Sunday, July 24, 2011

CALCULUS DOES DISTANCE

Near Earth's surface,

free falling object's velocity increases with time.


Thus, one can describe velocity as a function of time v(t).
 
v(t) = g × t

Simple Calculus

To describe distance traveled by a g-force,

integrate above velocity equation to

describe distance as a function of time.

d(t) =∫v(t) dt = g × t2/2
Above "Newtonian" equations work well for short distances.
Examples of G-force Distance
G-force
Time
G-force
Accel
Spot
Velocity
Cumm
Distance
tgVFind
1 sec
9.8065 m/sec²
9.8 m/sec
4.9 meter
1 min
35.03 km/min²
35 km/min
17.5 km
1 hour
78,919 mi/hr²
78,919 mi/hr
39,469 mile
1 day
0.49 AU/day²
0.49 AU/day
0.245 AU
tEquivalentg × tg × t2/2
However, interstellar travel needs another equation
to describe distance traveled by
Thought Experiment's (TE's) notional g-force spaceship.

Perhaps More Calculus .....Recall Einstein, c is always observed as consistent value, c, regardless of the observer's velocity.
Traditional values of c.
c = 299,792,458 m

sec
=299,792.5 km

sec
Convert to Astronomical Units (AUs) per second.
c = 299,792.5km

sec
×AU

149,597,870,700 m
=.002004 AU

sec
Convert to AUs per day.
c = .002004 AU

sec
×86,400 sec

day
=173.145 AU

day
Different durations of g-force produce different velocities; thus, observers might observe each other traveling in different speeds, but a few facts stay steadfast:
1. No one ever exceeds c, light speed.
2. All observers consistently measure photons at the same velocity, c= 173.14 AU/day.
3. One day's difference Δ is velocity increase due to one day of g-force acceleration. It's expressed as percent c (or equivalent decimal c).
Determine Daily Difference, Δ.
Δ = 0.49 AU/day

173.145 AU/day
=.00283=.283%
Previous work leads to following equation to model velocity of g-force spacecraft:
V(t) = (1 - (1 - Δ)t) c
where:
  • t is flight duration measured in days.
  • V(t) is velocity expressed as a function of time.
  • c is light speed, 299,792,458 m/sec. We choose to express c as 173.145 AU/day, an equivalent value.
  • Δ is initial daily velocity increase due to g-force acceleration. It's expressed as percent c (or equivalent decimal c). We choose the value of Δ = .2826%c = .002826 c based on previous analysis.





Following Method could approximate distances without calculus:
  1. Use Vt = (1 - (1 - Δ)t) c to determine spot velocities for a given quantity of days (t).Recall:
    Δ=.2826%c and c =173.145 AU/day
  2. Average velocity (AU/day) for each day (t). Determine  velocity midway between day's initial (Vt-1) and final (Vt) spot velocities:
    VAve = (Vt +Vt-1)/2
  3. Multiply this average velocity by time (one day) to approximate incremental distance during day t, dt= 1 day × VAve .
  4. Approximate cumulative distances by summing relevant incremental distances, Σ dt.
Interplanetary Ranges
G-force
Time
Spot
Velocity
Average VelocityIncr
Distance
Cumm Distance
tVtVavedtDt
0 day
0.0 AU/day
0.0 AU/day
0.0 AU
0.0 AU
1 day
0.49 AU/day
0.24AU/day
0.24 AU
0.24 AU
2 day
0.98 AU/day
0.735AU/day
0.735 AU
0.98 AU
3 day
1.46 AU/day
1.22 AU/day
1.22 AU
2.20 AU
t(1-(1-Δ)t)c(Vt +Vt-1)

2
1 day×VAve Σ dt
However a little calculus might work even better then brute force tabular method.
A Little Calculus......
d(t) =v(t)dt = c  ×(t - (1 - Δ)t

ln(1-Δ)
)
(Many thanks to Wolfram Integrator)
..........Goes a Long Way
Compute g-force distance: HOW?
Recall basic calculus, the integral, ∫, which is an elegant way of summing the area under the curve under consideration.

Calculus and Motion Problems.
Integration has many practical applications; perhaps the best known is for motion problems. For example, integrating a velocity function is known to yield distance.
For example:
abV(t)dt = d(b) - d(a)
is the general form of a velocity function (with time, t, as the independent variable) being integrated to produce distance traveled from time, a, to time, b.

Most common functions are of the form:
∫k × tn dt = k × tn+1/n+1

EXAMPLE: CONSTANT VELOCITY.  Most of us commonly ride automobiles at a constant velocity, such as 60 miles per hour, or cruise in aircraft at 500 nautical miles per hour (knots). Using the latter example:
0tk dt = k × t - 0
02500 NM/hr dt = 500 knots × 2 hrs - 0 = 1,000 NM.

