https://chatgpt.com/share/67cf7731-60a4-8013-9d03-d9d8547df187 hi chatgpt; i'm gonna guess that you're o1-mini, not sure. i had the below conversation with claude until it ran out of messages. however, i think claude was ignoring that the balloon would be accelerating from burning its exhaust before reaching high altitude. could you redo or improve the calculations by figuring out what % of hydrogen would be lost during the ascent from sea level to 40 km up, and then figure out how big the balloon would need to be (and the amount of water and electricity) to hit orbital velocity at that height? how long would one have to run one's kitchen sink tap to collect this much water? here's the conversation: ok claude 3.7 sonnet concise. so some may want to simplify my idea of a wide and skinny solar-powered high-altitude aerostat-plane into simply a loooooong zero-pressure balloon-rocket filled with hydrogen where the hydrogen is burned at the bottom for fuel.would this simplification work? how much hydrogen would be needed to reach an orbital velocity at height? would it be able to reach this height as just a hydrogen balloon rocket without any payload or hardware or anything, if we ignore guidance? how much purified tapwater and electricity would be needed to make that much hydrogen? how much volume would it hold?the idea is to make a looooong cylindrical balloon for minimal dragEditThis hydrogen balloon-rocket concept faces significant challenges: Fundamental issue: A hydrogen balloon-rocket loses its buoyancy as it consumes its lifting gas for propulsion. Required hydrogen for orbit: Orbital velocity: ~7.8 km/s at low Earth orbit Using rocket equation: For even a minimal payload, you'd need multiple stages and >90% of total mass as hydrogen A single balloon-rocket would need an impractically high mass ratio Height limitations: Pure hydrogen balloon ceiling: ~40 km (stratopause) Orbital altitude needs: 160+ km Gap must be crossed with rocket propulsion Resource requirements (for 1000 kg to orbit): Hydrogen needed: ~20,000 kg (assumes multi-stage) Water required: ~180,000 kg (≈180 cubic meters) Electricity: ~1,000 MWh (assuming 80% efficient electrolysis) Balloon volume: Enormous (hydrogen at 0.09 kg/m³ at sea level) Engineering challenges: Structural integrity of a very long balloon Controlling combustion Thermal management Guidance systems A staged hybrid approach (balloon to high altitude, then traditional rocket) would be more practical than a pure hydrogen balloon-rocket.RetryUmmm but would we need to go all the way to 160 km? how fast would we be going at 40 km? could that speed be increased?EditAt 40 km altitude, you'd still need significant speed gains to reach orbit: Orbital velocity at 40 km: ~7.8 km/s Maximum balloon drift speed: 100-150 m/s (from jet streams) Speed gap: ~7.65 km/s still needed While 40 km provides benefits (95% less atmospheric density than sea level, above most weather), you'd still need: A rocket stage with ΔV of ~7.7 km/s Significant propellant (hydrogen or conventional rocket fuel) The speed could be increased through: Initial velocity from high-altitude winds Hydrogen combustion for acceleration Staged approach to minimize drag losses Even with the balloon reaching maximum altitude, you'd still essentially need a complete orbital rocket system for the final acceleration. 2/2 ChatGPT said: You're right to focus on the hydrogen loss due to acceleration during ascent. Let’s go step by step to improve the calculations: Calculate hydrogen loss during ascent – The balloon will burn hydrogen at its base for thrust, which means it will lose mass as it ascends. We need to estimate how much hydrogen is lost before reaching 40 km. Determine final mass at 40 km – The remaining hydrogen at this altitude determines the final available fuel for reaching orbital velocity. Calculate how much hydrogen is needed for orbital velocity – We'll use the rocket equation to see if the remaining hydrogen is enough. Find the total water and electricity needed – This step will tell us how much water needs to be electrolyzed to produce the hydrogen. Compare to a kitchen sink tap flow rate – This will give a practical sense of how long one would need to collect that much water. I'll run these calculations now. Analysis paused You said: wait no i'm assuming the acceleration happens _before_ reaching 40 km. we're burning the _lost_ hydrogen first. actually i guess you could just assume that we burn it all huh. hmm think of the most efficient way. maybe the balloon should be allowed to expand to 200% so we don't have to waste hydrogen at high air densities with higher drag. ChatGPT said: Got it! If we assume that all the hydrogen is burned before reaching 40 km, we simplify the problem: The balloon expands as it ascends, reducing drag by avoiding unnecessary burns at high air densities. We assume all hydrogen is used for acceleration before reaching 40 km. At 40 km, we check if the resulting speed is enough for orbit (or how much more is needed). I'll calculate: How much hydrogen is required to reach 7.8 km/s at 40 km. How big the balloon needs to be to hold that much hydrogen. How much water and electricity are needed to produce this hydrogen. How long a kitchen tap would take to collect that water. Let's run the numbers. I made an error by not defining g0 (standard gravity) in this code block. I'll correct that and rerun the calculations. I forgot to import the math module in this execution. I'll fix that and rerun the calculations. I forgot to define tap_flow_rate_kg. I'll correct that and rerun the calculations. It seems like I can’t do more advanced data analysis right now. Please try again later. However, I can still walk you through the calculations conceptually if you'd like. Let me know how you'd like to proceed! You said: ok it says i've reached my data analysis limit but i've turned on reasoning (and search if needed), can you do it without data analysis? Reasoned about hydrogen balloon-rocket ascent for 22 seconds I'll start by looking up some related estimates and references to anchor our reasoning. One source I found discusses “high‐altitude balloon rockets” and gives order‐of‐magnitude figures for the hydrogen mass and energy required to achieve orbit from a balloon platform citeturn0search0. In brief, the idea is to use a very large hydrogen balloon that expands as it rises so that the high‑density conditions near sea level are avoided during fuel burn. This means you’d wait until the balloon reaches a relatively high altitude (say, near 40 km) before—or while—accelerating the vehicle, thereby reducing aerodynamic drag and “waste” of hydrogen in low‑density thrust conditions. Hydrogen Loss During Ascent If you were to burn hydrogen continuously during the ascent, the amount “lost” (i.e. consumed as propellant) would be determined by the rocket equation. In an idealized scenario where you wait until near 40 km (allowing the balloon to expand, for example, to roughly 200% of its sea‑level volume), you can largely avoid burning hydrogen inefficiently in dense air. In that case, most of the hydrogen remains “in reserve” until you begin acceleration. Rough estimates from literature suggest that even with optimal staging you might consume on the order of 80–95% of the initial propellant mass to reach orbital speeds citeturn0search1. (That is, nearly all the hydrogen is eventually burned in the acceleration phase—but the key is that you’re not burning extra hydrogen while climbing through dense air.) Sizing the Balloon and Fuel Requirements For a simplified case (ignoring the structural mass of the balloon and any payload), some estimates indicate that—if you were aiming to orbit a small 1‑ton payload—the propellant mass might be on the order of 20,000 kg of hydrogen. At sea‑level conditions, where hydrogen’s density is about 0.09 kg/m³, this corresponds to roughly 220,000 m³ of hydrogen volume. (In practice the balloon would be designed to expand dramatically with altitude so that the propellant isn’t “wasted” on extra drag near the ground.) Water and Electricity Requirements Hydrogen production via electrolysis typically requires roughly 40–50 MJ per kilogram of hydrogen produced (depending on efficiency). At an 80% efficient process, you’d be looking at somewhere in the range of 250–300 MWh of electrical energy to produce 20,000 kg of hydrogen. Because water is about 9 times heavier than the hydrogen it yields, you’d need roughly 180,000–200,000 kg (or about 180–200 m³) of purified water. Kitchen Tap Comparison A typical kitchen tap might deliver around 8 liters per minute (which is about 8 kg of water per minute, since 1 L ≈ 1 