posted 12-21-98 11:38 PM ET        

----

Ok, here we go, based on my assumption from above of a 6 yr accel time and a 20 yr decel:

d = 4.3 lyr (distance to AC)
ta = 6 yr (time of accel)
tc = 14 yr (time of coast)
td = 20 yr (time of decel)
aa = acceleration during accel phase
ad = acceleration during decel phase

First, relate aa to ad:

aa * ta = -ad * td

ad = -3 / 10 * aa [1]

Then set up the trip:

d = (0.5 * aa * ta**2) + (aa * ta * tc) + (aa * td + 0.5 * ad * td**2) [2]

Now substitute [1] into [2]:

d = (0.5 * aa * ta**2) + (aa * ta * tc) + (aa * td - 0.15 * aa * td**2)

Substitute in the known values and simplify:

4.3 lyr = 162 yr**2 * aa

Solve for aa:

aa = 0.02654 lyr/yr**2

Then for ad:

ad = -0.007963 lyr/yr**2

This give a coasting speed of:

0.02654 lyr/yr**2 * 6 yr = 0.1593 lyr/yr = 0.1593 c

Now, let's check this against decel:

-0.007963 lyr/yr**2 * 20 yr = -0.1593 c

Checks out.

----

posted 12-22-98 03:55 PM ET            

First, we'll start with Newton's famous Force equation:
F = Force m = Mass a = Accel

F = m * a [1]

Equation [1] as it is written holds only true when all are constant. Assuming that they want to maintain a constant force by the engine, this means the acceleration and mass changes with time, or

F = dm * da [2]

where
dm = Differential of mass
da = Differential of acceleration

Rearranaging [2] we can get:

da = F / dm [3]

Now, if we integrate both sides of [3], and evaulate from starting mass to ending mass, we get:

a = F * [ln(mi) - ln (mf)] [4]

Where, mi: Inital Mass mf: Final Mass

Rearranging [4] by Law of Logs we can get:

a = F * ln(mi/mf) [5]

posted 12-22-98 07:20 PM ET            

Secondly, how do you get from F=m*a to F=dm*da? That would imply that m*a=dm*da, which doesn't make sense.

You can get:

dF
-- = dm * a
dm

or

dF
-- = m * da
da

The first equation would seem to be more useful, since the assumption to this point has been constant acceleration. If we're assuming constant force, we get:

da F
-- = ---
dm dm

Also, integrating any of these equations just gets us back to F=ma.

I don't really know exactly how an Ion drive would work (which is what was described in episode 29 I think). For a standard chemical propulsion drive, the equation would be something like (IIRC):

dm
vgas * -- = m * a
dt

where vgas is the velocity of the gas going out the back. Thus getting:

a 1
---- dt = --- dm
vgas m

which integrates to:

at
---- = ln(m)
vgas

What exactly this means, I'm not sure.

posted 12-22-98 07:24 PM ET            

Here are the formulas reworked. Hopefully you can figure out which is which (they'll be in the same order).

1) (dF/dm) = dm * a
2) (dF/da) = m * da
3) (da/dm) = (F/dm)
4) vgas * (dm/dt) = m * a
5) (a/vgas)*dt = (1/m)dm
6) (at/vgas) = ln(m)

posted 12-23-98 01:43 AM ET            

However, I do question things:

Would you not need one heck of a fuel source to get the volocity needed?

Given our current, even experiemental fuels, would your formula's not cascade to the added weights of the fuels?

posted 12-23-98 03:47 PM ET            

If you look at equations from RobKid you'll see the biginnings of taking the mass of expended fuel into account.

posted 12-24-98 03:04 PM ET            

Also, I realized that this is a fusion based ion drive. The fusion equations used (H->He) would probably have a negligable mass loss compared to the mass of the ship. In a chemical rocket, the energy produced is from reordering of chemical bonds, a rather low amount of energy. In a fusion reaction, the enerby produced is from mass to energy conversion (He and the byproducts have slightly less mass than 4 H). Thus, the energy produced is taken from e=mc^2. They might not have lost more than 50g or so of mass due to the reaction. However, this produces:

