Home > Just Math
| Tunneling through a mine car (Posted on 2019-04-16) |
 |
Four safety engineers set out to inspect a newly cut tunnel through Mt. Popocaterpillar in the Andes. Each person walks at a different constant integer speed measured in meters per minute. In the tunnel there is a mine car which travels along a fixed track, automatically going from end to end at a fixed integer speed. When people board the car they may reverse its direction, but cannot change its speed.
At noon on Monday all four engineers start at the south end, while the mine car starts at the north end. The first (fastest) engineer meets the car, and takes it some distance north. The engineer gets out and continues going north, while the car resumes heading south. Then the second engineer meets the car and also takes it some distance north. Likewise for the third and fourth engineers. All the people, and the mine car, travel continuously with no pauses. The inspectors always go north. Each person enters and exits the car at an integral number of minutes.
All four engineers reach the end of the tunnel simultaneously. What is the earliest time this could happen?
soln
|
 | Comment 14 of 25 | 
|
The bottom line is that my result is 36 minutes with very advanced engineering and running rather than walking, but 46 minutes with more conventional engineering. Method: I described the method in my first posting: 7 parameters (4 Engineer walking speeds, the car's speed, the duration of E1’s 1st walk and drive) together serve to completely fix the simulation. I varied these parameters to get a minimum time. I show the math here. The output (abbreviated) is here. To run the simulation I choose the very maximal ranges:
Maximum tunnel length: 2 x Gotthard 57 km tunnel
Maximum foot speed: 28 mph running speed
Max car speed: 200 mph
(like a Maglev, but in a tunnel like Musk's futuristic
one for CA)
Maximum time in tunnel: 20 days
(Yikes. Fortunately all times are under an hour.)
The program ran in a couple of hours.
As the times got smaller, the tunnels,
the mine car and walkers became more fantastic. For example I did not post a solution with T=34 min, because the Engineers were moving too fast. Here are possible solutions:
---------
Tunnel 1:
Length: 2310 m, Time: 46 minutes: car speed: 105 m/min
Walking speeds: 49, 45, 45, 42, 35. E1 1st walk 15 min,
Car times 1, 4, 6, 10, min
--------
Tunnel 2:
Length: 18564 m, Time: 36 minutes: car speed:
2184 m/min (86 mph)
Walking speeds: 468 (18 mph!), 364, 182, 39 m/min
E1 1st walk 7 min
Car times 1, 3, 6, 8 min
--------------
Incidentally, I was able to reproduce Steve H.s result for
E1, E2 of 6 min. It was, in fact, very helpful to have
his result for debugging this simulation - many thanks!
The case of E1, E2, and E3 only (3 E's) has a minimal
time of 17 min:
---------------
Tunnel 1-3E:
Length: 540 m, Time: 17 minutes: car speed: 60 m/min