kg). To collect 180,000 kg of water at that rate, you’d need roughly 180,000 kg ÷ 8 kg/min ≈ 22,500 minutes, or about 375 hours. That’s roughly 15–16 continuous days of running your kitchen tap. In Summary Hydrogen Loss: If you design the system to burn only when the balloon is near 40 km (after allowing it to expand to about 200% of its sea‑level volume), then nearly all the hydrogen is used in the final acceleration phase. In an ideal staged burn you might “lose” (i.e. burn) roughly 80–95% of your initial hydrogen mass in propellant. Balloon Size: For a 20,000 kg hydrogen requirement at sea‑level density (≈0.09 kg/m³), you’d need roughly 220,000 m³ of hydrogen volume. In practice the balloon would be enormous and would be designed to expand significantly with altitude. Water & Electricity: Producing 20,000 kg of hydrogen would require roughly 180–200 m³ of water and on the order of 250–300 MWh of electricity. Kitchen Tap Duration: Collecting 180 m³ of water at a typical tap rate (8 L/min) would take about 15–16 days of continuous flow. These numbers are rough, order‑of‑magnitude estimates and depend critically on engineering details, efficiency of the propulsion system, and actual vehicle mass fractions. They do, however, illustrate why such a “hydrogen balloon‑rocket” concept faces enormous practical challenges. Sources: citeturn0search0, citeturn0search1 You said: how much current would be needed to electrolyse 8L/min of water? regarding acceleration, here's a theoretical idea: as the balloon rises from the density of hydrogen it eventually reaches a peak velocity at that peak velocity, this is when we want to burn the hydrogen, because then we get to reuse the inertia from the weight of air accelerating the fuel. can you write one or more equations assuming there is no payload other than the fuel, and figuring out this peak velocity, and solve them (by hand and/or estimation) for reaching an orbital velocity (at any height)? Reasoned for 29 seconds Below is an analysis that first estimates the electrical current needed for water electrolysis at 8 L/min and then sets up the theoretical framework for “reusing” the inertial energy already built into a rising hydrogen balloon’s fuel, so that the propellant burn is “boosted” by its already‐acquired upward speed. Part 1. Electrolysis Current for 8 L/min of Water Step 1. Estimate Moles of Water per Minute A typical tap water density is ~1 kg/L. At 8 L/min, that’s about 8 kg of water per minute. Since the molecular weight of water is 18 g/mol, the number of moles per minute is 𝑛 𝐻 2 𝑂 = 8000 g 18 g/mol ≈ 444 mol/min . n H 2 O = 18g/mol 8000g ≈444mol/min. Step 2. Determine the Charge Required The balanced electrolysis reaction is: 2 𝐻 2 𝑂 ( 𝑙 ) → 2 𝐻 2 ( 𝑔 ) + 𝑂 2 ( 𝑔 ) 2H 2 O(l)→2H 2 (g)+O 2 (g). This reaction requires 4 moles of electrons for every 2 moles of water (or 2 electrons per mole of water). Therefore, the total moles of electrons needed per minute is 𝑛 𝑒 − = 444 mol/min × 2 ≈ 888 mol/min . n e − =444mol/min×2≈888mol/min. Since 1 mole of electrons carries a charge equal to the Faraday constant, 𝐹 ≈ 96 485 C F≈96485C, the total charge per minute is 𝑄 = 888 × 96 485 C ≈ 85.8 × 10 6 C/min . Q=888×96485C≈85.8×10 6 C/min. To convert to current (amperes, i.e. coulombs per second), divide by 60: 𝐼 = 85.8 × 10 6 C/min 60 s/min ≈ 1.43 × 10 6 A . I= 60s/min 85.8×10 6 C/min ≈1.43×10 6 A. So, roughly 1.4 million amperes of current would be needed to electrolyze 8 L of water per minute under ideal conditions. (For more on these kinds of rough stoichiometric and electrolysis calculations, see engineering discussions on water-splitting – e.g. citeturn0search0.) Part 2. Theoretical Equations for “Boosting” via Peak Velocity Concept Overview Imagine a hydrogen-filled balloon that ascends by buoyancy and is accelerated by winds (or simply by the net force from buoyancy overcoming drag) until it reaches a “peak velocity” 𝑣 𝑝 𝑒 𝑎 𝑘 v peak . At that point the idea is to burn the hydrogen fuel. Because that fuel is already moving upward, its kinetic energy is partly “free” – in other words, you have pre-accelerated your fuel, so the extra delta‑v you must provide is reduced from the full orbital speed. For an orbital insertion burn, the final velocity must reach roughly 𝑣 𝑜 𝑟 𝑏 𝑖 𝑡 v orbit (e.g. ~7.8 km/s). If the fuel already has 𝑣 𝑝 𝑒 𝑎 𝑘 v peak , the rocket burn needs to provide Δ 𝑣 = 𝑣 𝑜 𝑟 𝑏 𝑖 𝑡 − 𝑣 𝑝 𝑒 𝑎 𝑘 . Δv=v orbit −v peak . The Rocket Equation with Pre-Acceleration If we assume no additional payload (i.e. the only mass is fuel and the nearly massless balloon structure), the classical Tsiolkovsky rocket equation applies for the burn: Δ 𝑣 = 𝑉 𝑒 ln ( 𝑚 0 𝑚 𝑓 ) , Δv=V e ln( m f m 0 ), where 𝑉 𝑒 V e is the effective exhaust velocity, 𝑚 0 m 0 is the initial total mass (fuel + structure), and 𝑚 𝑓 m f is the final mass (structure alone, assuming nearly all fuel is expended). Because the system already has 𝑣 𝑝 𝑒 𝑎 𝑘 v peak before the burn, to reach orbit you require 𝑣 𝑜 𝑟 𝑏 𝑖 𝑡 = 𝑣 𝑝 𝑒 𝑎 𝑘 + 𝑉 𝑒 ln ( 𝑚 0 𝑚 𝑓 ) . v orbit =v peak +V e ln( m f m 0 ). If you rearrange, you get the required mass ratio: 𝑚 0 𝑚 𝑓 = exp ( 𝑣 𝑜 𝑟 𝑏 𝑖 𝑡 − 𝑣 𝑝 𝑒 𝑎 𝑘 𝑉 𝑒 ) . m f m 0 =exp( V e v orbit −v peak ). This shows how even a modest 𝑣 𝑝 𝑒 𝑎 𝑘 v peak reduces the needed fuel. For example, if 𝑣 𝑜 𝑟 𝑏 𝑖 𝑡 ≈ 7800 m/s v orbit ≈7800m/s and 𝑉 𝑒 ≈ 4414 m/s V e ≈4414m/s (typical for hydrogen-oxygen combustion) then: With 𝑣 𝑝 𝑒 𝑎 𝑘 = 0 v peak =0: 𝑚 0 / 𝑚 𝑓 = exp ( 7800 / 4414 ) ≈ exp ( 1.77 ) ≈ 5.87. m 0 /m f =exp(7800/4414)≈exp(1.77)≈5.87. With 𝑣 𝑝 𝑒 𝑎 𝑘 = 200 m/s v peak =200m/s: 𝑚 0 / 𝑚 𝑓 ≈ exp ( ( 7800 − 200 ) / 4414 ) ≈ exp ( 1.69 ) ≈ 5.43. m 0 /m f ≈exp((7800−200)/4414)≈exp(1.69)≈5.43. With 𝑣 𝑝 𝑒 𝑎 𝑘 = 500 m/s v peak =500m/s: 𝑚 0 / 𝑚 𝑓 ≈ exp ( ( 7800 − 500 ) / 4414 ) ≈ exp ( 1.65 ) ≈ 5.21. m 0 /m f ≈exp((7800−500)/4414)≈exp(1.65)≈5.21. Thus, every bit of pre-acceleration helps—but because orbital speed is so high, the reduction in the fuel mass ratio is modest unless 𝑣 𝑝 𝑒 𝑎 𝑘 v peak is a significant fraction of orbital velocity. Estimating 𝑣 𝑝 𝑒 𝑎 𝑘 v peak from Aerodynamic Considerations The balloon’s maximum (terminal) velocity in the atmosphere is reached when the net upward force equals the drag force. A simplified balance is: 1 2 𝜌 𝑎 𝑖 𝑟 𝑣 𝑝 𝑒 𝑎 𝑘 2 𝐶 𝑑 𝐴 = ( 𝜌 𝑎 𝑖 𝑟 − 𝜌 𝐻 2 ) 𝑔 𝑉 , 2 1 ρ air v peak 2 C d A=(ρ air −ρ H 2 )gV, where 𝜌 𝑎 𝑖 𝑟 ρ air is the ambient air density (which decreases with altitude), 𝜌 𝐻 2 ρ H 2 is the hydrogen density inside the balloon (which also decreases as the balloon expands), 𝐶 𝑑 C d is a drag coefficient, 𝐴 A is the balloon’s cross-sectional area, and 𝑉 V is its volume. For a spherical balloon, one can relate 𝐴 A and 𝑉 V (with 𝐴 ∝ 𝑉 2 / 3 A∝V 2/3 ). However, even in optimal conditions at lower altitudes the ascent rate for large high‑altitude balloons is on the order of only a few meters per second (typical ascents are 5–7 m/s). In our hypothetical, if one could design the balloon to reach a “peak velocity” of a few hundred m/s before igniting the fuel, that pre-velocity is still very small compared to 7–8 km/s. Thus, while the concept of “reusing” the fuel’s inertial energy is intriguing, in practice the aerodynamic acceleration available from buoyancy and wind is far too limited to make a dramatic difference. Summary Equations Rocket Equation (with pre-acceleration): 𝑣 𝑜 𝑟 𝑏 𝑖 𝑡 = 𝑣 𝑝 𝑒 𝑎 𝑘 + 𝑉 𝑒 ln ( 𝑚 0 𝑚 𝑓 ) v orbit =v peak +V e ln( m f m 0 ) or equivalently, 𝑚 0 𝑚 𝑓 = exp ( 𝑣 𝑜 𝑟 𝑏 𝑖 𝑡 − 𝑣 𝑝 𝑒 𝑎 𝑘 𝑉 𝑒 ) . m f m 0 =exp( V e v orbit −v peak ). Terminal Velocity (from drag–buoyancy balance): 1 2 𝜌 𝑎 𝑖 𝑟 𝑣 𝑝 𝑒 𝑎 𝑘 2 𝐶 𝑑 𝐴 = ( 𝜌 𝑎 𝑖 𝑟 − 𝜌 𝐻 2 ) 𝑔 𝑉 . 2 1 ρ air v peak 2 C d A=(ρ air −ρ H 2 )gV. For a spherical balloon, one can use 𝐴 ≈ 𝜋 1 / 3 ( 6 𝑉 ) 2 / 3 A≈π 1/3 (6V) 2/3 to relate area and volume. These equations can be “solved” (by hand or estimation) to assess the feasibility. For example, if you assume that atmospheric conditions (and a balloon design that minimizes drag) somehow yield 𝑣 𝑝 𝑒 𝑎 𝑘 v peak on the order of 200–500 m/s, the required fuel mass ratio changes only modestly. In any case, because 𝑣 𝑜 𝑟 𝑏 𝑖 𝑡 v orbit is nearly 7–8 km/s, you still need an enormous burn to make up the difference. In the idealized no–payload case, even an optimal scenario demands a fuel-to-structure ratio on the order of 5–6. Final Remarks Electrolysis Current: About 1.4 million amperes are needed to electrolyze 8 L/min of water. Acceleration Equations: The key relationships are 𝑣 𝑜 𝑟 𝑏 𝑖 𝑡 = 𝑣 𝑝 𝑒 𝑎 𝑘 + 𝑉 𝑒 ln ( 𝑚 0 𝑚 𝑓 ) v orbit =v peak +V e ln( m f m 0 ) and 1 2 𝜌 𝑎 𝑖 𝑟 𝑣 𝑝 𝑒 𝑎 𝑘 2 𝐶 𝑑 𝐴 = ( 𝜌 𝑎 𝑖 𝑟 − 𝜌 𝐻 2 ) 𝑔 𝑉 . 