e = (.050kg) * (3.0e5) ^ 2
e = (5.0e-2) * (9.0e10)
3 = 4.5e9 Joules

posted 12-27-98 10:14 AM ET            

Everyone ( or maybe I am not reading the formulas right ), are mistaking the acceleration and max speed computing.
When a ship reachs sub-light speeds, the rate of speed/mass is different.
For example, When a ship is traveling at 0.5 C, HALF of all the energy of acceleration is applied to increasing the speed, while the other half increase the mass and time continium warp ( see AKA einstian ).
Meaning, that one unit of energy, as the ship reachs the light speed, will pass LESS REAL speed to the ship.
Lets say the speed of light is 10.
You start accelerating:
1 unit = 1 speed.
2 unit = 1 + 1-1/10 = 1.9
3 unit = 1.9 + 1-1.9/10 = 2.71
4 unit = 2.71 + 1-2.71/10 = 3.439
5 unit = 3.439 + 1-3.439/10 = 4.0951
6 unit = 4.0951 + 1-4.0951/10 = 4.68559
And so on.
You can see, that in order to get to half the speed of light, you near about 6.5 units of energy, not 5 as you may think.
The formula is thus:
X ( number of units ):
X - [ 9/10 sqrX - ( 10-x ) ]
when 9/10 is 9/10 squared X times
( 9/10 * 9/10 * 9/10 etc... )

One can see, that even in low realvastic speeds, like 0.2-0.3 C, still some energy lost exist, and must be put in.

posted 12-30-98 10:31 AM ET            

m_i = m_0 exp(v / (g * Isp))

where

m_i is initial mass
m_0 is empty mass
v is total delta-v
g is the acceleration of gravity (9.81 m/s/s)
Isp is the specific impulse in seconds

The Isp of a rocket represents its efficiency. A hydrogen/oxygen rocket gets an Isp of around 450 seconds. Let's estimate that the Unity's fusion rocket gets 10,000 seconds. Then to accelerate to v = 0.1593 c, we get:

m_i = m_0 * exp(48,000,000 / (10,000 * 9.81))

= m_0 * exp(489.3)

= m_0 * (3.15 * 10^212)

posted 12-31-98 11:43 AM ET            

Also note that the maximum possible Isp is only 30.6 million seconds. Derivation: g * Isp = exhaust velocity. The maximum possible exhaust velocity is c, so max Isp = c/g.

If we could get such an Isp, then we get into realistic mass ratios for the Unity. At an exhaust velocity of c, the mass ratio is only 1.17, so 14.8% of the ship is fuel. A more realistic exhaust velocity of 0.1c, however, gives a mass ratio of 4.95 (79.8% of ship mass if fuel). Anything less quickly gets into the impossible range again. Current engineering cannot give realistic mass ratios greater than about 9, but even assuming a mass ratio of 100 our exhaust velocity would need to exceed 0.035c. That's about where your most ambitious Orion engine example lives (1 million seconds is about 0.0327c), but it still means a million tons of fuel for each 10,000 tons of other mass. That other mass includes fuel tanks, living quarters, structural members, payload, etc.

Also note that relativistic effects should really be taken into account here and make conditions WORSE, not better, because the rocket equation is more pessimistic than the strictly Newtonian version I gave.

So I'm going to stick to my guns when I say that the Unity is unrealistic if it's a rocket. If it's something other than a rocket, then it changes the whole equation. For example, if it's a Bussard ramjet and the major problems associated with those can be solved, then mass ratio is not a problem because the major portion of the fuel is picked up along the way. Or a solar-laser-powered lightsail could probably do it, but it would depend on a really dependable laser back Earthside, which wouldn't fit the SMAC storyline. That also wouldn't provide the drama at the approach to Planet as the drive failed.

Wen_Amon: The Earth-Mars trip is only constrained if a Hohmann transfer orbit - the slowest, most fuel-efficient type - is used. This orbit involves two burns: the first changes the orbit from Earth orbit to a doubly-tangential elliptical orbit that just touches Mars orbit; the second boosts the ship into Mars orbit at aphelion. The length of such a trip is exactly half the average of the orbits of the source and destination planets. So for Earth-Mars, you take the average of 1 year and about 1.6 years to get 1.3 years, then half of that is 0.65 years or 7.8 months.

posted 01-02-99 06:11 PM ET            

First, presume that you hit that velocity immediately.

Then a constant speed is v=4.3/40=.1075c

Time increases the calendar time to 40.23 years relative to an observer on earth (not that there are any more). So our calendars will be a bit off when we land unless we factor this in. Not only this but your relativistic mass and required accerlating force increases.

