Some future update mod ideas/wishlist

The Astronomer

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#51
I never read up about the Raptor engine, but I do know it uses liquid fuel, so igniting the liquid fuel directly using a spark plug is out of the question. I guess that they first spray in an atomized gaseous version of the methane fuel ignited by a spark plug to induce a flame, before they spray in the liquid fuel.
Why don't you go read it up? I have no idea about how it actually works, either, and you seem more knowledgeable than me.

Either way, it looks like it's possible to create rocket engines ignited by spark plugs, right?
 
#52
Why don't you go read it up? I have no idea about how it actually works, either, and you seem more knowledgeable than me.
By the way I fucked up the explanation on spark plug use (the one below), I changed it up to avoid confusion of other readers. I kinda mixed up gasoline engines with diesel engines. XD
If you're using spark plugs, no.
[PS I changed the explanation, sorry. I mixed up a gas engine with a diesel engine XD]
The philosophy of car engines and rocket engines are very different, the reason cars can use spark plugs for gasoline (similar to kerosene) is because it atomized the fuel to gas which makes the ignition using a spark plug possible. For rocket engines however, fuel and oxidizer are sprayed into the chamber in a cold liquid state, this makes ignition using a spark plug impossible.

Hope you can understand, you could also ask your dad for more info about gasoline engines and how they work. Fun fact, do you know diesel engines don't use spark plugs at all? They use pressure ignition, by compressing the air-fuel mixture with the piston to generate enough heat and ignite it.
 
#54
Yup, it will use spark ignition
Last time I check, the Raptor engine is a methalox engine that uses spark plug, and it works.
Alright I looked into the video of the raptor engine's ignition phase, I just want to say I am not convinced about your claims of the use of spark plugs.
You can see from the video that upon ignition, you can see a green flame. From my knowledge of ignition methods, it is using some kind of boron-based pryoforic agent, hence the green flame at the beginning. From what i know, you don't need a spark plug when you are already using a pyroforic ignition system.
 

The Astronomer

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#55
Alright I looked into the video of the raptor engine's ignition phase, I just want to say I am not convinced about your claims of the use of spark plugs.
You can see from the video that upon ignition, you can see a green flame. From my knowledge of ignition methods, it is using some kind of boron-based pryoforic agent, hence the green flame at the beginning. From what i know, you don't need a spark plug when you are already using a pyroforic ignition system.
That one is from 2016 (old), and it's the only one I've seen which displays green flame. Finding more right now.
 

The Astronomer

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#60
Ok. It still doesn't prove me wrong though, he is using spark plugs to ignite a torch, and the torch's open flame ignites the fuel and oxidizer mixture. That means it still uses an open flame to ignite the engine, the torch is probably burning pressurized hydrogen.
It did say meth[al]ox torch igniters?
 
#61
It did say meth[al]ox torch igniters?
Can work too, we're using the same thing for aircraft engines. I guess that the torch is basically an aircraft's fuel injector that power a rocket's fuel injector. The aircraft fuel injector atomizes the fuel to make ignition possible with a spark plug, and the open flame generated can be used to ignite the liquid fuel.

Hey this is actually a pretty good idea, that eliminates the need to bring pyroforic agents, so as long as you have fuel, you can ignite.
 
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#65
Can work too, we're using the same thing for aircraft engines. I guess that the torch is basically an aircraft's fuel injector that power a rocket's fuel injector. The aircraft fuel injector atomizes the fuel to make ignition possible with a spark plug, and the open flame generated can be used to ignite the liquid fuel.

Hey this is actually a pretty good idea, that eliminates the need to bring pyroforic agents, so as long as you have fuel, you can ignite.
But this just brings me back to aircraft engineering, you can't just wrench on a fuel atomizing blowtorch and go for a tea break. A electrical ignition system will greatly add to the engine's complexity and cost.
 

The Astronomer

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#66
But this just brings me back to aircraft engineering, you can't just wrench on a fuel atomizing blowtorch and go for a tea break. A electrical ignition system will greatly add to the engine's complexity and cost.
Nice trade-off. My guess is correct, after all. Maybe smaller ships can opt to bring more ignitions, and larger ones can afford spark plugs.
 

