What is the climb rate of a Air/Raft?

[atpollard]

Your 2G air raft needs to use a declining amount of vertical thrust until it hits an orbital speed
your vertical component is (sin(45°)×2g)-G_{Localat elevation} the sin is roughly 0.7071, and thus the vertical is 1.4142-1 = 0.4142 G vertical at takeoff, and 1.4142-0.9= 0.5142 G vertical at LEO.

I was not using any “Grav technology” assumptions. I was just reading a c.1960 technical report on Ballistic Missiles and Thrust Angle of attack and read about how the optimal angle for cruise set the vertical component of the thrust vector to just offset gravity to achieve maximum horizontal acceleration. That caused me to reapply basic Pythagorean algebra to good old Traveller vector movement.

YOU do the math instead of me. Gravity is a vector downward at 1G at the surface of the earth (at launch). The Thrust/Weight ratio is 2, so the THRUST from the rocket exhaust is a Vector 2G out the back of the rocket, causing a 2G movement vector in the opposite direction.

CASE 1: The rocket launches straight up. 2G THRUST vertical up vector and 1G GRAVITY vertical down vector. What is the resulting vector?
I come up with 1G.

CASE 2: The rocket launches on an angle so that the Vertical component of the 2G thrust vector is 1G up and the total length of the thrust vector is 2G (just like a ballistic cruise missile). The Gravity vector is 1G downward. What is the HORIZONTAL COMPONENT of the 2G diagonal Thrust Vector? (the vertical component of the thrust vector offsets gravity so the craft moves horizontally.)
(Hint, according to Pythagoras, the sum of the squares of the vertical and horizontal sides is equal to the square of the angled hypotenuse.)
I come up with sqrt (3) G.

In case 1, the rocket accelerates 1G vertically and in case 2 the rocket accelerates sqrt (3)=1.73G horizontally.
Both cases will eventually reach Orbital Velocity (10 km/s). Case 2 will reach orbital velocity faster.
The observation is “straight up subtracting local gravity” (irrespective of 1G or 0.9G) is not the best model and Real Rockets only accelerate vertical (subtracting the full planetary gravity penalty) for the few seconds that it takes to clear the atmosphere (100 km?) … then they perform a “gravity roll” and accelerate horizontally at the “Case 2” rate for the majority of the launch.

[FYI: I think the launch angle is 30 degrees for 2G thrust to give 1G vertical and 1.73G horizontal.]

 

[atpollard]

Your 2G air raft needs to use a declining amount of vertical thrust until it hits an orbital speed
your vertical component is (sin(45°)×2g)-GLocalat elevation the sin is roughly 0.7071, and thus the vertical is 1.4142-1 = 0.4142 G vertical at takeoff, and 1.4142-0.9= 0.5142 G vertical at LEO.

Just to review your “45° liftoff” calculation:

Sin (45) = 0.7071
2G thrust x 0.7071 = 1.41 G (rounded) … both VERTICAL and HORIZONTAL
VERTICAL = 1.41G thrust - 1G gravity = 0.41 G VERTICAL acceleration.
HORIZONTAL = 1.41G thrust - 0G (ignoring atmospheric drag) = 1.41G HORIZONTAL acceleration.

Combining Vertical and Horizontal Vectors …
sqrt [(0.41)^2 + (1.41)^2] = sqrt (0.1681 + 1.9881) = sqrt (2.1562) = 1.4684 G acceleration!

 

[whartung]

In case 1, the rocket accelerates 1G vertically and in case 2 the rocket accelerates sqrt (3)=1.73G horizontally.
Both cases will eventually reach Orbital Velocity (10 km/s). Case 2 will reach orbital velocity faster.

Apollo 13 launched at ~72 degrees. That is, after launch (90 deg), it rolled to 72.

Chat GPT seemed to actually have a cogent explanation:

The Saturn V rocket, used during the Apollo program for crewed missions to the Moon, was designed to follow a specific trajectory during its launch. The launch trajectory is not a straight vertical ascent but involves intentional tilting or rolling maneuvers. This design has several reasons:

1. Gravity Turn:
   • The primary reason for the launch trajectory is to achieve a gravity turn. As the rocket ascends, it gradually tilts over to align itself with the desired orbit. This allows the rocket to take advantage of Earth's rotation and minimize the energy required to reach orbit.
2. Maximizing Velocity:
   • By initiating a gravity turn, the rocket is able to maximize its horizontal velocity. This is crucial for reaching orbit because achieving orbit is not just about reaching a certain altitude but also attaining the necessary horizontal velocity to stay in orbit.
3. Aerodynamic Forces:
   • The rocket's trajectory is influenced by aerodynamic forces. As the rocket gains speed and altitude, it needs to be tilted to counteract these forces and maintain control. The roll and pitch maneuvers are part of the process of managing aerodynamic forces.
4. Avoiding Excessive Atmospheric Drag:
   • Tilting the rocket during ascent helps minimize the effects of atmospheric drag. Going straight up for the entire ascent would subject the rocket to greater atmospheric resistance, making the launch less efficient.
5. Orbital Insertion:
   • The final trajectory is designed to achieve orbital insertion. The rocket's trajectory is carefully calculated to place the payload, such as the spacecraft or satellite, into the desired orbit around the Earth or on a trajectory to the Moon.