EXAMPLE: CONSTANT ACCELERATION. A more interesting integration might involve g-force acceleration. Let's consider two such examples. First example involves an object free falling from a great height toward the Earth for 10 seconds. Use traditional g=9.8 m/sec2.
010 g × t dt = g t2/2
010 9.8 m/sec2  (10 sec)2 dt
= 9.8 m/sec2 (10 sec)2/2 - 0 = 490 m

Second example involves our notional spacecraft g-force accelerating from Earth toward Jupiter for 2.5 days. Use TE's g=0.49 AU/day2.
02.5g × t dt = g t2/2
02.5.5 AU/day2 (2.5 day)2dt
=.5 AU/day2(2.5 day)2/2 -0 = 1.5625AU

Above examples show cases where distances can be easily validated via other computation methods.
However, the case under consideration
( V(t) = (1 - (1 - Δ)t) c)
needs more sophisticated integration
to determine the corresponding distance function.
New Formula for Distance
TE speculates that following equation might
model  g-force velocity over time (t).
V(t) = c (1 - (1 - Δ)t)
Substitute Remainder, R, for term: (1 - Δ).
V(t) = c (1 - Rt)
Integrate velocity equation
to determine the corresponding distance.
 ∫V(t) dt = d(t) = ∫c (1 - (1 - Δ)t) dt
Assume initial time to be zero;
thus, integral's lower limit = 0.
d(t) = ∫0t c (1 - Rt) dt
Integrate terms separately.
dt = ∫0tc dt  - ∫0tc Rt dt
Integrate and expand.
d(t) = [c × t - c × 0] - [cRt - cR0

lnR
]
Apply the "0" terms.  (Note:  R0 = 1.)
d(t) = ct -c Rt

lnR
+
c

lnR
Re-arrange as shown:
d(t) = ct +c(1 - Rt)

lnR
Substitute expression for velocity:  V(t)=c(1 - Rt)
d(t) = ct +V(t)

ln(1-Δ)
INTERPLANETARY RANGES
NewtonianTimeEinsteinian
d(t)V(t)tV(t)d(t)
AUAU/daydayAU/dayAU
0.24
0.49
1
0.49
0.24
0.98
0.98
2
0.98
0.98
2.20
1.47
3
1.46
2.20
3.91
1.96
4
1.95
3.90
6.11
2.45
5
2.43
6.09
8.80
2.93
6
2.91
8.76
11.98
3.42
7
3.39
11.91
15.65
3.91
8
3.87
15.54
19.80
4.40
9
4.35
19.65
24.45
4.89
10
4.83
24.24
611.25
24.45
50
22.82
583.89
g×t2/2g × ttc(1-Rt)
ct +V(t)

ln(1-Δ)
c = 173.15AU/day
LY = 63,241 AU
t1 - Δ = R = .99717
ln(R) = -.00283
After 50 days of g-force acceleration,
methods start to noticeable diverge.
For first ten days of g-force acceleration,
the Newtonian and Einsteinian methods show similar results.
For initial 10 days of g-force acceleration, Newtonian method approximates values determined by the Einsteinian. EXAMPLE: For tenth day, both methods show velocity of just under 5 AUs/day and a total distance of just over 24 AUs.

BEYOND INTERPLANETARY
Short range (10 days) chart contrasts enormously with a longer range chart.
After 500 days of g-force acceleration, chart shows greatly diverging values.
Well past the planets, we notice distinct differences between the two methods.
After one year (365 days) of g-force, Newtonian velocity exceeds c.  IMPOSSIBLE!!! Thought Experiment's solution shows g-force vessel at 64% c after one year of g-force.
NewtonianTimeEinsteinian
d(t)v(t)tv(t)d(t)
AUAU/daydayAU/dayAU
611.25
24.45
50
22.8
584
2,445
49.90
100
42.6
2,232
5,501
73.35
150
59.8
4,803
9,780
97.80
200
74.8
8,177
15,281
122.3
250
87.7
12,247
22,005
146.7
300
99.0
16,921
29,951
171.1
350
108.8
22,121
32,574
1.03 c
365
 111.6
 23,672
39,120
1.13 c
400
117.2
27,776
 49,511
1.27 c
 450
 124.6
 33,809
61,125
1.41 c
500
131.0
40,201
g×t2/2g × ttc(1-Rt)
ct  +V(t)

ln(R)
g=.489 AU/day²
LY = 63,241 AU
tc = 173.145AU/day
ln(R) = -.00283
Newton's equation is no good for larger values
above chart shows Newton's equation (g × t)
exceeding light speed after a year.
IMPOSSIBLE!!!
On the other hand, Thought Experiment assumes an Einsteinian version is adequately modeled by an exponential equation which shows at 500 days, a velocity of 131 AUs/day
and a total distance of about 40,000 AUs (≈ 2/3 LY).



VOLUME I: ASTEROIDAL
VOLUME II: INTERPLANETARY
VOLUME III: INTERSTELLAR

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