The Astronomer

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#67
It would be nice if there is a gravity assist tool that shows the planet's prograde and retrograde directions during the flyby. Should save lots of trouble.
 
#68
Nice trade-off. My guess is correct, after all. Maybe smaller ships can opt to bring more ignitions, and larger ones can afford spark plugs.
They all can use the same ignition system and enjoy the benefits of infinite relights, there's nothing to gain from having the smaller ship use pyroforic ignition and the other electrical.
 
#69
@Johnkurveen @Altaïr
Remember back when I said something about finding out a way to compensate for delta velocity change due to free stream pressure loss? Well I looked into it and did some math about it on paper.

IMG_20190303_162837.jpg IMG_20190303_162855.jpg

On the first image, since free stream pressure only affects thrust directly, I converted the Isp in the delta velocity equation to thrust divided by mass flow rate multiplied with the gravity constant, then I converted the thrust value to its full thrust equation to reveal the pressure factor. So far out of all the values in the equation, Po or Free Stream pressure is so far the only factor that changes, the rest in the thrust equation are constant.

On the second image, I am trying to figure out how to compensate for a changing delta velocity value due to pressure loss over altitude, I tried logarithmic and exponential functions to deduce the nature of the pressure ratio graph. I then came up with the final function of f(x)=0.2*2^x which was rather faulty, but later I noticed that delta velocity is not an addition system, so using integration to combine or compensate specific impulses of two different thrust values is completely wrong.

As you can see in the finalized (circled) formula which states delta thrust and delta mass, for delta thrust we know that it is affected by pressure, but pressure is affected by altitude and altitude gain is dependent on mass, so for delta mass? For this I thought about applying the time factor of the flight which can then be used to deduce the altitude of rocket, which then can allow us to find out the pressure, but all this depends on the payload of the rocket.

Look at it this way, depending on the payload of the rocket and Newton's 2nd law of F=ma, a rocket with greater payload may reach a lower altitude in for example 50 seconds, than a rocket with less payload, this means that the local pressure of the heavier rocket will be greater than the lighter rocket.

These writings still didn't help me in visualizing what factors I need nor the changes involved, I drew an entire graph which compares altitude, pressure ratio, thrust, seconds of specific impulse and time which can be seen below.

IMG_20190304_011653.jpg

So we move on the final image, you can see that I have compared Isp, thrust and pressure ratio on the y-axis against altitude on the x-axis, the time factor is also drawn into the x-axis but on top for better visualization. As you can see, I am using the RD-107 engine as an example, both thrust and Isp increases with decreasing rate against the pressure ratio that decreases with decreasing rate. Fuel consumption is stated as constant as mass flow rate never changes.

After I have drawn out all of the factors, I drew a line down to intersect all the lines of thrust, Isp, mass, pressure and altitude, and labelled the time line as time constant b or Tb. I then deduced these info to the two equations below, and noticed that with decreasing free stream pressure which leads to increasing thrust, the dry-wet mass ratio reduces as well, this means that each rocket has its own unique delta velocity gain, and in theory hydrogen rockets like the SLS will receive the least gain due to its smaller dry-wet mass ratio.

I am still yet to experiment on using the "area under the exponential curve" method to find the compensated dV value, and with so many factors I have to involve, I don't think it will be as easy as putting in an integration and x-function and be done with it. I think using a computer aided software to simulate this will make it much easier.

So what do you guys think?

P.S. if all of these mumbo jumbo sound like the ramblings of a drunk asshole, its because I am writing this at 3 in the morning.
 

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#70
@Johnkurveen @Altaïr
Remember back when I said something about finding out a way to compensate for delta velocity change due to free stream pressure loss? Well I looked into it and did some math about it on paper.

View attachment 14324 View attachment 14325

On the first image, since free stream pressure only affects thrust directly, I converted the Isp in the delta velocity equation to thrust divided by mass flow rate multiplied with the gravity constant, then I converted the thrust value to its full thrust equation to reveal the pressure factor. So far out of all the values in the equation, Po or Free Stream pressure is so far the only factor that changes, the rest in the thrust equation are constant.