As for the specific rolling motion you mentioned after liftoff, that could be part of the stabilization and control system. Rockets often have systems that induce a roll motion to help distribute heat more evenly across the vehicle and to prevent the rocket from deviating from its planned trajectory.

In summary, the launch trajectory of the Saturn V involves intentional tilting and rolling maneuvers to optimize the rocket's ascent, maximize velocity, and achieve the desired orbit or trajectory for the mission.

 

[atpollard]

Apollo 13 launched at ~72 degrees. That is, after launch (90 deg), it rolled to 72.

Chat GPT seemed to actually have a cogent explanation:

Yeah, working with the Kerbal system gave a hint at just how complex the launch trajectory and the details that contribute to an OPTIMAL path start to get. I was only making the observation that the common Traveller practice of just subtracting local gravity from the Thrust/Weight acceleration is an oversimplification in the other direction since it only applies to a brief initial stage of direct vertical ascent (the absolute worst moment in the flight path) and we are applying that penalty throughout the entire flight. Even 72 degrees will gain better than T/W-g acceleration towards the goal of achieving orbital velocity. The Saturn V started out with an acceleration of only 0.225 G going straight up. Using Traveller Math, we would keep it at a constant 0.225G acceleration all the way to orbit (and that is just plain wrong).

 

[Spinward Flow]

Yeah, working with the Kerbal system gave a hint at just how complex the launch trajectory and the details that contribute to an OPTIMAL path start to get.

Welcome to Rocket Surgery!

The Saturn V started out with an acceleration of only 0.225 G going straight up.

Yes ... but ... it NEEDED that 0.225G "going straight up" to lift off from the pad AT ALL and begin gaining altitude.
WITHOUT that net 0.225G "going straight up" from a stationary launch on the ground at the start ... You Are Not Going To Space Today™.

The moral of the story being, if you can't get off the ground(!) ... you're going to have a hard time reaching orbit.
I know ... who knew?

Insert obligatory "your mileage may vary, of course" caveat here.

 

[atpollard]

Yes ... but ... it NEEDED that 0.225G "going straight up" to lift off from the pad AT ALL and begin gaining altitude.
WITHOUT that net 0.225G "going straight up" from a stationary launch on the ground at the start ... You Are Not Going To Space Today™.

Agreed ... but if you don't "kick it into high gear" and advance beyond 0.225 G "going straight up" ... Your Sounding Rocket Is Coming Straight Down Today™

The moment you change that angle from 90 degrees, you instantly gain acceleration without increasing the G-rating (T/W) of the craft!

 

[Spinward Flow]

The moment you change that angle from 90 degrees, you instantly gain acceleration without increasing the G-rating (T/W) of the craft!

???

You need to start drawing, on grids, showing the vector math that you're talking about, how you think that works.
Show Your Work.
Prove what you say is correct, visually, using (digital) graph paper.

 

[coliver988]

???

You need to start drawing, on grids, showing the vector math that you're talking about, how you think that works.
Show Your Work.
Prove what you say is correct, visually, using (digital) graph paper.

think he already did here https://www.travellerrpg.com/index....he-climb-rate-of-a-air-raft.43934/post-661407

 

[Spinward Flow]

think he already did here

I see numbers in that post.
What I do NOT see is ...

You need to start drawing, on grids, showing the vector math that you're talking about, how you think that works.
Show Your Work.
Prove what you say is correct, visually, using (digital) graph paper.

 

[atpollard]

???

You need to start drawing, on grids, showing the vector math that you're talking about, how you think that works.
Show Your Work.
Prove what you say is correct, visually, using (digital) graph paper.

Done.


Note that EVERY time the ship moves from 90 degrees, the TOTAL ACCELERATION increases!
That was my point (an observation discovered by people conducting Ballistic Missile tests ... don't shoot them too steep UP, just enough to balance gravity for MAXIMUM ACCELERATION!

[As you noted, less than gravity and it lays on the ground ... which is bad.]
However, just subtracting g from acceleration is INCORRECT (except for the first few seconds of a launch as you clear the buildings).