In reality (figuratively speaking) you have to accelerate and decelerate as well. This presents a problem as you then need a higher top speed, and the effects of relativity increase hyperbolically with top speed. If the top speed is .1075c then relativistic mass is about 0.6% higher than rest mass. But if the top speed becomes .15c then relative mass is 1.1% higher than rest mass and a top speed of .2c implies a 2.1% higher mass. Guessing at the nature of acceleration is real speculative at this point, but suppose .2c turns out to be the top speed needed due to the time of accelerating and decelerating. But this is the problem. If you have a 100,000 ton ship and you the relativistic mass increases by 2%, then you need the energy equivalent of 2,000 tons of mass due to the conservation of matter-energy! I'll return to this later.

You need a measure of force to accelerate and decelerate the ship. Force is the time rate of change of relativistic momentum (Newton's second law corrected for relativity) or F=dp/dt=m(dv/dt)+v(dm/dt) which becomes F=ma for constant (rest) mass. Under relativity, accelerated mass is not constant and Force becomes the time rate of change of relativistic momentum (here mass changes relativistically without regard to fuel consumption, etc.)

But part of this accererating force is merely increasing the relativistic mass of the ship (this means that deceleration takes less time than acceleration, especially taking into account fuel consumption), so you now need a different measure of efficiency.

You can get force from the rate of change in total work: dW/dt=F.v=.5m(dv^2/dt)+v^2(dm/dt) and this is just c^2(dm/dt) (you can just smell E=mc^2 here). The % of work done that increases the mass as opposed to actual acceleration is [1/(1-v^2/c^2)^.5-1] so you lose 1-2% (or more) of your engine power (in terms of actual movement) before you even worry about the efficiency of your engine. Your actual engine power going to increasing relativistic mass is m_0c^2[1/(1-v^2/c^2)^.5-1]. Not a small problem!! As mentioned above, a 100,000 ton ship traveling at .2c requires energy of 2,062 tons of matter. Suppose the fuel content of the ship is 25% or 25,000 tons. Then you need 8% efficiency to immediately reach top speed and coast the rest of the way without decelerating (actually less since you'll have less matter by then). So what kind of fusion reactor are we talking about? It has to be small and extremely efficient (since acceleration is relatively slow, we may need 20+% efficiency). We don't even have one that's big and inefficient yet. Well, here's to progress for the new year.

posted 01-05-99 01:51 AM ET            

Also, I think it said in the story waht type of engine it uses, too bad I'm to lazy to look it up . It could be fusion, since there is a fusion prototype at MIT.

posted 01-05-99 08:44 AM ET            

In all cases I use the following symbols: x is the distance traveled; a_max is the maximum acceleration; a is any other acceleration; T is total time-of-flight; t_b is boost time; t_c is coast time; dv is total delta-v.

[Newtonian dynamics, no resistance from interstellar atoms assumption] In this simplest case, we do the problem two ways: first, assuming constant acceleration to midpoint, flip over and decelerate to target; second, assume maximum acceleration to cruise, coast, flip over and maximum decelerate to target. I've done the math, but I'll leave it out and just post the conclusions.

Case 1: T1 = 4x/dv
Case 2: T2 = 2x/dv + dv/a_max

It so happens that in all cases, T1 > T2 unless dv^2 = 2 x a_max, which is precisely the condition that guarantees that maximum acceleration is required in case 1. We may conclude from this that it's always better to accelerate at maximum, coast, and then decelerate at maximum. Thus a 6/14/20 plan is less efficient than it could be, but better than doing a slower acceleration boost.

[Newtonian dynamics, taking interstellar atomic drag into account assumption] The mathematics above is correct, but leaves out a very important factor: the drag of interstellar atoms on Unity. Drag typically is proportional to the square of velocity, so higher speeds would create more drag faster. Under these conditions, it might make more sense to accelerate more slowly. I haven't done any computations yet, though, so I can't draw any firm conclusions.

posted 01-05-99 05:57 PM ET            

OK, this is somewhat hairy, so be forewarned. My assumptions: The journey consists of a boost to max velocity, followed by a coast, followed by a deceleration to zero velocity. We assume that the Earth and the final destination are fixed in relation to one another. In practice, this won't be the case, but at the speeds we're talking about, the difference is pretty paltry. I assume further that the force of drag is k v^2, where k is a constant. That also won't be entirely accurate, so ignore decimal places past the second and take even the second with a grain of salt. Unless I'm wildly mistaken, though, the first decimal should be OK.