Walking speeds: 30, 28, 20; E1 1st walk time 6 min,
Car times 1, 2, 5 min -------------- program:
program tunn
implicit none
integer Dmax,D,fsmax,csmax,cs,Trecord,Tmax,
1 fs1,fs2,fs3,fs4,ft1,ct1,ct1max,posint,nonneg,
2 irat,ihalfrat,id,iexp,texp,T2,T3,T4,ttot
real xcs,xct1,ct2,ct3,ct4,xft1,xfs1,xfs2,xfs3,xfs4,
2 pin1,pin2,pin3,pin4,pout1,pout2,pout3,pout4,
3 pcc1,pcc2,pcc3,pcc4,T,rat,del
Dmax = 114000 ! [meters] two times Gotthard 57 km tunnel length
fsmax = 700 ! [meters/min] foot speed running at 28 mph
csmax = 5000 ! [meters/min] car at 200 mph
Tmax = 28800 ! [min] 20 days
Trecord = 100000 ! [min] time to beat
iexp=3
texp=10**iexp
pcc1=0
print 2222
2222 format('Time Tunnel Car Wlk fs1 pcc1 pin1',
1' ct1 pout1 fs2 pcc2 pin2 ct2 pout2 ',
2' fs3 pcc3 pin3 ct3 pout3 fs4 pcc4 pin4',
3' ct4 pout4 ttot',/,170('-'))
do 7 D = 20,Dmax
id=D/texp
c if (texp*id.eq.D)print*,D,' out of ',Dmax
do 6 ft1 = 1,Tmax
xft1=ft1
rat=D/xft1
irat=rat
del=abs(rat-irat)
if(del.gt.0.001)go to 6
ihalfrat=irat/2-1
do 5 fs1=4,ihalfrat
xfs1=fs1
pin1=ft1*fs1
cs=irat-fs1
xcs=cs
ct1max = (D - pin1)/cs -5
do 4 ct1 = 1, ct1max
xct1=ct1
pout1 = pin1 + ct1 * cs
T = ft1 + ct1 + (D - pout1)/xfs1
if (T.gt.Tmax.or.T.gt.Trecord) go to 4
do 3 fs2 = 3, fs1-1
xfs2=fs2
pcc2 = (ft1 + ct1) * fs2
pin2 = pcc2 + (pout1 - pcc2) * xfs2/(xfs2+xcs)
if(posint(pin2).eq.0) go to 3
ct2=(D-T*xfs2)/(xcs-xfs2)
if(posint(ct2).eq.0)go to 3
pout2 = pin2 + ct2*cs
if(pout2.gt.D)go to 3
T2 = pin2/xfs2 + ct2 + (D - pout2)/xfs2
do 2 fs3 = 2, fs2-1
xfs3=fs3
pcc3 = (pin2/xfs2 + ct2)*fs3
pin3 = pcc3 + (pout2 - pcc3) * xfs3/(xfs3+xcs)
if(posint(pin3).eq.0) go to 2
ct3=(D-T*xfs3)/(xcs-xfs3)
if(posint(ct3).eq.0)go to 2
pout3 = pin3 + ct3*cs
if(pout3.gt.D)go to 2
T3 = pin3/fs3 + ct3 + (D - pout3)/xfs3
do 1 fs4 = 1, fs3-1
xfs4=fs4
pcc4 = (pin3/xfs3 + ct3)*xfs4
pin4 = pcc4 + (pout3 - pcc4) * xfs4/(xfs4+xcs)
if(posint(pin4).eq.0) go to 1
ct4=(D-T*xfs4)/(xcs-xfs4)
if(nonneg(ct4).eq.0)go to 1
pout4 = pin4 + ct4*cs
if(pout4.gt.D)go to 1
T4 = pin4/fs4 + ct4 + (D - pout4)/xfs4
if(T.le.Trecord)then
c print*,'T = ',T
Trecord = T
print 777, T,D,cs,ft1,
1 fs1,pcc1,pin1,xct1,pout1,
2 fs2,pcc2,pin2, ct2,pout2,
3 fs3,pcc3,pin3, ct3,pout3,
4 fs4,pcc4,pin4, ct4,pout4,T2,T3,T4
777 format(f4.1,x,i6,x,i5,x,i3,x,
1 4 (i4,x,3(f7.1,x),f8.1,1x),3(i2,x) )
endif
1 enddo
2 enddo
3 enddo
4 enddo
5 enddo
6 enddo
7 enddo
end
integer function posint(x)
implicit none
integer ix
real x,del
posint=0
if(x.le.0)return
ix=x
del=abs(x-ix)
if(del.gt.0.001)return
posint=1
return
end
integer function nonneg(x)
implicit none
integer ix
real x,del
nonneg=0
if(x.lt.0)return
ix=x
del=abs(x-ix)
if(del.gt.0.0000000001)return
nonneg=1
return
end
Edited on August 3, 2020, 11:22 pm
|
|
Please log in:
Forums (0)
Newest Problems
Random Problem
FAQ |
About This Site
Site Statistics
New Comments (15)
Unsolved Problems
Top Rated Problems
This month's top
Most Commented On
Chatterbox:
|
