Post/Author/DateTime	Post
Masterful_Norg 04-11-04, 05:24 PM	Let me begin by giving a quote from a thread of mine in the epic level handbook forums:Originally posted by Doomhawk Now, first, to get some numbers out of the way. Sorry about having to use metric, I don't know the value of the G constant in the English system. Outer space = 1,000 miles up (according to my encyclopedia - 99% of the atmosphere is within 50 miles of the surface, however). That's 1,600 km, or 1,600,000 m. An arrow weighs .15 lbs. (.068181 kg). Let's assume that the bow is stretched 4 feet before being fired (approximately the distance if you assume a bow-firing pose, with the firing hand pulled all the way back to your shoulder). That's 1.2 meters. In order to reach a height of 1.6 million meters, the arrow has to have sufficient potential energy (I'm ignoring air resistance because I'm only in first-year physics and don't know how to calculate air resistance). The potential energy difference between the surface of the earth, 6,400,000 m from the center, and outer space, 8,000,000 meters from the center of the earth, is equal to (G constant)(Mass of earth)(Mass of arrow)/6,400,000m - (G constant)(Mass of earth)(Mass of arrow)/8,000,000m Which is (6.67*10^-11)(6*10^24)(.068181)*[1/6,400,000 - 1/8,000,000] =2.73*10^13 x 3.125*10^-8 =853,125 J So the arrow has to have 853,125 J of kinetic energy upon departing from the bow in order to reach outer space. If we treat the bow as a spring, then all of the spring's elastic potential energy will be converted into the arrow's kinetic energy. Elastic potential energy is E=.5kx^2, where k is spring constant and x is distance. The bow was stretched 1.2 meters, so that's your x, and E is equal to 853,125, as shown above. 853125=.5*k*4^2 853125*2/16=k k=106640 N/m When the bow is stretched the full four feet, it is exerting a force equal to F=kx. F=106640*1.2 =127968 N A kilogram is 2.2 pounds, and 9.8 newtons are a kilogram, so 4.45 newtons are a pound. That means the bow is exerting 28,727 pounds of force on the arrow when it is pulled. I assume that the maximum 'pull' a character can exert on a bow is equal to his light load (which is probably reasonable since he can only use his arm strength, not his legs or back). So the minimum Strength score necessary in order to have a light load of at least 28,727 pounds is 59. So assuming the character has a sufficiently strong bow, he would need a strength score of 59 in order to get an arrow into orbit. [As an aside, the arrow's velocity upon leaving the bow would be 16,508 feet per second, or 3.126 miles per second.] This arrow would, however, fall back to Earth after attaining a height of 1,000 miles... if you wanted the numbers for the arrow to never fall back to Earth (that is, escape velocity) then too bad, because I'm too lazy.
Masterful_Norg 04-11-04, 05:26 PM	Now here is the purpose of the thread. I would like to make new simple rules for designing bows, for enhancing the range of thrown weapons, and for determining how far an object of any weight can be thrown by a character with any given strength score. One of my conscerns is figuring out how to learn the actual speed of a projectile so that I can instance its location during mid flight at any point in the round. (This is to work hand in hand with my new initiative system.)
Masterful_Norg 04-11-04, 05:53 PM	Deleted
bootor 04-11-04, 06:41 PM	Vx=Velocity in the X axis Vy=Velocity in the Y axis Vin=Initial Velocity Vt=Total Velocity Sx=x position Sy=y position A=acceleration of gravity 9.81m/s or 32.2ft/s t=time theta=Initial angle of projectile Vx=Vin - sin(theta)*Vin Vy=sin(theta)*Vin - A*t Vt=sqrt(Vx^2+Vy^2) angle=tan(Vy/Vx) Sx=Vin*t-sin(theta)*Vin*t Sy=sin(theta)*Vin*t - .5*A*(T^2) Assuming you neglect air resistance Vx is always constant after the projectile is released. Also pounds mass is different the pounds force, as in the pull of a bow (lbf=32.2*lbm)
Masterful_Norg 04-11-04, 07:40 PM	Now I assume that the above posted range specifications are for people who have strength scores of 10. I have no intention to make this very complex. If I wanted a hard and fast rule I could just say that every +1 bonus to strength gives you one extra range increment and that a projectile can move its full range in 3 seconds. For every size catagory larger you are then specified for the weapon you gain one additional range increment, for every catagory lower you lose 1 range increment. Anything not meant to be thrown has a range increments of 5 and loses one range increment for every 10 pounds (round up to the nearest increment of ten) it is heavier than 5. This means that a person with a strength score of 10 would be able to throw something weighing 40-45 pounds a distance of 5', which makes some sense. And a person with a strength score of 50 would be able to throw the 40 pound object 105 feet, which makes sense. The same person with a strength of 50 would be able to throw a 980 pound object 25 feet. A person with a strength score of 50 has a light load of 8,512 pounds. A person with a strength score of 10 has a light load of 33 pounds. A throwing axe for a character with a strength score of 10 weighs 3 pounds (roughly 1/10 of the character's light load capacity) which that character can throw 25 feet. A character with a strength score of 50 can throw a 980 pound (roughly 1/8 the characters light load) object 25 feet. This system cannot work because it doesn't coincide with weapon sizing and a character will eventually be able to throw more weight than he can carry. Maybe someone could toss me rules for how the hulking hurler works? :D
Masterful_Norg 04-11-04, 07:49 PM	Thank you bootor. I must admit that your method will work admirably if we all have 1 minute of time between each arrow shot or a scientific calculator on hand. Do you have a way of converting this to simplified D&D physics? For example. One reaches terminal velocity after falling 120' in the D&D world. I'm trying to figure a method of determining how far someone can throw an object based on how much they can carry that works while being simple at the same time.
bootor 04-11-04, 09:10 PM	Ok Assumtions: Arrow is shot straight i.e. no y Velocity Component. Character height is 5 ft. Therefor Sy= -.5*A*(T^2) Sy=height(5ft) A=G(-32.2ft/sec) T^2=sqrt(5/-.5/-32.2) T=.55727 sec in flight Since there is no y component of initial velocity each shot will be in the air for T time regardless of all other factors except size then apply size modifier to time. i.e large character T=1.114 s, small character T=.2786 s. Now Sx=Vin*T(.55727 sec) Vin=A*T F=M*A A=F/M Vin=F*T/M for each weapon the change in T is so small it can be considered constant again for each weapon M is always constant so, the velocity is based the max force a person can apply and also increases linearly. Now Using a character with a strength of 10 as a base for all weapons base increment for any weapon take the base distance of a weapon. divide it by (33 light load for str 10) and multiply it by the light load of user. example long bow shot by str 20 normal range 100 light load 133 new range 100/33*133 = 400 ft since height also determines the time the projectile is in the air the distance should change, but since load is changed based on size it about evens its self out in the end. to find the position of the projectile in mid flight coming soon...
Masterful_Norg 04-11-04, 10:05 PM	Thank you very much!:D I appreciate your help!