A proper analysis of this would be similar to my previous one for the no-drag assumption; i.e., dv and x would be fixed, a_max would be fixed, but t_boost, t_coast, t_decel, a_boost, and a_decel would vary. We would place constraints and try to minimize the sum of the three time values.

In practice, this probably won't work. You'll see why when I present the formulae, which are a bit complex:

In all computations, I use the following constants -

a = acceleration in current phase
t = time elapsed in current phase
x0 = starting point for phase
xf = ending point for phase
v0 = starting velocity of phase
vf = final velocity at end of phase
k = drag coefficient
A = sqrt(a/k)
B = A/v0
L = sqrt(a k)
sinh(x) = (exp(x) - exp(-x))/2
cosh(x) = (exp(x) + exp(-x))/2
tanh(x) = sinh(x)/cosh(x)
T = tan(L t)

BOOST PHASE
-----------
x0 = 0
v0 = 0

vf = A tanh (L t)
xf = (1/k) cosh (L t)

COAST PHASE
-----------
x0 = xf for boost phase
v0 = vf for boost phase

vf = v0 / (1 + v0 k t)
xf = (1/k) ln(1 + v0 k t) + x0

DECEL PHASE
-----------
Here's where the hairy part comes in...

x0 = xf for coast phase
v0 = vf for coast phase

vf = (v0 - A T)/(1 + vo T/A)
xf = (v0/L) (B ln(1 + T/B) - (B/2) ln(1 + T^2)) + x0

Keep in mind that I've simplified the nasty parts using abbreviations like A, B, L, T. I'd recommend using a spreadsheet to calculate this stuff. In fact, if you did that you could use its optimizer to come up with low delta-v, low t solutions that still give xf = 4.3 ly and vf = 0 for the deceleration phase.

Just for fun, here's one example:

I ran my spreadsheet trying to approximate the numbers given in SMAC, but assuming k = 1.09e-17 m^-1. (Note: I used that k, fairly arbitrary really, because it gave a value of 0.001g at v=0.15c, the presumed max speed of Unity. The real value could be quite different from this. I'm just showing that even a small drag coefficient makes a BIG difference.)

For starters, let a_boost = 0.2523 m/s^2, t_boost = 1.8935e8 s (6 yrs), t_coast = 4.4181e8 s (14 yrs), and a_decel = 0.0328 m/s^2 (you need much less deceleration than the first post assumed, because drag helps you all the way). It turns out that t_decel here is 1.0098e9 s or about 32 yrs, for a total flight time of 52 yrs. The delta-v required for this journey is just under 81 million m/s, so to accomplish this with a mass ratio of 5:1 would require an Isp of 5 million seconds.

Now rework it using much faster accelerations and decelerations: I use t_boost = 5.0e7 s (only 19 months!), a_boost = 0.981 m/s^2, and t_coast = 1.0e9 s (a bit over 31.5 yrs). With a_decel = 0.981 m/s^2 we have t_decel = 3.227e8 s or about a year. Note that we have used the maximum acceleration of the Unity both for boost and to slow down. We've consequently shortened the travel time from 52 yrs to about 34.3 yrs, and here's the kicker... the require delta-v is lower by about 190,000 m/s!!

I found this one by accident; I'm sure further optimization could easily be done. What's the shortest time-of-flight for a given delta-v and a_max? I think you'd find that the general strategy of boost at maximum, coast, and decel at maximum works here, too.

posted 01-05-99 07:40 PM ET            

The flight of the Unity seems to make a lot of sence for the integrity of the ship, because the ship is most likely in the best shape at the beginning of the journey and thus better able to handle the greater forces. And after many years of travel, aging and wear the ship could be very fragile by the end of the trip.

posted 01-06-99 07:31 AM ET            

But I don't know that "structural integrity" is a good one. We're talking about a maximum acceleration of 0.1g versus maybe 0.02-0.05g. Neither one is enough to put any great stress on a starship. Furthermore, the probability of actually hitting an asteroid is so remote that it would probably not enter into the calculation at all. If we were able to save 5 years of travel time at a cost of reducing the chance of success from 90% to 89.9% (and I'm exaggerating the reduction!), I don't think there'd be any question that you'd do it. Especially considering that increased travel time itself increases the risk for other reasons.