On the second image, I am trying to figure out how to compensate for a changing delta velocity value due to pressure loss over altitude, I tried logarithmic and exponential functions to deduce the nature of the pressure ratio graph. I then came up with the final function of f(x)=0.2*2^x which was rather faulty, but later I noticed that delta velocity is not an addition system, so using integration to combine or compensate specific impulses of two different thrust values is completely wrong.

As you can see in the finalized (circled) formula which states delta thrust and delta mass, for delta thrust we know that it is affected by pressure, but pressure is affected by altitude and altitude gain is dependent on mass, so for delta mass? For this I thought about applying the time factor of the flight which can then be used to deduce the altitude of rocket, which then can allow us to find out the pressure, but all this depends on the payload of the rocket.

Look at it this way, depending on the payload of the rocket and Newton's 2nd law of F=ma, a rocket with greater payload may reach a lower altitude in for example 50 seconds, than a rocket with less payload, this means that the local pressure of the heavier rocket will be greater than the lighter rocket.

These writings still didn't help me in visualizing what factors I need nor the changes involved, I drew an entire graph which compares altitude, pressure ratio, thrust, seconds of specific impulse and time which can be seen below.

View attachment 14331

So we move on the final image, you can see that I have compared Isp, thrust and pressure ratio on the y-axis against altitude on the x-axis, the time factor is also drawn into the x-axis but on top for better visualization. As you can see, I am using the RD-107 engine as an example, both thrust and Isp increases with decreasing rate against the pressure ratio that decreases with decreasing rate. Fuel consumption is stated as constant as mass flow rate never changes.

After I have drawn out all of the factors, I drew a line down to intersect all the lines of thrust, Isp, mass, pressure and altitude, and labelled the time line as time constant b or Tb. I then deduced these info to the two equations below, and noticed that with decreasing free stream pressure which leads to increasing thrust, the dry-wet mass ratio reduces as well, this means that each rocket has its own unique delta velocity gain, and in theory hydrogen rockets like the SLS will receive the least gain due to its smaller dry-wet mass ratio.

I am still yet to experiment on using the "area under the exponential curve" method to find the compensated dV value, and with so many factors I have to involve, I don't think it will be as easy as putting in an integration and x-function and be done with it. I think using a computer aided software to simulate this will make it much easier.

So what do you guys think?

P.S. if all of these mumbo jumbo sound like the ramblings of a drunk asshole, its because I am writing this at 3 in the morning.
Wow, looks interesting but that's a lot of text to understand at once!
You're trying to evaluate the loss due to the difference of pressure right?
What do you call "free stream pressure"? Is it the gas pressure at the end of the nozzle? I'm not used to english terms... :confused:
The method is interesting, but is it still relevant to talk about delta-V when the launcher is still in the atmosphere? The launcher already fights gravity and drag in those conditions, and looses delta-V because of this (and also because of the pressure difference as you stated it of course). But I understand what you're trying to achieve.
I suppose there's no way to calculate the delta-V loss formally indeed, this depends on too much things. I already tried to do that for relatively simple situations (no atmosphere, small altitude variations, so g can be considered constant...), but this is already a nightmare. I suppose numerical integration will be the only option for that.
 
#71
Wow, looks interesting but that's a lot of text to understand at once!
You're trying to evaluate the loss due to the difference of pressure right?
What do you call "free stream pressure"? Is it the gas pressure at the end of the nozzle? I'm not used to english terms... :confused:
The method is interesting, but is it still relevant to talk about delta-V when the launcher is still in the atmosphere? The launcher already fights gravity and drag in those conditions, and looses delta-V because of this (and also because of the pressure difference as you stated it of course). But I understand what you're trying to achieve.
I suppose there's no way to calculate the delta-V loss formally indeed, this depends on too much things. I already tried to do that for relatively simple situations (no atmosphere, small altitude variations, so g can be considered constant...), but this is already a nightmare. I suppose numerical integration will be the only option for that.
Yeah, mainly pressure. About the free stream pressure, it is most likely the air pressure of a specific altitude the rocket is in. The basic delta velocity equation in my opinion is only fully reliable under prefect vacuum conditions.