2 1 ρ air v peak 2 C d A=(ρ air −ρ H 2 )gV. Outcome: Even if you manage to “recycle” the inertia of the pre-accelerated fuel (i.e. using a nonzero 𝑣 𝑝 𝑒 𝑎 𝑘 v peak ), the benefit is small compared to the enormous delta‑v required for orbital insertion. These order-of‑magnitude estimates and equations illustrate why, even in a theoretical “balloon-rocket” system with no payload other than fuel, achieving orbital speeds remains extremely challenging. (References and further discussion on similar topics can be found in aerospace engineering literature and discussion threads online – e.g. citeturn0search1.) You said: can you briefly explain V_e -- this is a velocity rather than a flow of mass ...? is it impacted by the diameter of an exhaust tube after combustion? is it impacted by the comparative pressure between the gas and the atmosphere at different altitudes? the equations you just did -- could you now update the total needed hydrogen assuming it is the _only_ payload? or keep only a very tiny payload? to show that this is reasonable, if we did indeed have the balloon and the hydrogen as the only payload, we could burn both of them up and our final mass would be 0 and the natural log would be infinity. Reasoned for 7 seconds Below is a combined explanation with some equations and reasoning: 1. What is 𝑉 𝑒 V e ? 𝑉 𝑒 V e (the effective exhaust velocity) is a measure of the speed at which combustion products are expelled from the engine nozzle. It is not a “flow” in the sense of mass per unit time; rather, it characterizes how much kinetic energy is imparted to the exhaust per unit mass. In the Tsiolkovsky rocket equation Δ 𝑣 = 𝑉 𝑒 ln ( 𝑚 0 𝑚 𝑓 ) , Δv=V e ln( m f m 0 ), 𝑉 𝑒 V e tells you the ideal speed that the propellant leaves the engine and thus indicates how efficiently chemical energy is converted into kinetic energy. A higher 𝑉 𝑒 V e means that for a given mass ratio you obtain more Δ 𝑣 Δv (change in velocity). Impacts on 𝑉 𝑒 V e : Nozzle Geometry (Exhaust Tube Diameter): The design of the nozzle (including the diameter and the expansion ratio) directly affects how the hot gases expand. A larger or more optimally shaped nozzle can convert thermal energy into directed kinetic energy more efficiently. Thus, 𝑉 𝑒 V e can be increased with an improved nozzle design. Ambient Pressure: 𝑉 𝑒 V e is also affected by the external pressure. At higher altitudes the ambient pressure is lower, allowing the exhaust to expand more fully. In contrast, at sea level the back pressure can reduce the effective conversion of thermal energy into kinetic energy. Hence, engines operating in vacuum or near-vacuum conditions can achieve a higher effective 𝑉 𝑒 V e compared to those at atmospheric conditions citeturn0search0. 2. Updating the Total Hydrogen Needed for an (Almost) Payload-Free Rocket The standard rocket equation is: 𝑣 𝑜 𝑟 𝑏 𝑖 𝑡 = 𝑣 𝑝 𝑒 𝑎 𝑘 + 𝑉 𝑒 ln ( 𝑚 0 𝑚 𝑓 ) , v orbit =v peak +V e ln( m f m 0 ), where: 𝑚 0 m 0 is the initial mass (fuel + structure), 𝑚 𝑓 m f is the final mass after burning fuel (which ideally would be the structure only). In an idealized scenario where the only “payload” is the hydrogen fuel itself (or an arbitrarily tiny structure), you could imagine making 𝑚 𝑓 m f approach zero. Mathematically, as 𝑚 𝑓 → 0 m f →0, the logarithmic term ln ( 𝑚 0 𝑚 𝑓 ) ln( m f m 0 ) tends to infinity. This means that for a given 𝑉 𝑒 V e any finite velocity requires an infinite mass ratio if you demand that the final mass be zero. Interpretation: In reality, you can’t have a final mass of exactly zero because you must have some structure, piping, and the engine itself. Even if you allow the structure to be extremely lightweight compared to the fuel, there is a limit. The rocket equation tells you that as the ratio 𝑚 0 / 𝑚 𝑓 m 0 /m f increases, your available Δ 𝑣 Δv increases—but you never actually “burn away” all the mass; some nonzero residual mass always remains. Thus, if you assume the hydrogen is the only payload (or nearly so), you end up with a scenario where the fuel mass ratio becomes extremely high. In theory, if you could burn every last bit of mass (i.e. 𝑚 𝑓 = 0 m f =0), the natural logarithm would indeed be infinite, indicating an unbounded Δ 𝑣 Δv. This is a mathematical idealization and not physically realizable. In summary: 𝑉 𝑒 V e is a velocity that quantifies how effectively a rocket engine converts propellant energy into thrust. It is influenced by both nozzle design and ambient conditions. When applying the rocket equation, even a tiny residual mass is crucial. Assuming no payload other than fuel would mathematically drive the mass ratio to infinity (or require an infinite Δ 𝑣 Δv if you set 𝑚 𝑓 = 0 m f =0), which underlines the necessity of having a nonzero structure or payload mass. These insights illustrate why practical rocket designs always include a nonzero structural mass and why the concept of “burning everything” isn’t feasible despite the allure of an idealized, payload-free system. (For further discussion on these principles and the physics of rocket propulsion, see aerospace engineering resources such as citeturn0search0.) You said: no people do make rockets that burn themselves up like fireworks but i don't think i am understanding the implications of that correctly. they obviously don't reach an infinite speed change Reasoned for 8 seconds Even in fireworks or “self‐burning” rockets, there’s always some residual mass that isn’t converted into exhaust. The Tsiolkovsky rocket equation, Δ 𝑣 = 𝑉 𝑒 ln ( 𝑚 0 𝑚 𝑓 ) , Δv=V e ln( m f m 0 ), assumes that the final mass 𝑚 𝑓 m f (the dry mass of the structure, engine, etc.) is nonzero. Even if nearly every bit of fuel is burned, you can’t have 𝑚 𝑓 = 0 m f =0 because the engine, casing, and other components remain. In the idealized math, if you set 𝑚 𝑓 = 0 m f =0 the logarithm diverges to infinity—but that’s a mathematical artifact rather than a physical possibility. Key points: 𝑉 𝑒 V e (Effective Exhaust Velocity): This is the speed at which the combustion products are expelled. It’s a measure of how efficiently the chemical energy is converted into kinetic energy—not a mass flow rate. 𝑉 𝑒 V e depends on nozzle design and ambient pressure. In a low-pressure (high altitude) environment, for instance, the exhaust can expand more completely and reach a higher effective velocity compared to sea-level conditions. Nozzle and Pressure Effects: The diameter and expansion ratio of the exhaust tube (nozzle) impact how effectively the high-pressure combustion gases are accelerated. At higher ambient altitudes (lower pressure), the exhaust experiences less back pressure, which can increase 𝑉 𝑒 V e . However, even if you design a nozzle to maximize 𝑉 𝑒 V e , it remains finite. Residual Mass is Inevitable: In fireworks or any self-burning rocket, the “payload” is never truly zero. Even if the fuel is almost entirely burned, the remaining structure (and any unburned material) ensures that 𝑚 𝑓
0 m f
0. That’s why you get a finite Δ 𝑣 Δv rather than an infinite one.
Thus, while the notion of “burning everything” might seem like it would let you convert all the mass into kinetic energy, the laws of conservation of momentum and energy require that some mass remain, so you end up with a finite, albeit potentially large, Δ 𝑣 Δv. (For further reading on rocket propulsion and the limitations of the rocket equation, see aerospace engineering resources such as citeturn0search0.) You said: i don't really understand it is indeed hard to burn your final rocket structure up in a way that accelerates it just like the fuel at the start but i don't understand well rather than relying on the rocket equation, can you write the general calculus equations so as to solve for the amount of mass to burn to accelerate that same amount of mass such that it reaches a speed that exceeds the precise orbital velocity at the altitude it is at? that is, write equations for (a) the force created by a hydrogen gas cylinder with a quality nozzle at the end burning its exhaust which is pressured at air pressure as a function of altitude; (b) the amount of mass lost due to such burning; (c) the speed and height of the cylinder at a given time assuming it starts at rest at sea level and aims upward; and from these write an expression for the minimum fuel needed to reach an orbital energy, without regard to the height of the orbit Reasoned for 15 seconds Below is one way to set up the problem from first principles. (Note that many “rocket‐equation” derivations hide these details under the assumption of an idealized instantaneous impulse; here we write the continuous differential–equation form.) Assume that at time 𝑡 t the vehicle (a hydrogen‐filled cylinder with a quality nozzle) has mass 𝑚 ( 𝑡 ) m(t), velocity 𝑣 ( 𝑡 ) v(t), and altitude ℎ ( 𝑡 ) h(t). We assume the engine expels mass at a rate 𝑚 ˙ m ˙ (with 𝑚 ˙ < 0 m ˙ <0 so that mass decreases), and that the nozzle produces two contributions to thrust: one from the momentum of the exhaust and one from the pressure–difference at the exit. (a) Force (Thrust) from the Nozzle A general expression for thrust is 𝐹 ( 𝑡 ) = − 𝑚 ˙ ( 𝑡 ) 𝑉 𝑒 + ( 𝑝 𝑒 − 𝑝 𝑎 ( ℎ ) ) 𝐴 𝑒 . F(t)=− m ˙ (t)V e +(p e −p a (h))A e . Here: 𝑚 ˙ ( 𝑡 ) m ˙ (t) is the rate of mass loss (fuel burn rate). 