 

[Spinward Flow]

Note that EVERY time the ship moves from 90 degrees, the TOTAL ACCELERATION increases!

That is (properly speaking) an incorrect interpretation.
For that to be true, the 2G thrust would have to be increasing.

To better explain what I mean, you're trying to say that a+b=c is what is happening, not a²+b²=c² ... which is a very different proposition.

Things get even more complicated when you're trying to assert that a(-1)+b=c is all you care about, while losing sight of the fact that what is actually happening is (a-1)²+b²=c² in your math, where a>1 is required AT ALL TIMES the entire way "up" along your trajectory (until reaching orbital velocity, more or less).

This is why I have kept asserting that if you want to lift off from a 1.0-1.5G surface gravity world, you basically "need" a 2G (thrust) acceleration maneuver drive ... and a 1G maneuver drive will not have enough "oomph" to even get you off the ground to start gaining altitude.

Traveller RAW vastly oversimplifies these vector math complexities by simply saying (in CT Striker, for example) you need enough thrust to counter gravity (usually assumed as 1G in Striker construction rules) and then anything left over becomes available for lateral thrust/flight speed in a horizontal direction. It's why an air/raft with what amounts to a 1.1G thrust rating uses 1G to counter gravity, leaving 0.1G for lateral motion control so the air/raft "flies" at ~100kph.

 

[Grav_Moped]

The moment you change that angle from 90 degrees, you instantly gain acceleration without increasing the G-rating (T/W) of the craft!

???

You need to start drawing, on grids, showing the vector math that you're talking about, how you think that works.
Show Your Work.
Prove what you say is correct, visually, using (digital) graph paper.

You do not gain actual acceleration. (That goes up as fuel burns off and thrust/weight goes up.) You do gain some velocity, as gravity is no longer a direct subtraction from thrust when determining velocity. With enough vertical momentum, you can incline the thrust axis well away from vertical.

 

[Grav_Moped]

where a>1 is required AT ALL TIMES the entire way "up" along your trajectory (until reaching orbital velocity, more or less).

No, a does not have to be >1 at all times. You're discounting the accumulated vertical vector. Just by eyeball, once you're about halfway up to orbital altitude you can start using all your thrust for lateral acceleration. You'll coast (with an upward velocity component that decreases due to the downward acceleration of gravity, to zero, hopefully just as you reach orbital altitude) the rest of the way to orbital altitude.

 

[atpollard]

Things get even more complicated when you're trying to assert that a(-1)+b=c is all you care about, while losing sight of the fact that what is actually happening is (a-1)2+b2=c2 in your math

Every single BLUE example in the graphic that you requested is (a-1)2+b2=c2.
I first presented it mathematically and you requested it graphically. Then I obliged and presented it graphically and you tell me the graphics need to be calculated mathematically.

I simply resolved a 2G vector into its vertical and horizontal components (presented in RED in the graphic) then subtracted GRAVITY from the vertical component of the vector, and then added the resulting vertical and horizontal vectors (presented in BLUE) back together to calculate the resulting acceleration vector for the ship. Feel free to apply a2+b2=c2 to ANY of the right triangles in the graphic and check my math.

If you are not traveling straight up, then a 2G thrust on a 1G world will result in greater than 1G acceleration!
Up to a maximum of 1.73G at a 30 degree angle when the vertical component will exactly balance gravity and the ship will accelerate sideways (like a 1960s cruise missile).

 

[atpollard]

You do not gain actual acceleration.

Those vectors are ACCELERATION that is being added and subtracted and resolved into vertical and horizontal components … not velocity or displacement. Changing the angle changes the vertical and horizontal component of the resulting acceleration vector (shown in RED on the graph). Gravity is an acceleration vector that is only subtracted from the Vertical component. Adding the resulting vertical and horizontal acceleration vectors (shown in BLUE) yields the resulting Acceleration vector for the ship. Look at the graph, as the angle moves further from 90 degrees, the resulting acceleration vector increases (from 1.0G at 90 degrees to 1.239 G at 60 degrees).

If you work out 30 degrees, you will find the Vertical component of a 30 degree 2G vector is 1G and the Horizontal component is 1.73 G. Subtracting 1G for gravity from the Vertical component yields 0G vertical and 1.73G horizontal. Adding those vectors together yields 1.73G acceleration SIDEWAYS! Your ship is on its way to making a 1.73 G transfer orbit from the SURFACE to LEO.

 

[Spinward Flow]

Your ship is on its way to making a 1.73 G transfer orbit from the SURFACE to LEO.

