Books being read:
Odd Thomas by Ken Follett, a ghost story like Stephen King and X Files
Bleak House by Charles Dickens, being a template for so many TV miniseries, if they only copy the humorous ironies Dickens incorporates into his minor characters.
An Introduction to Medieval Europe: 300 - 1500 by James Westphal Thompson and Edgar Nathaniel Johnson one of Mom's many tomes.
Immediate plans to visit Fran later this month. Many preparations need to be under gone:
Packing
Monday, February 11, 2008
Saturday, February 2, 2008
new blog entries moved
I'm not blogging here anymore. It will remain as an archive for 2007. My new blogging is done on http://thoreausautumn.blogspot.com
Sunday, December 30, 2007
What is the mathematical equation of an orbit or falling down?
by jim m Member since:
September 21, 2007
Total points:
2721 (Level 4)
Best answers 30percent 487 answers
Member Since: September 21, 2007
Total Points: 2721 (Level 4)
Points earned this week: 30 For a 6400 km radius Earth, and g force of 10 meter/second^2, if you shot your self straight from a cannon
to a distance of 1km then by the curvature of the Earth that means you would be √(6400^2 +1)-6400 or 7.83 e-5km or 7.83cm
above the horizon. Your time of flight falling through 7.83 cm would be √(.1566m/10m/s^2)
or 1/8 of a second. Approximating a cruise missile type orbit,
of 40000 km around the Earth, you would complete that in
1 hour, 23-1/2 minutes. 8km /sec. Compare that with rotation
of the Earth which corresponds to 40000km/86,400s or .463
km/sec at the Equator. So if Earth spun itself about 17 times
faster, we and everything would float weightless at the
Equator
Just wanted to edit my answer and add a couple of things.
(First of all Newton saw that the apple and bodies in freefall are a special case of
bodies in orbit, since attraction is directed toward the center of mass and were all the mass of earth to be concentrated in one central point falling bodies would have orbits as well, which would describe elliptical orbits with the top of trajectory being the apogee. It's just that surface of the planet is in the way, and bodies describe parabolas instead).
If you use the centrifugal definition of acceleration V^2/R=a=g=10m/s^2,
V=√(10*6.4e6)=8km/s for a circular orbit
Since 6400km is an altitude of 30 km, you can calculate
intercontinental ballistic ranges for a variety of suborbital
velocities, resulting radial and tangential accelerations.
radial a = v^2/R - GM./R^2; G=6.67e-11, M=6.0e24
A missile horizontally inserted at 30km altitude with velocity
< 8km/s will range out by resultant of
a = V^2/R-10.
t=sqrt(2*3e4/a)
RangeDown=V*t
If the missile as inserted to the horizontal entry point by
artillery shot, then double to get the ballistic range..
If the missile is boosted at the insertion point to > 8km/s
the missile enters an elliptical orbit. Perigee, or Ro, is the insertion point, or 6400 km (more later)
Because we are starting with velocity and perigee, the complex differential equations are reduced so one obtains the semimajor axis and eccentricity as a function of initial velocity and distance:
semimajor axis a=
R*(2-R*V^2*(GM)^-1)
eccentricity e=
sqrt(M*a-(RV)^2/G)*(M*a)^-1)
23 hours ago - Edit - Delete
Source(s):
http://www.enchantedlearning.com/subject..
September 21, 2007
Total points:
2721 (Level 4)
Best answers 30percent 487 answers
Member Since: September 21, 2007
Total Points: 2721 (Level 4)
Points earned this week: 30 For a 6400 km radius Earth, and g force of 10 meter/second^2, if you shot your self straight from a cannon
to a distance of 1km then by the curvature of the Earth that means you would be √(6400^2 +1)-6400 or 7.83 e-5km or 7.83cm
above the horizon. Your time of flight falling through 7.83 cm would be √(.1566m/10m/s^2)
or 1/8 of a second. Approximating a cruise missile type orbit,
of 40000 km around the Earth, you would complete that in
1 hour, 23-1/2 minutes. 8km /sec. Compare that with rotation
of the Earth which corresponds to 40000km/86,400s or .463
km/sec at the Equator. So if Earth spun itself about 17 times
faster, we and everything would float weightless at the
Equator
Just wanted to edit my answer and add a couple of things.
(First of all Newton saw that the apple and bodies in freefall are a special case of
bodies in orbit, since attraction is directed toward the center of mass and were all the mass of earth to be concentrated in one central point falling bodies would have orbits as well, which would describe elliptical orbits with the top of trajectory being the apogee. It's just that surface of the planet is in the way, and bodies describe parabolas instead).
If you use the centrifugal definition of acceleration V^2/R=a=g=10m/s^2,
V=√(10*6.4e6)=8km/s for a circular orbit
Since 6400km is an altitude of 30 km, you can calculate
intercontinental ballistic ranges for a variety of suborbital
velocities, resulting radial and tangential accelerations.
radial a = v^2/R - GM./R^2; G=6.67e-11, M=6.0e24
A missile horizontally inserted at 30km altitude with velocity
< 8km/s will range out by resultant of
a = V^2/R-10.
t=sqrt(2*3e4/a)
RangeDown=V*t
If the missile as inserted to the horizontal entry point by
artillery shot, then double to get the ballistic range..