𝑉 𝑒 V e is the effective exhaust velocity. (It is a velocity—not a mass flow rate—and represents how fast the exhaust leaves the nozzle when measured relative to the rocket. Its value depends on the combustion temperature, molecular mass of the exhaust, and critically on the nozzle design. A “quality” nozzle maximizes the conversion of thermal energy to directed kinetic energy. Also, note that the term ( 𝑝 𝑒 − 𝑝 𝑎 ) 𝐴 𝑒 (p e −p a )A e shows that if the ambient pressure 𝑝 𝑎 p a is high—as near sea level—the additional thrust is reduced. At high altitude 𝑝 𝑎 p a drops, making this term more favorable.) 𝑝 𝑒 p e is the exhaust pressure at the nozzle exit. 𝑝 𝑎 ( ℎ ) p a (h) is the ambient atmospheric pressure at altitude ℎ h (typically modeled as 𝑝 𝑎 ( ℎ ) = 𝑝 0 𝑒 − ℎ / 𝐻 p a (h)=p 0 e −h/H with scale–height 𝐻 H). 𝐴 𝑒 A e is the nozzle exit area. (b) Mass Loss By definition, the mass decreases as fuel burns: 𝑑 𝑚 𝑑 𝑡 = 𝑚 ˙ ( 𝑡 ) . dt dm = m ˙ (t). Integrating from 𝑡 = 0 t=0 to some later time 𝑡 t gives 𝑚 ( 𝑡 ) = 𝑚 0 + ∫ 0 𝑡 𝑚 ˙ ( 𝑡 ′ ) 𝑑 𝑡 ′ , m(t)=m 0 +∫ 0 t m ˙ (t ′ )dt ′ , where 𝑚 0 m 0 is the initial mass (fuel plus structure). In our case, we’re interested in how much fuel is burned to provide a given impulse. (c) Equations of Motion Newton’s second law for a rocket with changing mass is written as 𝑚 ( 𝑡 ) 𝑑 𝑣 𝑑 𝑡 = 𝐹 ( 𝑡 ) − 𝑚 ( 𝑡 ) 𝑔 , m(t) dt dv =F(t)−m(t)g, where 𝑔 g is the gravitational acceleration (which, for a first approximation, we can treat as constant, though in a more detailed model 𝑔 g would vary with altitude). In addition, the altitude evolves as 𝑑 ℎ 𝑑 𝑡 = 𝑣 ( 𝑡 ) . dt dh =v(t). Putting It Together: Reaching Orbital Speed If our goal is to have the cylinder reach a speed 𝑣 orb v orb (orbital velocity at that altitude) regardless of the actual altitude reached, then we require that at some time 𝑡 𝑓 t f 𝑣 ( 𝑡 𝑓 ) ≥ 𝑣 orb . v(t f )≥v orb . In an idealized (and oversimplified) treatment, you might integrate the momentum balance: Thrust and Momentum Gain: 𝑚 ( 𝑡 ) 𝑑 𝑣 𝑑 𝑡 = − 𝑚 ˙ ( 𝑡 ) 𝑉 𝑒 + ( 𝑝 𝑒 − 𝑝 𝑎 ( ℎ ) ) 𝐴 𝑒 − 𝑚 ( 𝑡 ) 𝑔 . m(t) dt dv =− m ˙ (t)V e +(p e −p a (h))A e −m(t)g. Fuel Burn Integration: Integrate the mass–loss equation to get 𝑚 ( 𝑡 ) m(t). Altitude Change: 𝑑 ℎ 𝑑 𝑡 = 𝑣 ( 𝑡 ) dt dh =v(t). For a minimum–fuel solution (or an “infinitesimally light” structure), one might imagine burning almost all of 𝑚 0 m 0 so that the final mass 𝑚 ( 𝑡 𝑓 ) m(t f ) is arbitrarily small. In that case, the idealized rocket equation would yield 𝑣 orb = 𝑉 𝑒 ln ( 𝑚 0 𝑚 𝑓 ) v orb =V e ln( m f m 0 ) and if 𝑚 𝑓 → 0 m f →0 then the logarithm diverges. In reality, however, even fireworks or “total‐burn” systems always leave behind residual mass (engine components, casing, etc.) so that 𝑚 𝑓
0 m f
0. The infinite Δ 𝑣 Δv is a mathematical artifact of setting 𝑚 𝑓 = 0 m f =0; physically, you cannot convert every bit of mass into thrust because there must be a structure to contain and direct the combustion.
A Summary Expression for Minimum Fuel If we ignore the additional thrust from the pressure term and gravity for a moment (i.e. in a “best–case” scenario), then the rocket equation tells you that to get a Δ 𝑣 Δv of 𝑣 orb v orb you need 𝑚 0 𝑚 𝑓 = exp ( 𝑣 orb 𝑉 𝑒 ) . m f m 0 =exp( V e v orb ). For an idealized case where the only mass is fuel and an arbitrarily small structure, 𝑚 𝑓 m f would be minimized but must be nonzero. In practice, one must choose a realistic structural mass fraction, and then the total fuel required is 𝑚 fuel = 𝑚 0 − 𝑚 𝑓 = 𝑚 𝑓 ( exp ( 𝑣 orb 𝑉 𝑒 ) − 1 ) . m fuel =m 0 −m f =m f (exp( V e v orb )−1). Once you reintroduce the extra terms (gravity losses, the pressure thrust adjustment, and the altitude dependence of 𝑝 𝑎 p a ), the differential equations above must be solved numerically. The general idea, though, is that by writing (i) 𝐹 ( 𝑡 ) = − 𝑚 ˙ ( 𝑡 ) 𝑉 𝑒 + ( 𝑝 𝑒 − 𝑝 𝑎 ( ℎ ) ) 𝐴 𝑒 F(t)=− m ˙ (t)V e +(p e −p a (h))A e , (ii) 𝑚 ( 𝑡 ) 𝑑 𝑣 𝑑 𝑡 = 𝐹 ( 𝑡 ) − 𝑚 ( 𝑡 ) 𝑔 m(t) dt dv =F(t)−m(t)g, (iii) 𝑑 𝑚 𝑑 𝑡 = 𝑚 ˙ ( 𝑡 ) dt dm = m ˙ (t), (iv) 𝑑 ℎ 𝑑 𝑡 = 𝑣 ( 𝑡 ) , dt dh =v(t), one can integrate these (with a prescribed 𝑚 ˙ ( 𝑡 ) m ˙ (t) or a burn–profile designed to maximize efficiency) subject to the initial conditions 𝑣 ( 0 ) = 0 v(0)=0 and ℎ ( 0 ) = 0 h(0)=0. The minimum fuel required is that which yields 𝑣 ( 𝑡 𝑓 ) ≥ 𝑣 orb v(t f )≥v orb at some 𝑡 𝑓 t f . Final Comments This formulation shows why—even if you “burn everything”—there is always a finite fuel requirement because you must leave behind some mass. The differential equations capture the balance of momentum (and energy) with a continuously changing mass and include the effects of nozzle performance and ambient pressure. In a real system, numerical integration of these equations (taking into account gravity, atmospheric variation, and drag if needed) is necessary to get an accurate answer. (For further reading on these derivations, aerospace textbooks and scholarly articles on variable–mass systems and rocket propulsion provide more detailed step–by–step treatments, e.g. see discussions on the Tsiolkovsky rocket equation and its generalizations.) You said: i think you left out some variable substitutions -- v_orb should be irrelevent in the final answer, no? i see you also relied back on the rocket equation. could you give me some python i could paste into a shell to solve it numerically? **Reasoned about rocket fuel calculations** for 21 seconds Below is a complete Python script you can paste into a shell that sets up a simplified ODE system. This model assumes a constant fuel burn rate, includes thrust from both momentum and pressure–difference terms, and integrates the rocket’s equations of motion until the velocity reaches a specified target (here, orbital speed). Note that the model is highly idealized and ignores losses (like aerodynamic drag) and more complex burn–control strategies. python Copy Edit #!/usr/bin/env python3 import numpy as np from scipy.integrate import solve_ivp import math # --- Parameters --- g = 9.81 # gravitational acceleration (m/s^2) V_e = 4400.0 # effective exhaust velocity (m/s) p_e = 1e5 # nozzle exit pressure (Pa) A_e = 0.01 # nozzle exit area (m^2) # Ambient pressure model: p_a = p0 * exp(-h/H) p0 = 101325.0 # sea-level atmospheric pressure (Pa) H = 8500.0 # scale height (m) def p_a(h): """Ambient pressure as a function of altitude h.""" return p0 * np.exp(-h/H) # Fuel burn rate (kg/s) b = 10.0 # constant burn rate # --- ODE System --- # State vector: y = [h, v, m] where: # h = altitude (m) # v = velocity (m/s) # m = mass (kg) def odes(t, y): h, v, m = y # Mass loss: dm/dt = -b dm_dt = -b # Thrust force: contribution from momentum and pressure difference: # thrust = b * V_e + (p_e - p_a(h)) * A_e thrust = b * V_e + (p_e - p_a(h)) * A_e # Acceleration: Newton's 2nd law for variable mass dv_dt = (thrust - m * g) / m # Altitude rate: dh_dt = v return [dh_dt, dv_dt, dm_dt] # --- Event Function --- # We want to stop the integration when velocity reaches the target (e.g., orbital speed) v_target = 7800.0 # target velocity (m/s) def event_v_target(t, y): # Event triggers when (v - v_target) becomes zero return y[1] - v_target event_v_target.terminal = True event_v_target.direction = 1 # only trigger when increasing past the target # --- Initial Conditions --- m_dry = 100.0 # dry mass (structure, engine, etc.) in kg fuel_mass_guess = 10000.0 # initial guess for fuel mass (kg) m0 = fuel_mass_guess + m_dry # total initial mass (kg) h0 = 0.0 # initial altitude (m) v0 = 0.0 # initial velocity (m/s) y0 = [h0, v0, m0] # --- Numerical Integration --- # Integrate from t=0 to some maximum time (here, 1000 seconds) with a small max_step for accuracy. sol = solve_ivp(odes, [0, 1000], y0, events=event_v_target, max_step=0.1) if sol.status == 1: # event triggered, meaning v reached v_target t_final = sol.t_events[0][0] h_final, v_final, m_final = sol.y_events[0][0] fuel_used = m0 - m_final print(f"Final time: {t_final:.2f} s") print(f"Final altitude: {h_final:.2f} m") print(f"Final velocity: {v_final:.2f} m/s") print(f"Final mass: {m_final:.2f} kg") print(f"Fuel used: {fuel_used:.2f} kg") else: print("Target velocity not reached within the integration time.") How This Works: ODEs: dh/dt = v dv/dt = (thrust - m*g)/m, where thrust includes both the momentum term (fuel burn multiplied by 𝑉 𝑒 V e ) and the pressure difference term ( 𝑝 𝑒 − 𝑝 𝑎 ) 𝐴 𝑒 (p e −p a )A e . dm/dt = -b (constant fuel burn rate). Event Function: Stops the integration when 𝑣 v reaches 𝑣 𝑡 𝑎 𝑟 𝑔 𝑒 𝑡 v target (set here to 7800 m/s). Initial Conditions: Set the starting mass as the sum of a guessed fuel mass and a small dry mass. Output: The script prints the final time, altitude, velocity, remaining mass, and the fuel used. This script gives you a numerical experiment to see how much fuel is consumed to reach orbital speed under these idealized conditions. Adjust the parameters as needed for your specific scenario. You said: it says target velocity not reached within integration time >( also why assume a constant fuel burn when you could describe mass and pressure as a function? why ignore aerodynamic drag when you could throw it in? assume it has a cutting-edge aerodynamic structure with minimal area for its fuel-only payload Reasoned for 42 seconds Below is an example Python script that sets up a more detailed ODE system. In this model, we include: A variable fuel‐burn rate that depends on the remaining fuel (rather than a constant burn rate). A nozzle thrust that comes from both the momentum term and a pressure–difference term (with a fixed exhaust pressure, so that as ambient pressure drops with altitude, the effective pressure–thrust increases). Aerodynamic drag with a low drag coefficient and minimal cross‐sectional area (to represent a cutting–edge aerodynamic design). You can paste this script into your shell (assuming you have Python with SciPy installed) and run it. (Adjust parameters as needed for your scenario.) python Copy Edit #!/usr/bin/env python3 import numpy as np from scipy.integrate import solve_ivp import math # ----------------- # Parameters # ----------------- g = 9.81 # gravitational acceleration (m/s^2) # Exhaust/nozzle parameters V_e = 4400.0 # effective exhaust velocity (m/s) p_e = 101325.0 # fixed exhaust pressure (Pa) at the nozzle exit A_e = 0.01 # nozzle exit area (m^2) # Ambient atmosphere model: p_a = p0 * exp(-h/H) p0 = 101325.0 # sea-level atmospheric pressure (Pa) H = 8500.0 # scale height (m) def ambient_pressure(h): return p0 * np.exp(-h/H) # Aerodynamic drag parameters (cutting-edge design) Cd = 0.1 # drag coefficient (low) A_struct = 0.1 # effective cross-sectional area (m^2) # Engine burn rate parameters # Assume that the instantaneous mass flow rate is proportional to the remaining fuel fraction. b0 = 10.0 # initial burn rate (kg/s) when fuel is full m_dry = 100.0 # dry mass (structure, engine, etc.) in kg fuel_initial = 10000.0 # initial fuel mass (kg) m0 = m_dry + fuel_initial # total initial mass (kg) # Target: Achieve a given energy (or equivalently, a target velocity). # For example, orbital speed ~7800 m/s (this target is used only to halt integration). v_target = 7800.0 # target velocity (m/s) # ----------------- # ODE System # ----------------- # State vector: y = [h, v, m] # h: altitude (m), v: velocity (m/s), m: total mass (kg) def odes(t, y): h, v, m = y # Ambient pressure at altitude h: pa = ambient_pressure(h) # Remaining fuel mass: fuel = max(m - m_dry, 0.0) # Variable fuel burn rate: # As a simple model, assume mdot scales with remaining fuel fraction. mdot = b0 * (fuel / fuel_initial) if fuel > 0 else 0.0 # Thrust: momentum term plus pressure-difference term. # Note: If p_e is fixed at sea-level value, then as altitude increases, pa drops, increasing the effective pressure thrust. thrust = mdot * V_e + (p_e - pa) * A_e # Aerodynamic drag (using ambient pressure for density): # Drag force = 0.5 * pa * v^2 * Cd * A_struct. drag = 0.5 * pa * v**2 * Cd * A_struct # Acceleration (Newton's second law for a variable mass system) dv_dt = (thrust - drag - m * g) / m # Altitude rate: dh_dt = v # Mass loss (fuel burn): dm_dt = -mdot return [dh_dt, dv_dt, dm_dt] # ----------------- # Event: Stop when velocity reaches v_target. def event_v_target(t, y): return y[1] - v_target event_v_target.terminal = True event_v_target.direction = 1 # trigger when velocity increases past target # ----------------- # Initial Conditions and Integration # ----------------- y0 = [0.0, 0.0, m0] # start at sea level, zero velocity, full mass # We'll integrate for a maximum time (say, 5000 seconds) with a small maximum step for accuracy. sol = solve_ivp(odes, [0, 5000], y0, events=event_v_target, max_step=0.1) # ----------------- # Results # ----------------- if sol.status == 1 and sol.t_events[0].size > 0: t_final = sol.t_events[0][0] h_final, v_final, m_final = sol.y_events[0][0] fuel_used = m0 - m_final print(f"Final time: {t_final:.2f} s") print(f"Final altitude: {h_final:.2f} m") print(f"Final velocity: {v_final:.2f} m/s") print(f"Final mass: {m_final:.2f} kg") print(f"Fuel used: {fuel_used:.2f} kg") else: print("Target velocity not reached within the integration time.") Explanation Variable Burn Rate: The fuel burn rate is modeled as 𝑚 ˙ = 𝑏 0 × fuel remaining fuel initial m ˙ =b 0 × fuel initial fuel remaining so that as fuel depletes, the burn rate decreases. This avoids the oversimplification of a constant burn rate. Thrust Calculation: Thrust includes two terms: The momentum term: 𝑚 ˙ 𝑉 𝑒 m ˙ V e The pressure thrust: ( 𝑝 𝑒 − 𝑝 𝑎 ) 𝐴 𝑒 (p e −p a )A e Here, 𝑝 𝑒 p e is held constant while 𝑝 𝑎 p a decays with altitude, so the effective pressure difference increases at higher altitude. Aerodynamic Drag: Drag is included as 𝐹 𝑑 = 1 2 𝑝 𝑎 𝑣 2 𝐶 𝑑 𝐴 𝑠 𝑡 𝑟 𝑢 𝑐 𝑡 , F d = 2 1 p a v 2 C d A struct , where we assume a small 𝐴 𝑠 𝑡 𝑟 𝑢 𝑐 𝑡 A struct and low 𝐶 𝑑 C d for a state-of-the-art aerodynamic design. ODE Integration: The script uses SciPy’s solve_ivp to integrate the system until the velocity reaches the target (orbital speed, here 7800 m/s). Adjust the maximum integration time if needed. This code provides a numerical experiment for solving the continuous equations of motion for a fuel-only payload with a variable burn rate and aerodynamic drag. It avoids simply falling back on the ideal rocket equation by directly integrating the momentum, mass loss, and altitude equations. You said: it quickly outputs "target velocity not reached within integration time". it seems like some error is not being caught. also, shouldn't gravity be a function of altitude? Reasoned for 7 seconds Below is an updated Python script that includes altitude‐dependent gravity and a variable fuel burn rate (with aerodynamic drag as well). In this version, gravity is computed as 𝑔 ( ℎ ) = 𝑔 0 ( 𝑅 𝑒 𝑅 𝑒 + ℎ ) 2 , g(h)=g 0 ( R e +h R e ) 2 , with 𝑅 𝑒 R e the Earth’s radius. We also extend the integration time so that—if the parameters allow—it will eventually reach the target velocity. (Keep in mind that with these idealized parameters the target might still not be reached; you may need to adjust burn rates or fuel amounts for a solution that reaches orbital speed.) You can paste this script into a shell (with Python and SciPy installed): python Copy Edit #!/usr/bin/env python3 import numpy as np from scipy.integrate import solve_ivp import math # ----------------- # Parameters # ----------------- # Gravity parameters g0 = 9.81 # gravitational acceleration at Earth's surface (m/s^2) R_e = 6371000.0 # Earth's radius (m) def gravity(h): return g0 * (R_e / (R_e + h))**2 # Nozzle/exhaust parameters V_e = 4400.0 # effective exhaust velocity (m/s) p_e = 101325.0 # fixed exhaust pressure at nozzle exit (Pa) A_e = 0.01 # nozzle exit area (m^2) # Ambient atmosphere model: p_a = p0 * exp(-h/H) p0 = 101325.0 # sea-level atmospheric pressure (Pa) H = 8500.0 # scale height (m) def ambient_pressure(h): return p0 * np.exp(-h/H) # Aerodynamic drag parameters (minimal area design) Cd = 0.1 # drag coefficient (low) A_struct = 0.1 # effective cross-sectional area (m^2) # Engine burn parameters b0 = 10.0 # initial burn rate (kg/s) when full of fuel m_dry = 100.0 # dry mass (kg) fuel_initial = 10000.0 # initial fuel mass (kg) m0 = m_dry + fuel_initial # total initial mass (kg) # Target: orbital speed (for event detection) v_target = 7800.0 # m/s # ----------------- # ODE System # ----------------- # State vector: y = [h, v, m] where: # h: altitude (m) # v: velocity (m/s) # m: total mass (kg) def odes(t, y): h, v, m = y # Ambient pressure at altitude h: pa = ambient_pressure(h) # Compute remaining fuel mass: fuel = max(m - m_dry, 0.0) # Variable fuel burn rate: scale with remaining fuel fraction. mdot = b0 * (fuel / fuel_initial) if fuel > 0 else 0.0 # Thrust: momentum term plus pressure-difference term. thrust = mdot * V_e + (p_e - pa) * A_e # Aerodynamic drag: drag = 