Travel math | Star Frontiers

Herein I do the math related to acceleration and Void speed. Just for fun.<div> </div><div>Let:</div><div>1 hex = 1×107 m</div><div>1 g = 9.8 m·s-2</div><div>c = 3×108 m·s-1</div><div>1 turn = 600 s</div><div>1 day = 7.2×104 s</div><div> </div><div>First, we calculate the relationship between 1 ADF and 1 g:</div><div> </div><div>1 ADF = 1 hex·turn-2 = (1×107 m)(600 s)-2 = (1×107 m)(3.6×105 s2)-1 =~ (28 m·s-2)(9.8 m·s-2)-1g =~ 2.9 g</div><div> </div><div>So the assertion in the rule books that 1 ADF is 1 g is incorrect; 1 ADF is nearly 3 g.</div><div> </div><div>Since the size of a hex is arbitrarily set at 10,000 km across, we can feel free to select a different arbitrary size and call it a "short hex" (sh), in which 1 ADF is 1 g:</div><div> </div><div>d = vt = at2</div><div>d[1turn@1g] = (1 g)(1 turn)2 = (9.8 m·s-2)(600 s)2 = (9.8 m·s-2)(3.6×105 s2) = 3.528×106 m = 3528 km</div><div> </div><div>Therefore a short hex is 3528 km. You can say "about 3500 km" if you like. This is the new size of a hex in the board game, in which one "short ADF" is equal to 1 g. You're just scaling down the speeds of everything in the game to where you read 1 ADF as equal to 1 g.</div><div> </div><div>Starships traveling between systems spend most of their time accelerating or decelerating at 1 g. Any trip of X light years is supposed to take 0.5·X days of 1 g acceleration and 0.5·X days of 1 g deceleration to reach the destination. Furthermore, astrogators must spend 0.5·X days plotting a jump, unless they perform risk-jumping. To get to a destination faster, a ship must both accelerate faster than 1 g and risk-jump.</div><div> </div><div>Jumping through the Void is supposed to take place at 0.01c. This doesn't work:</div><div> </div><div>v = at</div><div>t = va-1</div><div>t[Void] = (0.01 c)(1 g)-1 = (0.01)(3×108 m·s)(9.8 m·s-2)-1 = (3×106 m·s)(9.8 m·s-2)-1 =~ 3.06×105 s = 4.25 day</div><div> </div><div>Therefore at 1 g a ship will reach 1% of the speed of light in about 4.25 days, making the shortest trip in the Frontier 8.5 days. This is clearly not the case, as shown on the Frontier map.</div><div> </div><div>So how fast are we going if we travel 1 day at 1 g?</div><div> </div><div>v[1day@1g] = (1 g)(1 day) = (9.8 m·s-2)(7.2×104 s) = (7.056×105 m·s-1)(600 s·turn-1) = (4.2336×108 m·turn-1)(1 sh)(3.528×106 m)-1 = 120 sh·turn-1</div><div> </div><div>We're traveling 120 short hexes per turn. This is equal to</div><div> </div><div>v[1day@1g] = (7.056×105 m·s-1)(3×108 m·s-2)-1c = (2.352×10-3)c = 0.002352c</div><div> </div><div>Thus, 1 day at 1 g brings you to about 0.2% of light speed, a far cry from the 1% of c in the book.</div><div> </div><div>Therefore, it is necessary to drop the 1% of c idea and implement a "jump drive" or "Void field" device to be activated at the requisite 1 day 1 g acceleration per light year.</div><div> </div><div>Summary:</div><div>Change hex size to 3528 km ("short hex").</div><div>1 short ADF = 1 g.</div><div>Traveling to a different system requires total acceleration for a number of days equal to (distance in LY) ÷ ADF ÷ 2.</div><div>Traveling to a different system requires a number of days of astrogation calculations equal to (distance in LY) ÷ 2, or less if you're risk-jumping.</div><div>When requisite speed is achieved, Void-jumping device must be activated to enter Void and deactivated to leave it.</div><div>Deceleration takes (distance in LY) ÷ ADF ÷ 2 days. The astrogator must make his calculations already knowing what the ADF on the far side of the jump will be (or else the ship will arrive too far or too close to the system and have to travel for longer to get there).</div><div>Void-speed is not related to the speed of light.</div>

Pinned

Off

Locked

Off

Forum category

Game Talk