A lion and a lion tamer are enclosed within a circular cage. If they move at the same speed but are both restricted by the cage, can the lion catch the lion tamer? (Represent the cage by a circle, and the lion and lion tamer as two point masses within it.)
(In reply to
Numeric integration: it takes a while to get close by Charlie)
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While both the lion and the tamer seem to be both at 100 feet from the center, the lion is not quite that far--it just rounds to that distance--so he has a more inside track, more than offsetting his needing a radial velocity detracting from his tangential velocity.
The program didn't stop at the first 2.0 separation, as that was rounded down from a slightly larger separation.
DEFDBL A-Z
pi = ATN(1) * 4
CLS
spd = 15 ' ft/sec
rad = 100 ' ft
r = 80: th = 60: tamex = r * COS(pi * th / 180): tamey = r * SIN(pi * th / 180)
r = 30: th = 150: lionx = r * COS(pi * th / 180): liony = r * SIN(pi * th / 180)
dt = .00001#
ysep = tamey - liony: xsep = tamex - lionx
sep = SQR(ysep * ysep + xsep * xsep)
tamer = SQR(tamex * tamex + tamey * tamey)
ptst = -1
DO
tst = t * 1
IF INT(tst) > INT(ptst) THEN
lionr = SQR(lionx * lionx + liony * liony)
PRINT USING "####.# ####.# ####.# ####.# ####.# ###.# ####.# ####.#"; t; lionx; liony; tamex; tamey; sep; tamer; lionr
ct = ct + 1
IF ct MOD 40 = 0 THEN DO: LOOP UNTIL INKEY$ > "": PRINT
END IF
ptst = tst
tamer = SQR(tamex * tamex + tamey * tamey)
ysep = tamey - liony: xsep = tamex - lionx
sep = SQR(ysep * ysep + xsep * xsep)
liondy = dt * spd * ysep / sep: liondx = dt * spd * xsep / sep
liony = liony + liondy: lionx = lionx + liondx
IF tamer >= rad THEN
IF tamer > rad THEN tamer = rad
tameth = ATN(tamey / tamex) * 180 / pi
IF tamex < 0 THEN tameth = tameth + 180
tamedir = tameth - 90
tamedx = spd * dt * COS(tamedir * pi / 180)
tamedy = spd * dt * SIN(tamedir * pi / 180)
tamex = tamex + tamedx: tamey = tamey + tamedy
ELSE
tamey = tamey + liondy: tamex = tamex + liondx
END IF
t = t + dt
LOOP UNTIL sep <= 2
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Posted by Charlie
on 2013-06-11 13:43:28 |