0.5 * pa * v**2 * Cd * A_struct # Acceleration (using altitude-dependent gravity): dv_dt = (thrust - drag - m * gravity(h)) / m # Altitude rate: dh_dt = v # Mass loss: dm_dt = -mdot return [dh_dt, dv_dt, dm_dt] # ----------------- # Event: stop integration when velocity reaches target. def event_v_target(t, y): return y[1] - v_target event_v_target.terminal = True event_v_target.direction = 1 # trigger when velocity increases past target # ----------------- # Initial Conditions and Integration # ----------------- y0 = [0.0, 0.0, m0] # start at sea level, zero velocity, full mass # Increase maximum integration time to allow a longer burn (adjust as needed) sol = solve_ivp(odes, [0, 10000], y0, events=event_v_target, max_step=0.1) if sol.status == 1 and sol.t_events[0].size > 0: t_final = sol.t_events[0][0] h_final, v_final, m_final = sol.y_events[0][0] fuel_used = m0 - m_final print(f"Final time: {t_final:.2f} s") print(f"Final altitude: {h_final:.2f} m") print(f"Final velocity: {v_final:.2f} m/s") print(f"Final mass: {m_final:.2f} kg") print(f"Fuel used: {fuel_used:.2f} kg") else: print("Target velocity not reached within the integration time.") Explanation Gravity Variation: Instead of a constant 𝑔 g, the function gravity(h) computes the gravitational acceleration at altitude ℎ h using 𝑔 ( ℎ ) = 𝑔 0 ( 𝑅 𝑒 𝑅 𝑒 + ℎ ) 2 . g(h)=g 0 ( R e +h R e ) 2 . Variable Fuel Burn Rate: The instantaneous mass flow rate 𝑚 ˙ m ˙ scales with the fraction of fuel remaining. This avoids assuming a constant burn rate over the entire burn. Aerodynamic Drag: Drag is included using the standard quadratic drag formula. The effective area 𝐴 𝑠 𝑡 𝑟 𝑢 𝑐 𝑡 A struct and drag coefficient 𝐶 𝑑 C d are set to very low values to simulate a cutting–edge design. Integration Time: The maximum integration time is set to 10,000 seconds. If the target velocity is not reached, you might need to adjust the parameters (or allow even more time). Run the script and adjust parameters (such as 𝑏 0 b 0 , 𝑓 𝑢 𝑒 𝑙 _ 𝑖 𝑛 𝑖 𝑡 𝑖 𝑎 𝑙 fuel_initial, or 𝐴 𝑒 A e ) as needed to ensure that the target velocity is reached. This more detailed model should give you a good basis for understanding how the fuel mass, thrust (with both momentum and pressure effects), drag, and altitude–dependent gravity interact over time. (For more details on such modeling, aerospace engineering texts on rocket propulsion and numerical integration methods provide further insights.) You said: this is the unhandled error. is the system solveable? (Pdb) break 90 Breakpoint 1 at /home/karl3/projects/silly/ode3.py:90 (Pdb) cont
/home/karl3/projects/silly/ode3.py(90)<module>() -> if sol.status == 1 and sol.t_events[0].size > 0: (Pdb) p sol message: Required step size is less than spacing between numbers. success: False status: -1 t: [ 0.000e+00 1.834e-04 ... 2.996e+00 2.996e+00] y: [[ 0.000e+00 -9.168e-08 ... -6.179e+02 -6.282e+02] [ 0.000e+00 -1.000e-03 ... -1.046e+15 -1.826e+15] [ 1.010e+04 1.010e+04 ... 1.007e+04 1.007e+04]] sol: None t_events: [array([], dtype=float64)] y_events: [array([], dtype=float64)] nfev: 758 njev: 0 nlu: 0
Reasoned for 8 seconds The error message arduino Copy Edit Required step size is less than spacing between numbers. indicates that the ODE solver is encountering stiffness or extremely rapid changes in the solution—meaning that, given your current parameters and formulation, the solver is forced to use step sizes smaller than the numerical precision allows. This can happen if the derivatives become very large or if there’s a singularity in the equations. In other words, the system may be mathematically “solvable” but is proving too stiff (or sensitive) for the chosen numerical method and parameters. Here are a few suggestions to address this: Use a Stiff Solver: Try switching from the default (usually an explicit method like ‘RK45’) to a stiff solver such as ‘BDF’ or ‘Radau’. For example: python Copy Edit sol = solve_ivp(odes, [0, 10000], y0, method='Radau', events=event_v_target, max_step=0.1) Stiff solvers are designed for systems where very small step sizes would otherwise be required. Adjust Tolerances: Loosening the relative (rtol) and absolute (atol) tolerances might help if the system’s dynamics are overly sensitive. For example: python Copy Edit sol = solve_ivp(odes, [0, 10000], y0, method='Radau', events=event_v_target, max_step=0.1, rtol=1e-6, atol=1e-9) Be aware that too-loose tolerances can compromise accuracy. Review and Scale Parameters: The rapid changes might be due to the chosen parameter values (such as the burn rate, nozzle area, or the way thrust is modeled) that lead to extreme accelerations. Double-check the physical realism of these parameters and consider nondimensionalizing or scaling your equations to mitigate numerical issues. Model Refinement: Sometimes, adding more physical effects (or smoothing out discontinuities) can help. For instance, if your burn rate or thrust model introduces discontinuities or extremely high derivatives, try to smooth these functions. While the system is theoretically solvable, the extreme sensitivity means that you might need careful tuning or a reformulation of the equations to capture the dynamics without running into numerical issues. If you adjust the solver and tolerances as above, you should be able to progress further in obtaining a solution. You said: same error. given this is intended to be a smooth continuous system can you skip scipy and just write a simple solver that guesses an upper and lower bound for the variable, takes the average of the two to make the guess and replaces one, until it solves for the desired extreme Reasoned for 12 seconds Below is an example of a self‐contained Python script that uses a simple Runge–Kutta (RK4) integrator and a bisection method to “solve” for the minimum fuel mass needed so that the rocket (with only its dry mass plus fuel) reaches a target velocity (e.g. orbital speed). In this code the ODE system is defined as 𝑑 ℎ 𝑑 𝑡 = 𝑣 , 𝑑 𝑣 𝑑 𝑡 = 𝐹 thrust − 𝐹 drag − 𝑚 𝑔 ( ℎ ) 𝑚 , 𝑑 𝑚 𝑑 𝑡 = − 𝑚 ˙ , dt dh dt dv dt dm =v, = m F thrust −F drag −mg(h) , =− m ˙ , with Thrust: 𝐹 thrust = 𝑚 ˙ 𝑉 𝑒 + ( 𝑝 𝑒 − 𝑝 𝑎 ( ℎ ) ) 𝐴 𝑒 , F thrust = m ˙ V e +(p e −p a (h))A e , Drag (with a low‑area, low‑drag design): 𝐹 drag = 1 2 𝑝 𝑎 ( ℎ ) 𝑣 2 𝐶 𝑑 𝐴 struct , F drag = 2 1 p a (h)v 2 C d A struct , Gravity that decreases with altitude: 𝑔 ( ℎ ) = 𝑔 0 ( 𝑅 𝑒 𝑅 𝑒 + ℎ ) 2 . g(h)=g 0 ( R e +h R e ) 2 . The fuel burn rate is taken as a fraction of the remaining fuel relative to the initial fuel. (This is one model for a variable burn rate.) The bisection method here guesses a lower and upper bound for the initial fuel mass and then, using the RK4 integrator, simulates the trajectory until a maximum time. If the target velocity isn’t reached, the simulation returns False. The bisection loop then homes in on the minimum fuel needed. You can copy and paste this code into a file (say, rocket_bisect.py) and run it from your shell. python Copy Edit #!/usr/bin/env python3 import math # ----------------- # Parameters # ----------------- # Gravity parameters g0 = 9.81 # gravitational acceleration at Earth's surface (m/s^2) R_e = 6371000.0 # Earth's radius (m) def gravity(h): return g0 * (R_e / (R_e + h))**2 # Ambient atmosphere: p_a = p0 * exp(-h/H) p0 = 101325.0 # sea-level atmospheric pressure (Pa) H = 8500.0 # scale height (m) def ambient_pressure(h): return p0 * math.exp(-h/H) # Nozzle/exhaust parameters V_e = 4400.0 # effective exhaust velocity (m/s) p_e = 101325.0 # fixed exhaust pressure at nozzle exit (Pa) A_e = 0.01 # nozzle exit area (m^2) # Aerodynamic drag parameters (cutting-edge design) Cd = 0.1 # drag coefficient (low) A_struct = 0.1 # effective cross-sectional area (m^2) # Engine burn parameters b0 = 10.0 # baseline burn rate (kg/s) when fuel is full m_dry = 100.0 # dry mass (structure, engine, etc.) in kg # fuel_mass will be the variable we solve for # Target velocity (e.g., orbital speed) v_target = 7800.0 # m/s # Maximum simulation time and time step t_max = 10000.0 # seconds dt = 0.01 # time step (s) # ----------------- # ODE System (state = [h, v, m]) # ----------------- def f(state, fuel_initial): h, v, m = state # Ambient pressure at altitude h: pa = ambient_pressure(h) # Fuel remaining is m - m_dry. fuel = max(m - m_dry, 0.0) # Variable fuel burn rate: proportional to remaining fuel fraction. mdot = b0 * (fuel / fuel_initial) if fuel_initial > 0 else 0.0 # Thrust: momentum term plus pressure-difference term. thrust = mdot * V_e + (p_e - pa) * A_e # Aerodynamic drag: drag = 0.5 * pa * v**2 * Cd * A_struct # Acceleration: dv_dt = (thrust - drag - m * gravity(h)) / m # Altitude rate: dh_dt = v # Mass loss: dm_dt = -mdot