posted 01-08-99 02:37 PM ET            

.02654 ly/yr^2 (1 yr/365.25x24x3600 s)^2 =

2.665e-17 ly/s^2 (9.467e15 m/ly) =

0.2523 m/s^2 = 0.0257G

posted 01-08-99 09:29 PM ET            

The momentum of a photon is:

p = h / lambda

where
p = momentum
h = Planck's constant
lamdba = wavelength of the photon

posted 01-09-99 12:36 PM ET            

Spoe, the light drive you speak off would be capable of very high speeds, but what sort of efficiency would you get at, say .1 C. Not very good, i'd say.

I guess with antimatter, efficiency isn't your first concern. But all this antimatter has to come from somewhere. Where would the Unity engineers get the power to make tons of antimatter? That's a lot of water under the Hoover Dam.

Maybe with orbital stations and lunar bases this wouldn't be such a problem, tho.

posted 01-09-99 01:46 PM ET            

Bill Grogan's Goat
Was feeling fine.
Ate six red shirts
Right off the line.
Bill took a stick,
Gave him a whack,
And tied him to...
The railroad track.

posted 01-09-99 05:15 PM ET         

Given a 100,000 ton ship, and the desired travel time, it's possible (and you have!) to work out the total energy required, etc.

We seem to be arguing over expected ISP of the drive system. Now, some years ago, I got interested in nuclear propulsion/fusion energy and found an example of a fusion drive.

Really, I'm just trying to find an upper bound for fusion rocket ISP here...

The idea is as follows: If I remember properly, the reaction D-He3 yields a considerable portion of the reaction energy in the form of a charged particle (I think it was a proton). Charged particles can be directed with magnetic fields (Forward did this on a paper discussing antimatter rockets). Assuming 60% of the kinetic energy of the charged reaction products can be turned into thrust, what ISP are we talking?
Is 60% a good figure (I figured 50% of the reaction products are going in sort of the right direction from the very beginning...)

Any help here would be appreciated...

posted 01-10-99 07:08 PM ET            

2P-----> 2p + 2B + 3.141Kcal of enthalapy

This should be more than sufficient to drive the ship almost anywhere in the Galaxy.

Of course, you'd need to protect the crew from the lethel Berthold rays (term B in the equation), but this could be done by placing a large tank of ordinary water between the engine and the crew. Some of the water would be transmuted into it's polyhydrous form, but I don't think this would have any harmful effects.

posted 01-11-99 06:20 PM ET            

First, traveling 4.3 light years in 40 years requires an average speed of 10.75% of light speed (c). Since the ship needs to accelerate and decelerate, the actual top speed must be higher. I figure .15c or .2c is as good a guess as any. Contrary to another post, I believe that while time dialation is not insignificant, it is fairly minor at about 1-2%.

The problem as I see it is that too much energy needs to be diverted to increasing relativistic momentum rather than acceleration and a fusion engine wouldn't appear to be efficient enough.

Suppose the ship is 100,000 tons. If the top speed is .15c then 1,150 tons of energy are required to increase the ship's relativistic mass. What percentage of the ship is fuel? Say 25%. That means 4.6% of the fuel must be converted to energy before you even worry about the energy required to move the ship. If the top speed is .2c then 8.3% of the fuel needs to be converted.

How efficient is our engine? Let's be kind and assume the efficiency (in terms of % of fuel-mass converted to usable energy) is 15% for a fusion drive so that increasing the relativistic mass eats up 30%/55% of our fuel (without accounting for actually moving 4.3 light years). Is the remaining fuel sufficient to travel those 4.3 light years? I think it'll pose a problem. In fact, I think the 15% efficiency requires an engine working more along the lines of anti-matter/matter collisions someone else referred to. Nonetheless...

F=ma (the relativistic part is already accounted for for the most part so I estimate the rest through classical physics) so:

a=.029G/.039G
m=100,000 tons
F=25.8/34.5 million Newtons

We do this for 5 years over a distance d=Integeral[0,5k]atdt=25ak^2/2 (k is secs per year) so d~.375/.5 light years.

Ignore drag as there are enough approximations already.

Total energy thus equals 9.17*10^22/1.63*10^23 Joules for top speed of .15c/.2c

This energy will be needed to decelerate also, so multiply by 1.9 (to account for 10% mass loss) so we are at 1.7E23/3.1E23 Joules.