If the missile is boosted at the insertion point to > 8km/s
the missile enters an elliptical orbit. Perigee, or Ro, is the insertion point, or 6400 km (more later)
Because we are starting with velocity and perigee, the complex differential equations are reduced so one obtains the semimajor axis and eccentricity as a function of initial velocity and distance:
semimajor axis a=
R*(2-R*V^2*(GM)^-1)
eccentricity e=
sqrt(M*a-(RV)^2/G)*(M*a)^-1)
23 hours ago - Edit - Delete
Source(s):
http://www.enchantedlearning.com/subject..
The first 1000 digits of π
3.1415926535 8979323846 2643383279 5028841971 6939937510 5820974944 5923078164 0628620899 8628034825 3421170679 8214808651 3282306647 0938446095 5058223172 5359408128 4811174502 8410270193 8521105559 6446229489 5493038196 4428810975 6659334461 2847564823 3786783165 2712019091 4564856692 3460348610 4543266482 1339360726 0249141273 7245870066 0631558817 4881520920 9628292540 9171536436 7892590360 0113305305 4882046652 1384146951 9415116094 3305727036 5759591953 0921861173 8193261179 3105118548 0744623799 6274956735 1885752724 8912279381 8301194912 9833673362 4406566430 8602139494 6395224737 1907021798 6094370277 0539217176 2931767523 8467481846 7669405132 0005681271 4526356082 7785771342 7577896091 7363717872 1468440901 2249534301 4654958537 1050792279 6892589235 4201995611 2129021960 8640344181 5981362977 4771309960 5187072113 4999999837 2978049951 0597317328 1609631859 5024459455 3469083026 4252230825 3344685035 2619311881 7101000313 7838752886 5875332083 8142061717 7669147303 5982534904 2875546873 1159562863 8823537875 9375195778 1857780532 1712268066 1300192787 6611195909 2164201989
Friday, December 21, 2007
cycle 3
Day 4-7 of Cyclacel 116 cycle 3.
No untoward events to report. My blood panel last clinic was good. The doctors loved it. I feel fine except for the usual weakness and soreness in my hip. The weakness peaked on Friday evening when the pain of hobbling to set up my recliner and then struggle to get in it was enough to bring tears to my eyes, especially with pugs jumping on my lap.
After dosing I sleep until noon. I notice a bone spur on my hip backside at the amputation site and an eruption of itch; might be telling me not to lay flat, but to elevate the back at an angle and take weight off of that spot. Recalling my severed nerve which makes my leg feel like a leaden appendage, I can only wonder what the pain would be like with normal nerves, or whether a rerouting of synapses has occurred, leading to the sensations I now feel in my hip when I overdo it
Dosage times T, W, Th, F, S, Su, M
=12 pm, 6:30 am, 6 am, 8am, 8:42am, 9am, 8:15am.
No untoward events to report. My blood panel last clinic was good. The doctors loved it. I feel fine except for the usual weakness and soreness in my hip. The weakness peaked on Friday evening when the pain of hobbling to set up my recliner and then struggle to get in it was enough to bring tears to my eyes, especially with pugs jumping on my lap.
After dosing I sleep until noon. I notice a bone spur on my hip backside at the amputation site and an eruption of itch; might be telling me not to lay flat, but to elevate the back at an angle and take weight off of that spot. Recalling my severed nerve which makes my leg feel like a leaden appendage, I can only wonder what the pain would be like with normal nerves, or whether a rerouting of synapses has occurred, leading to the sensations I now feel in my hip when I overdo it
Dosage times T, W, Th, F, S, Su, M
=12 pm, 6:30 am, 6 am, 8am, 8:42am, 9am, 8:15am.
Friday, December 14, 2007
Wednesday, December 12, 2007
Off portion of Cycle 2 sum-up
If I had doubts about continuing on this therapy in the darkest night of pain and discomfort, those doubts have vanished. Only the vaguest memory remains of those awful times. My impression of the experience is that my body is taken aback by the insult of the pills the first two or three days culminating in sick days on Thursday and Friday night. This sickness finds its way into certain areas of my hip, leg and abdominal muscles. I get these gout-like symptoms in my feet, arms, different joints once in a while. Its been concurrent with my cardiomyopathy. So on top of the culmination, I get gout to, making me totally debilitated.
By Saturday the spell is broken and little by little, through pain medication, extra hydrocodone, Lidocaine patches, and ibuprofen for unusual pain, I get through it. Remarkably I breeze through the last two doses.
Knowing the course of my sickness through the seven days I ready for next time.
By Saturday the spell is broken and little by little, through pain medication, extra hydrocodone, Lidocaine patches, and ibuprofen for unusual pain, I get through it. Remarkably I breeze through the last two doses.
Knowing the course of my sickness through the seven days I ready for next time.
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