Hence why you need to have a 2G drive in order to achieve a 1.73G transfer.
1.73 < 2.00

Now, watch what happens when you reduce the Thrust vector (the 2G arrow) while keeping gravity at a consistent 1G. As thrust goes down, you'll find that the "toppling angle" away from 90º from vertical gets smaller and smaller. As thrust/acceleration from maneuver drive goes down, the "drag" of gravity becomes an increasingly large factor.

So you've discovered that the "balance point" for thrust angle from a 2G drive in a 1G gravity field is 30º, allowing you to have zero acceleration vertically (meaning constant vertical velocity) while continuing to accelerate sideways.

Try dropping the maneuver drive thrust rating down from 2G to 1.5G ... then 1.2G ... then 1.1G ... of maneuvering power and see what happens. This kind of knowledge is applicable to Kerbal Space Program for anyone playing that, to give you notions about thrust to weight ratios for rockets in that game. The difference there of course is that Kerbal is using the Rocket Equation while Traveller "bypasses" most of that equation with "magical reactionless anti-grav thruster plates" made of high tech bolted onto ship hulls.

The fun thing about these kinds of engineering/physics problems is that there can be some peculiar edge cases that emerge which then show up in real world rocketry ... such as things like the way to make a SpaceX Starship+Booster more fuel efficient at "throwing" mass into orbit is to increase thrust. This is because gravity is "pulling the rocket down" every second, so the more upward thrust you have the faster you get to orbit and the less time you need to spend "fighting gravity" to go up higher ... less time spent equals less fuel burn means a more efficient rocket. This is why SpaceX continues to iterate and advance their Raptor engine technology to achieve higher thrust outputs in order to "waste less fuel fraction" in getting to orbit, because of how the Rocket Equation works to calculate these things when dealing with chemical propulsion (fusion powered anti-grav kinda breaks a lot of assumptions).

My point being that trying to lift off from the ground using 1.1G of maneuver drive is going to require 10x longer to accelerate to orbit from a 1G world than using 2G of maneuver drive would ... because (1.1-1)*10=(2-1) ... even before getting into things like ascent angles to split the acceleration vector between vertical and horizontal components (which is really "pilot work" ).

 

[Grav_Moped]

Those vectors are ACCELERATION that is being added and subtracted and resolved into vertical and horizontal components … not velocity or displacement. Changing the angle changes the vertical and horizontal component of the resulting acceleration vector (shown in RED on the graph). Gravity is an acceleration vector that is only subtracted from the Vertical component. Adding the resulting vertical and horizontal acceleration vectors (shown in BLUE) yields the resulting Acceleration vector for the ship. Look at the graph, as the angle moves further from 90 degrees, the resulting acceleration vector increases (from 1.0G at 90 degrees to 1.239 G at 60 degrees).

If you work out 30 degrees, you will find the Vertical component of a 30 degree 2G vector is 1G and the Horizontal component is 1.73 G. Subtracting 1G for gravity from the Vertical component yields 0G vertical and 1.73G horizontal. Adding those vectors together yields 1.73G acceleration SIDEWAYS! Your ship is on its way to making a 1.73 G transfer orbit from the SURFACE to LEO.

Yes. No vertical component to the vector though, since its 1g vertical component is cancelled by gravity (or so, I'm just doing this in my head). That is, it's accelerating horizontally into to an orbit at zero altitude.

(Unless some time was previously spent accelerating vertically beforehand.)

 

[Krikkitone]

Yes. No vertical component to the vector though, since its 1g vertical component is cancelled by gravity (or so, I'm just doing this in my head). That is, it's accelerating horizontally into to an orbit at zero altitude.

(Unless some time was previously spent accelerating vertically beforehand.)

However…as its orbit gets velocity, the orbit itself effectively negates some of the 1g and you start moving farther from the surface while maintaining the 1.7g horizontally.

 

[atpollard]

However…as its orbit gets velocity, the orbit itself effectively negates some of the 1g and you start moving farther from the surface while maintaining the 1.7g horizontally.

Let's INTEGRATE the equation! (Ok ... NOT!)

 

[aramis]

Done.
View attachment 4200

Note that EVERY time the ship moves from 90 degrees, the TOTAL ACCELERATION increases!
That was my point (an observation discovered by people conducting Ballistic Missile tests ... don't shoot them too steep UP, just enough to balance gravity for MAXIMUM ACCELERATION!

[As you noted, less than gravity and it lays on the ground ... which is bad.]
However, just subtracting g from acceleration is INCORRECT (except for the first few seconds of a launch as you clear the buildings).

All of these are missing the downward gravity acceleration. You need to add the 1 unit down to every step.