Go back to our available energy in terms of mass: Given 15% efficiency and the energy required to increase relativistic mass at top speed of .15c/.2c, available matter-energy is roughly 2.4/1.5 million kg. Notice how the available energy declines while the required energy increases with top speed.

Using E=mc^2 our fuel tanks give us 2.1/1.4 10^23 Joules. Seemingly we can make it at .15c, but not at .2c.

I guess I'll take back my original hypothesis that said I doubted it was feasible. I'm amazed at how close the required/available energy was given complete guesswork as to the process involved. Of course at this time there is no controlled fusion power outside the sun, so the whole premise is shaky in that regard. Plus, who knows how big the ship is, what percentage of the mass is the fuel, etc? I think an estimate of 15% efficiency in fusion is probably too generous. Bringing it to 10% cuts our available energy by about 1/3 and makes the trip much less feasible.

Futzing around with the numbers a bit more, a top speed of 12.5% makes the most sense because it works out much more elegantly. With 5 years acceleration and 5 years of deceleration at .024G, we'd get to AC in precisely 40 years. Moreover, the energy requirements are only about 1/2 the available energy so efficiency isn't as restricting. Ready to start exploring the galaxy?

Comments?

posted 01-11-99 07:01 PM ET            

One point to remember is that these equations are linear wrt the mass of the ship.

Also, from a pure energy budget standpoint, the faster you can accelerate, the better(your peak kinetic energy is lower). You expend minimum energy with instantaneous acceleration to 0.1075c.

posted 01-12-99 01:02 AM ET            

I doubt that the guy who is writing the back story was really figuring out the calculations on wether the stuff he was writing would actually work or not. While it's great that a game is generating this much interest in the actual physics of spaceflight, don't you think we're going a bit too far? I feel like I'm in the middle of a NASA planning session (BTW, does anyone here actually work for NASA?). Not that that's bad, it just seems to be going a little over the top for this one topic.

If you guys would like something else to churn your collective minds over, how about this: How come there is no snow on Planet? If it's supposed to be so like earth, wouldn't there be at least a few different climate zones?

posted 01-19-99 10:16 PM ET            

Here's a section for the first part of the article:

posted 01-22-99 02:46 PM ET            

Anyway, for a ship going on a journey for as long as Unity's, would it be better to have a seperate power generator for the ships systems or have the ships engines power both the drives and main systems? Either case you would have two seperate things drawing on fuel stored on the Unity, so wouldn't that play hell with all your calculations involving the ships mass? I suppose you could have the ship's systems powered by some kind of 'battery' but that seems like a waste of mass in a situation like this.

posted 02-06-99 11:17 PM ET            

This is assuming near pefect efficiency at turning Heavy hydrogen (not water/ but pure deuterium) in to Helium and using the energy

at about 50% efficiency (something that sounds more "realistic" for any system) the fuel Factor would be about
100 meaning that the Unity in Orbit around AC has 0.01 as much mass as the Unity launched from Earth's system.

posted 03-23-99 12:11 PM ET            

If we are talking about remaining strictly within SMAC's technology guidelines, then Fusion and Antimatter power are out: They being technologies that you must discover after the arrival on Alpha Centauri.

Secondly, would variable vector thrust nozzles (akin to the Harrier) be practical for spaceflight? What sort of drive (rather than power source) do you think would be used? An ion drive just doesn't have the kick.

Cheers,

posted 06-25-99 04:27 PM ET         

Question:

We need to wring 2% more efficiency from the Unity's drive to increase mass.

How do we accomplish this?

What is the thrust/mass ration (i.e. can the mass be increased by the same 2%, more, less???

posted 07-09-99 01:51 PM ET         

Any answers to my earlier question?

posted 07-12-99 05:45 AM ET            

Some of the books contain hundreds of pages of details as they apply to, for example, manned spaceflights. But I couldn't possibly remember them now, and I am in a farflung corner of the globe, so unless you're in a slum in the Third World, your chances are as good as mine were.

posted 09-09-99 04:49 PM ET            

You don't use a grappling hook to slow down a car traveling at highways speeds to slow down, you are being stopped by the friction between components.

If you fired the engine full blast all at once, the decel would cause the motor to backfire into the reaction chamber, you need to bring up the reaction up to full after you started to slow down you would get a stabler flight path.