But what do you mean by "relevant to talk about delta-V when the launcher is still in the atmosphere"? The delta V budget is spent throughout the entire burn time, so that means from surface to orbit no matter the atmospheric conditions, its just that calculating for atmospheric conditions compensation for delta V gain are far more complicated.
 

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#73
But what do you mean by "relevant to talk about delta-V when the launcher is still in the atmosphere"?
He means that unless you have a standardised flight profile (same launch site altitude, same flight path etc) and the weather is identical at every launch, you're going to have to re-calculate the exact Dv required based upon those conditions everytime you hit go and the amount of variables that will affect the calculations is so horrifying that he just goes off a ballpark figure of roughly x amount of delta v to achieve orbit and then works out what he has left to achieve whatever task he's doing afterwards more exactly.

Not to say it's not worth doing, but you're going to need some computer action to help you out there especially if you're doing this in a real world setting.
 
#74
Wow, looks interesting but that's a lot of text to understand at once!
You're trying to evaluate the loss due to the difference of pressure right?
What do you call "free stream pressure"? Is it the gas pressure at the end of the nozzle? I'm not used to english terms... :confused:
The method is interesting, but is it still relevant to talk about delta-V when the launcher is still in the atmosphere? The launcher already fights gravity and drag in those conditions, and looses delta-V because of this (and also because of the pressure difference as you stated it of course). But I understand what you're trying to achieve.
I suppose there's no way to calculate the delta-V loss formally indeed, this depends on too much things. I already tried to do that for relatively simple situations (no atmosphere, small altitude variations, so g can be considered constant...), but this is already a nightmare. I suppose numerical integration will be the only option for that.
Alright I looked into using the "area under an exponential graph" method, which led me to another problem, during my days in math class the teachers only taught us to implement this method for problems like changing speed or acceleration in exponential graphs, where the only way of finding a area under such a graph is best done by integration.

The problem with this is that the two values of the x and y axis must end up with a usable value, take for example a speed and time graph, finding the area for a speed-time graph will get us distance, and finding the area for an acceleration-time graph will get us speed. However for this scenario? Finding the area under a pressure-altitude graph won't get us anything officially usable. So due to this I have to ditch this method.

Another problem with the delta V formula is it cannot give you a range, but a specific value, which means finding out the gain or loss of delta V must be done per specific altitude and be plotted onto a graph in order to properly visualize the changes. So if you refer to the first image below, I have found the functions for pressure change for two altitude regimes; troposphere and post-tropopause, as each regime has their unique pressure, density and temperature change rates.

IMG_20190304_172153.jpg

After I have found the two functions, I substituted them into the delta V equation for each of them. In the image below, you can see two equations with their free stream pressure replaced by the functions, one labelled "troposphere compensation"
and the other "Post-Tropopause compensation" respectively.

IMG_20190304_172230.jpg

I would love to hear your opinion on this.
 
#75
He means that unless you have a standardised flight profile (same launch site altitude, same flight path etc) and the weather is identical at every launch, you're going to have to re-calculate the exact Dv required based upon those conditions everytime you hit go and the amount of variables that will affect the calculations is so horrifying that he just goes off a ballpark figure of roughly x amount of delta v to achieve orbit and then works out what he has left to achieve whatever task he's doing afterwards more exactly.

Not to say it's not worth doing, but you're going to need some computer action to help you out there especially if you're doing this in a real world setting.
This little project of mine is to find out the stuff I need to find an altitude compensated delta velocity value. Yes it may look ludicrous, but it may actually help us understand which launch vehicle of specific fuel types are the most efficient under specific conditions. For example a 100 ton to LEO heavy lift rocket using highly efficient but low density hydrogen fuel may be less efficient overall, compared to a heavy lift rocket with the same capability but burning less efficient but high density fossil/alkane/hypergolic fuel.