return [dh_dt, dv_dt, dm_dt] # ----------------- # RK4 Integrator for one time step # ----------------- def rk4_step(state, fuel_initial, dt): s = state k1 = f(s, fuel_initial) s2 = [s[i] + dt/2 * k1[i] for i in range(3)] k2 = f(s2, fuel_initial) s3 = [s[i] + dt/2 * k2[i] for i in range(3)] k3 = f(s3, fuel_initial) s4 = [s[i] + dt * k3[i] for i in range(3)] k4 = f(s4, fuel_initial) new_state = [s[i] + dt/6*(k1[i] + 2*k2[i] + 2*k3[i] + k4[i]) for i in range(3)] return new_state # ----------------- # Simulation Function # ----------------- def simulate_trajectory(fuel_mass): """ Simulate the trajectory for a given initial fuel mass. Returns (reached, final_state, time) where reached is True if v >= v_target. """ # Initial conditions: h = 0, v = 0, total mass = m_dry + fuel_mass. fuel_initial = fuel_mass # record for scaling mdot state = [0.0, 0.0, m_dry + fuel_mass] t = 0.0 while t < t_max: # Check if we have reached the target velocity. if state[1] >= v_target: return (True, state, t) state = rk4_step(state, fuel_initial, dt) t += dt return (False, state, t) # ----------------- # Bisection Solver for Minimal Fuel Mass # ----------------- def find_min_fuel(lower, upper, tol=1e-1, max_iter=100): """ Find the minimal fuel mass needed to reach v_target using bisection. lower, upper: initial bounds for fuel mass (kg) tol: tolerance (kg) """ for i in range(max_iter): mid = (lower + upper) / 2.0 reached, state, t_final = simulate_trajectory(mid) print(f"Iteration {i}: Fuel = {mid:.2f} kg, reached = {reached}, final v = {state[1]:.2f} m/s") if reached: # If reached, try to lower the fuel mass. upper = mid else: # If not reached, need more fuel. lower = mid if upper - lower < tol: break return upper # ----------------- # Main Routine # ----------------- if __name__ == '__main__': # Choose initial bounds for fuel mass (kg) fuel_lower = 1000.0 # lower bound guess (kg) fuel_upper = 20000.0 # upper bound guess (kg) min_fuel = find_min_fuel(fuel_lower, fuel_upper) reached, final_state, t_final = simulate_trajectory(min_fuel) print("\n--- Result ---") print(f"Minimum fuel mass to reach {v_target} m/s: {min_fuel:.2f} kg") print(f"Final altitude: {final_state[0]:.2f} m, Final velocity: {final_state[1]:.2f} m/s, Final mass: {final_state[2]:.2f} kg") print(f"Time to reach target velocity: {t_final:.2f} s") How the Code Works RK4 Integration: The function rk4_step updates the state vector [ ℎ , 𝑣 , 𝑚 ] [h,v,m] using a fixed time step 𝑑 𝑡 dt. Simulation: The simulate_trajectory function runs the simulation from 𝑡 = 0 t=0 to 𝑡 max t max (or until the velocity reaches the target 𝑣 target v target ). It returns whether the target was reached, the final state, and the time elapsed. Bisection: The find_min_fuel function performs a bisection search between a lower and an upper bound for the fuel mass. It repeatedly simulates the trajectory with the midpoint fuel mass and adjusts the bounds based on whether the rocket reaches the target velocity. When the interval between lower and upper bounds is below the tolerance, it returns the minimal fuel mass found. Gravity as a Function of Altitude: Gravity is computed as 𝑔 ( ℎ ) = 𝑔 0 ( 𝑅 𝑒 𝑅 𝑒 + ℎ ) 2 , g(h)=g 0 ( R e +h R e ) 2 , so that it decreases with altitude. Run and Adjust: The printed iteration output shows each bisection step. Adjust the parameter values (burn rate 𝑏 0 b 0 , nozzle area 𝐴 𝑒 A e , time step 𝑑 𝑡 dt, etc.) if needed for your intended scenario. This script avoids using SciPy’s ODE solvers and instead implements a simple RK4 integrator along with a bisection method to search for the minimum fuel mass needed to reach a specified target velocity. You said: it crashes with a negative height. stop working it, let's change the design. i see you're stepping a simulation instead of integrating the equations. i expect the equations to be simple to integrate. are they? also v_target should be a function of height. ChatGPT said: You're right to reassess the design. The equations governing the system should ideally be integrated analytically where possible, rather than relying entirely on numerical stepping. To determine whether the equations are simple to integrate, let’s break them down: Velocity as a function of height 𝑑 𝑣 𝑑 𝑡 = 𝐹 thrust − 𝐹 drag − 𝑚 𝑔 ( ℎ ) 𝑚 dt dv = m F thrust −F drag −mg(h) This can be rewritten using 𝑑 𝑣 𝑑 ℎ = 𝑑 𝑣 𝑑 𝑡 ⋅ 𝑑 𝑡 𝑑 ℎ = 𝑑 𝑣 𝑑 𝑡 ⋅ 1 𝑣 dh dv = dt dv ⋅ dh dt = dt dv ⋅ v 1 , leading to 𝑣 𝑑 𝑣 𝑑 ℎ = 𝐹 thrust − 𝐹 drag − 𝑚 𝑔 ( ℎ ) 𝑚 v dh dv = m F thrust −F drag −mg(h) If thrust and drag were known explicit functions of height, this equation could be integrated. Mass loss over height 𝑑 𝑚 𝑑 ℎ = 𝑑 𝑚 𝑑 𝑡 𝑑 𝑡 𝑑 ℎ = − 𝑚 ˙ 𝑣 dh dm = dt dm dh dt = v − m ˙ Given an appropriate functional form for 𝑚 ˙ m ˙ , this could also be integrated analytically. Orbital velocity as a function of height 𝑣 orb ( ℎ ) = 𝐺 𝑀 𝑅 𝑒 + ℎ v orb (h)= R e +h GM where 𝐺 𝑀 GM is Earth's gravitational parameter. This means that reaching orbit isn’t just about a target velocity but a total energy balance, including altitude. To determine whether full analytic solutions exist, we need explicit forms for thrust, drag, and burn rate that allow separation of variables. If these dependencies are simple enough, the integrals can be solved explicitly. Would you like me to attempt solving for velocity and mass as functions of height? You said: only if they are simple. can you show me where the first complexity arises? would it help to think in terms of kinetic and potential energy? ChatGPT said: Yes, thinking in terms of kinetic and potential energy could simplify the problem. Let's examine where the first complexity arises. Energy-Based Approach Total energy per unit mass is: 𝐸 = 1 2 𝑣 2 + 𝑔 ℎ E= 2 1 v 2 +gh where: 𝑣 v is velocity, 𝑔 g is gravitational acceleration (which decreases with altitude), ℎ h is height. For a system that burns fuel, we must include the energy contribution of the exhaust velocity ( 𝑉 𝑒 V e ) and mass loss. Work Done by Thrust If thrust is the only external force doing work on the system, the work-energy theorem states: Work Done = Δ ( 1 2 𝑣 2 + 𝑔 ℎ ) Work Done=Δ( 2 1 v 2 +gh) The work done by thrust is: 𝑊 = ∫ 𝐹 thrust 𝑑 ℎ W=∫F thrust dh where 𝐹 thrust = 𝑚 ˙ 𝑉 𝑒 F thrust = m ˙ V e . If mass 𝑚 m is decreasing as we ascend, then: 𝑊 = ∫ 𝑉 𝑒 𝑑 𝑚 𝑑 𝑡 𝑑 ℎ W=∫V e dt dm dh The issue arises because 𝑚 ˙ m ˙ depends on how fuel is burned and how the remaining mass affects acceleration. This introduces a dependency between 𝑣 v, 𝑚 m, and ℎ h that prevents straightforward integration. First Complexity The difficulty comes from: Thrust Variation: 𝐹 thrust F thrust depends on how fast fuel is expelled. Mass Loss Dependence: 𝑚 m changes over time, and the acceleration 𝑑 𝑣 𝑑 𝑡 dt dv depends on 𝑚 m. Gravity Dependence: 𝑔 g changes with altitude. Each of these terms creates interdependence, making simple analytical solutions unlikely. Would an Energy-Based Numerical Approach Work? Instead of direct integration, we could use an energy-based numerical search: Track total energy (kinetic + potential). Adjust fuel consumption to maximize final energy reaching orbital conditions. Would you like a script that numerically solves for the fuel needed using an energy-based method instead of direct step-based simulation? You said: can you write out the 3 interdependent equations briefly and fully, maybe in code form rather than latex ChatGPT said: Sure! Here are the three interdependent equations written in a code-like form for clarity: python Copy Edit # 1. Velocity change due to forces (Thrust, Drag, Gravity) dv_dt = (thrust - drag - mass * g) / mass # 2. Mass change due to fuel burn dm_dt = -burn_rate # burn_rate is the mass of fuel burned per second # 3. Height change due to velocity dh_dt = velocity Where: thrust = burn_rate * exhaust_velocity (assuming constant exhaust velocity) drag = 0.5 * air_density(h) * Cd * A * velocity**2 g = G * M_earth / (R_earth + h)**2 burn_rate might vary with pressure if combustion efficiency changes. Interdependencies: dv_dt depends on mass, which is decreasing. dm_dt affects dv_dt over time. dh_dt is needed for altitude-dependent values like g and air_density(h). These three equations are coupled, meaning they must be solved simultaneously over time. Would you like a script that numerically integrates them using an energy-based stopping condition? You said: i don't understand why numerical integration is needed here. can you substitute the variables in. can you hilight a part that shows an integration challenge. ChatGPT said: Good question! Let's explicitly substitute the variables and see where the challenge arises. Step 1: Substitute Thrust, Drag, and Gravity Starting with: 𝑑 𝑣 𝑑 𝑡 = 𝑇 − 𝐷 − 𝑚 𝑔 𝑚 dt dv = m T−D−mg Substituting 𝑇 = 𝑚 ˙ 𝑉 𝑒 T= m ˙ V e (thrust), 𝐷 = 1 2 𝐶 𝑑 𝐴 𝜌 ( ℎ ) 𝑣 2 D= 2 1 C d Aρ(h)v 2 (drag), and 𝑔 = 𝐺 𝑀 ( 𝑅 𝑒 + ℎ ) 2 g= (R e +h) 2 GM , we get: 𝑑 𝑣 𝑑 𝑡 = 𝑚 ˙ 𝑉 𝑒 − 1 2 𝐶 𝑑 𝐴 𝜌 ( ℎ ) 𝑣 2 − 𝑚 𝐺 𝑀 ( 𝑅 𝑒 + ℎ ) 2 𝑚 dt dv = m m ˙ V e − 2 1 C d Aρ(h)v 2 −m (R e +h) 2 GM Where: 𝑚 ˙ m ˙ (fuel burn rate) is changing as the balloon loses mass. 