DETERMINE DUE 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

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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

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) )	..........Goes a Long Way

(Many thanks to Wolfram Integrator)

Above "Newtonian" equations work well for short distances. Examples of G-force Distance 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 = 299,792.5 km 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 at different speeds, but a few facts stay steadfast: 1. Nothing ever exceeds c, light speed. 2. All observers at any speed consistently measure photons at the same velocity, c = 173.14 AU/day = 299,792.5 km/sec 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%

New Formula for Velocity 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: 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 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 Multiply this average velocity by time (one day) to approximate incremental distance during day t, dt= 1 day × VAve . Approximate cumulative distances by summing relevant incremental distances, Σ dt. Interplanetary Ranges G-force Time Spot Velocity Average Velocity Incr Distance Cumm Distance t Vt Vave dt Dt 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

Compute g-force distance: HOW? Recall basic calculus, the integral, ∫, 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: ∫ab  V(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: ∫0t    k dt = k × t - 0 ∫02 500 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. ∫02.5 g × t dt = g t2/2 ∫02.5 0.5 AU/day2 (2.5 day)2 dt =.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

Newtonian Time Einsteinian d(t) V(t) t V(t) d(t) AU AU/day day AU/day AU 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 2 g × t t c(1-Rt) ct + V(t) ln(1-Δ) c = 173.15AU/day LY = 63,241 AU t 1 - 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 values resemble Einsteinian values. EXAMPLE: For tenth day of g-force acceleration, both methods show velocity about 5 AUs/day and a total distance of about 24 AUs.

Newton's equation is no good for larger values; above chart shows Newton's equation (g × t) exceeding light speed (c) 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).

BEYOND INTERPLANETARY Short range (10 days) chart contrasts starkly with a longer range chart. After 500 days of g-force acceleration, velocity charts show enormous divergence.

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.

Newtonian Time Einsteinian d(t) v(t) t v(t) d(t) AU AU/day day AU/day AU 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 2 g × t t c(1-Rt) ct + V(t) ln(R) g=.489 AU/day² LY = 63,241 AU t c = 173.145AU/day ln(R) = -.00283

SLIDESHOW

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VOLUME 0: ELEVATIONAL
VOLUME I: ASTEROIDAL
VOLUME II: INTERPLANETARY
VOLUME III: INTERSTELLAR

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