an overflight mission as does the U2 or SR-70, but at very high altitudes at a moments notice anywhere in the world. It can land anywhere then ferry itself back to any nearby booster airport for a return trip. There are several other missions as well.

Future development would expand its capability to a LEO vehicle, resuming shuttle like flights.

But, to prove its feasibility, I need to know what my fuel loads for various flight profiles would be, so the airframe designs can be defined. Then the business case can be either developed, or go I to work on some other pipe-dream. I got a bunch.

jgb
 
  • #18
I hate to be the guy who tells you your idea isn't going to work, but I really think it's not going to work. If your craft is able to go from New York to Singapore sub-orbitally with minimal input from jet assist, it's going to have to attain probably 90% of orbital speed. That's on the order of 7 km per second. To do that with H2/O2, you need on the order of 10 to 20 pounds of fuel per pound of structure, depending on initial launch statistics. You are planning on doing it with much less efficient fuels, which will add even more needed fuel. The final nail in the coffin is that you want to slow it down by rocket also. That means that you will need an additional 20+ pounds of fuel per pound of structure AND pound of fuel for the retro firing. That adds up to over 400 pounds of fuel per pound of structure and payload. The craft would need to be absolutely gargantuan.
 
  • #19
Enigma,

That is why I need the formulas (in non-calculus form) to prove/disprove the idea.

Yes, I know the craft will be big, I need to figure out how big. If its anywhere near 400 lb/lb (and I don't have issue with that number) then it's onto something else. But I suspect (gut feel) it's not that high, based on the fact that the first 50,000 ft and 500 mph is airplane mode with a drop-off 1st stage, and the reentry will have some aero drag (just not high enough to warrant an exotic heat shield) and the final flight will be in a very low wing loaded airplane. I suspect that even at flight idle, the 2 GE90-115b's may be more than enough power.

As for fuel, the thought was to burn H2O2 with jet fuel, not just use H2O2 alone, except for RCS. That should raise the isp somewhat and reduce the fuel load.

That's also why I need the formulas. Everybody is dancing around that, notwithstanding you pointed me towards some good formulas, but they are of very little use to me in their generic form.

jgb
 
  • #20
deltaV=-Isp*g0*ln(MF) where g0 is 9.8m/sec^2, MF is the mass fraction (final mass divided by initial mass), and Isp is the fuel specific impulse.

You need delta V to be around 7000. I'll ballpark Isp to be 280. That's probably high for RP1 and H2O2. Re-arranging to solve for mass fraction gives .078. One divided by that gives you how many pounds of fuel for each pound of structure - just about 13.

Square that because you need to do the same burn twice. I'm getting 164 pounds of fuel per pound of structure. Less than my ballpark, but I'm giving hefty allowances for delta V and Isp.

You'll be hard pressed to get a fuel tank able to hold that much fuel, regardless of how big it is.

No calculus involved.

Also, the 500 mph you get from your first stage is basically insignificant. To do a suborbital trajectory to the other side of the earth, you need to be going 16000 mph give or take.
 
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  • #21
Well I guess it's onto something else.

I designed an airframe to hold 27million lbs of fuel that weighs in at just over 240,000 lbs empty, including engines, landing gear and payload. Its just a bit bigger than an Airbus A380. This is larger than what I expected it to be, by about 20+% in size, and the weight is a bit understated by my approximations. I guess it should be closer to 300,000 lbs OEW.

But 27m lbs equates to about 45 GE90-115b engines for takeoff, and that's the show stopper.

If anyone is interested, I will post some JPG's of the design. They are rough, very preliminary drawings of the airframe, fuel tanks and payload module, with an A380 for comparison. I used Sketchup to get the tank volumes and surface areas to do the volumetric and structural weight approximations.

Thanks Enigma, even though you shot it down, but I had to know. No hard feelings.

jgb
 

1. What is the formula for calculating the maximum altitude of a sub-orbital flight?

The formula for calculating the maximum altitude of a sub-orbital flight is: h = R * (1 + (v^2 / (2 * g * R))) where h is the maximum altitude, R is the radius of the Earth, v is the initial velocity of the spacecraft, and g is the acceleration due to gravity.

2. How do you calculate the required velocity for a successful sub-orbital flight?

The required velocity for a successful sub-orbital flight can be calculated using the formula: v = sqrt((2 * g * R * (sqrt((h / R) + 1))) where v is the required velocity, g is the acceleration due to gravity, and h is the desired maximum altitude.

3. Can you explain the role of the rocket equation in sub-orbital flight?

The rocket equation, also known as the Tsiolkovsky rocket equation, is used to calculate the change in velocity (delta-v) of a spacecraft based on the mass ratio of the spacecraft and the exhaust velocity of its engines. In sub-orbital flight, this equation is important in determining the amount of fuel needed for the spacecraft to reach the required velocity and altitude.

4. How do you calculate the trajectory of a sub-orbital flight?

The trajectory of a sub-orbital flight can be calculated using the equations of motion, which take into account the initial velocity, acceleration due to gravity, and the angle of ascent. Depending on the specific flight path, other factors such as air resistance may also need to be considered in the calculation.

5. What is the difference between sub-orbital and orbital flight?

Sub-orbital flight refers to a flight path that does not reach the velocity required to enter into orbit around the Earth. The spacecraft follows a ballistic trajectory, reaching a maximum altitude before returning to the Earth. Orbital flight, on the other hand, requires a much higher velocity to enter into a stable orbit around the Earth, and does not return to the Earth's surface unless intentionally deorbited.

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