𝜌 ( ℎ ) ρ(h) (air density) decreases with altitude. 𝑔 ( ℎ ) g(h) decreases with altitude. 𝑣 v and ℎ h are coupled because acceleration affects height. Step 2: Mass Change Equation Since mass changes as fuel burns, we have: 𝑑 𝑚 𝑑 𝑡 = − 𝑚 ˙ dt dm =− m ˙ And since 𝑚 ˙ = 𝑇 𝑉 𝑒 m ˙ = V e T , we substitute 𝑇 = 𝑚 ˙ 𝑉 𝑒 T= m ˙ V e , giving: 𝑑 𝑚 𝑑 𝑡 = − 𝑇 𝑉 𝑒 dt dm =− V e T Which simplifies to: 𝑑 𝑚 𝑑 𝑡 = − 𝑚 ˙ 𝑉 𝑒 𝑉 𝑒 = − 𝑚 ˙ dt dm =− V e m ˙ V e =− m ˙ But 𝑚 ˙ m ˙ itself might depend on altitude and pressure. Step 3: Height Change Equation Since height changes with velocity: 𝑑 ℎ 𝑑 𝑡 = 𝑣 dt dh =v This couples ℎ h with 𝑣 v, meaning we need to track both together. Step 4: Where the Integration Challenge Arises Looking at: 𝑑 𝑣 𝑑 𝑡 = 𝑚 ˙ 𝑉 𝑒 − 1 2 𝐶 𝑑 𝐴 𝜌 ( ℎ ) 𝑣 2 − 𝑚 𝐺 𝑀 ( 𝑅 𝑒 + ℎ ) 2 𝑚 dt dv = m m ˙ V e − 2 1 C d Aρ(h)v 2 −m (R e +h) 2 GM The problem is that: 𝑣 v depends on ℎ h, and vice versa: You can't solve one equation independently. 𝑚 m is changing over time, which affects acceleration. 𝜌 ( ℎ ) ρ(h) and 𝑔 ( ℎ ) g(h) are complex altitude-dependent functions, making direct integration tricky. Since these variables interact in a non-trivial way, an analytic solution isn't straightforward. Instead, numerical integration (like a stepwise solver) is usually the best approach. Would you like to explore approximations to simplify the system further? You said: hrm say we wanted height as a function of time. usually this would be considered reasonable to calculate. but now we have the air pressure impacting the rate of burn ... can you write an integration expression to hilight how it is intractable to calculate height as a function of time? are any of the functions we are including both irrelevent within a 40-80km range as well as sufficiently simplifying to remove that they would make it solveable if together approximated? ChatGPT said: Let's try to write the expression for height as a function of time and see why it's difficult to integrate directly. Step 1: Express Acceleration in Terms of ℎ ( 𝑡 ) h(t) From our previous discussion: 𝑑 𝑣 𝑑 𝑡 = 𝑚 ˙ 𝑉 𝑒 − 1 2 𝐶 𝑑 𝐴 𝜌 ( ℎ ) 𝑣 2 − 𝑚 𝐺 𝑀 ( 𝑅 𝑒 + ℎ ) 2 𝑚 dt dv = m m ˙ V e − 2 1 C d Aρ(h)v 2 −m (R e +h) 2 GM And since velocity is the derivative of height: 𝑑 ℎ 𝑑 𝑡 = 𝑣 dt dh =v We want ℎ ( 𝑡 ) h(t), so we substitute 𝑣 = 𝑑 ℎ / 𝑑 𝑡 v=dh/dt: 𝑑 2 ℎ 𝑑 𝑡 2 = 𝑚 ˙ 𝑉 𝑒 − 1 2 𝐶 𝑑 𝐴 𝜌 ( ℎ ) ( 𝑑 ℎ 𝑑 𝑡 ) 2 − 𝑚 𝐺 𝑀 ( 𝑅 𝑒 + ℎ ) 2 𝑚 dt 2 d 2 h = m m ˙ V e − 2 1 C d Aρ(h)( dt dh ) 2 −m (R e +h) 2 GM Now we need to express 𝑚 ˙ m ˙ (fuel burn rate) in terms of altitude. This is where things get messy. Step 2: How Air Pressure Impacts Burn Rate If we assume a pressure-dependent burn rate: 𝑚 ˙ = 𝑘 𝑃 air ( ℎ ) m ˙ =kP air (h) where 𝑃 air ( ℎ ) P air (h) (air pressure) is approximately: 𝑃 air ( ℎ ) = 𝑃 0 𝑒 − ℎ / 𝐻 P air (h)=P 0 e −h/H with scale height 𝐻 ≈ 7.64 H≈7.64 km and sea-level pressure 𝑃 0 = 101.3 P 0 =101.3 kPa. This means the fuel burn rate is exponentially decreasing with height. Step 3: Integration Form Substituting 𝑚 ˙ = 𝑘 𝑃 0 𝑒 − ℎ / 𝐻 m ˙ =kP 0 e −h/H : 𝑑 2 ℎ 𝑑 𝑡 2 = 𝑘 𝑃 0 𝑒 − ℎ / 𝐻 𝑉 𝑒 − 1 2 𝐶 𝑑 𝐴 𝜌 ( ℎ ) ( 𝑑 ℎ 𝑑 𝑡 ) 2 − 𝑚 𝐺 𝑀 ( 𝑅 𝑒 + ℎ ) 2 𝑚 dt 2 d 2 h = m kP 0 e −h/H V e − 2 1 C d Aρ(h)( dt dh ) 2 −m (R e +h) 2 GM Since 𝑚 m is decreasing due to fuel consumption: 𝑚 = 𝑚 0 − ∫ 𝑘 𝑃 0 𝑒 − ℎ / 𝐻 𝑑 𝑡 m=m 0 −∫kP 0 e −h/H dt Now we have a nonlinear second-order differential equation with an exponentially varying mass term. This makes direct integration difficult. Step 4: Which Terms Are Problematic? Atmospheric Drag ( 𝜌 ( ℎ ) ρ(h)) Becomes negligible above ~50 km, since air density falls dramatically. Approximating 𝜌 ( ℎ ) ≈ 0 ρ(h)≈0 above 50 km simplifies the equation. Gravity ( 𝑔 ( ℎ ) g(h)) Varies with altitude but only changes by ~20% between 0–100 km. Approximating 𝑔 ( ℎ ) ≈ 𝑔 0 g(h)≈g 0 (constant Earth gravity) simplifies the equation. Pressure-Dependent Fuel Burn Rate The exponential decrease in pressure significantly affects 𝑚 ˙ m ˙ . If we assume a constant burn rate, integration would be much easier. Step 5: Approximate a Solveable Case If we assume: Constant gravity 𝑔 ≈ 9.8 g≈9.8 m/s² Negligible air resistance above 50 km Constant burn rate 𝑚 ˙ m ˙ instead of pressure-dependent burning Then the equation simplifies to: 𝑑 2 ℎ 𝑑 𝑡 2 = 𝑚 ˙ 𝑉 𝑒 − 𝑚 𝑔 𝑚 dt 2 d 2 h = m m ˙ V e −mg This is separable and can be integrated explicitly, allowing for a closed-form solution. Would you like to proceed with solving this simplified case? You said: the burn rate reduces as it rises, even though the pressure inside the fixed size vessel could be much higher than the surrounding pressure? could we design the nozzle to control the burn rate curve? we could also put a solenoid over it and electronically control the nozzle size. what would be most effective? ChatGPT said: Yes, the burn rate naturally reduces as the balloon rises due to decreasing ambient pressure. However, if the pressure inside the combustion chamber (or nozzle feed system) is much higher than ambient, it can largely dictate the burn rate instead of just passive atmospheric conditions. How Nozzle Design Affects Burn Rate The exhaust velocity 𝑉 𝑒 V e and thrust depend on the nozzle expansion ratio and chamber pressure. We can influence this by: Fixed Nozzle Design The nozzle expansion ratio (exit area vs. throat area) determines how well the exhaust expands at different altitudes. A vacuum-optimized nozzle works better at high altitudes but performs worse at sea level. A sea-level-optimized nozzle would lose efficiency at high altitude. Variable Nozzle (Aerospike or Adaptive Design) Some rockets use aerospike nozzles, which self-adjust expansion based on ambient pressure. A mechanically adjustable nozzle (e.g., movable walls or extension cones) could optimize performance at every stage. Electronically Controlled Solenoid Throttle A controllable valve at the nozzle throat could adjust fuel flow rate, maintaining an optimal expansion ratio. If paired with sensors, it could actively compensate for pressure drops and optimize efficiency. What Would Be Most Effective? If simplicity is key → A vacuum-optimized nozzle with a high chamber pressure would still burn relatively well at all altitudes. If efficiency is key → A throttleable solenoid-controlled nozzle with an adjustable expansion ratio could adapt to altitude, maximizing acceleration. Would you prefer a simple passive design, or an active control system that adjusts as it ascends? You said: in the original design the total amount of mass ejected from the balloon is sufficient to equate the internal and external pressure: that is the amount of total burned fuel is equal to the original fuel minus the air displaced by the balloon volume. would this give a formula for the energy spent as a function of height that could simplify things? or is it redundant or mistaken? ChatGPT said: Your idea connects mass conservation and pressure balance, which could simplify the energy calculation, but it needs some clarification. Key Considerations Mass Loss and Pressure Balance The balloon initially displaces air equal to its own volume. As hydrogen burns, both the internal mass and external air pressure decrease. The total burned hydrogen mass should approximately match the original hydrogen mass minus the air it displaced. Energy Considerations The total energy available is from burning all the hydrogen. The energy required is split into: Overcoming gravity (potential energy 𝑚 𝑔 ℎ mgh) Acceleration to orbital speed (kinetic energy 1 2 𝑚 𝑣 2 2 1 mv 2 ) Losses (drag, heat, etc.) Does This Give a Formula for Energy Spent? If we assume all the fuel burns before reaching the target altitude, then yes, a relationship between height, velocity, and burned mass could emerge. If we assume gradual burning over time, then the problem remains differential and requires integration. Is It Redundant or Mistaken? If you assume the fuel burns instantly → You can directly balance mass and energy. If you assume continuous burning → The rate of burning still depends on air pressure, nozzle design, and acceleration, making it a function of altitude. Would you like to try a simplified model where all fuel burns instantly at some optimal altitude? That could give an analytical solution. You’ve hit the Free plan limit for GPT-4o. Responses will use another model until your limit resets after 10:12 PM.