
#73
Jan2105, 03:38 PM

Astronomy
Sci Advisor
PF Gold
P: 22,800

Back in post #71 I listed some rough sizes, including these force and power benchmarks.
weight of 50 kg sack of cement E40 power of a 160 watt lightbulb E49 I am thinking of the force E40 as a "sack" force benchmark and imagine a 50 kg weight on a pulley descending at speed E9 (which is 2/3 mph, or a billonth of the speed of light) and as it descends it does work, like turning a spindle, maybe even generating electricity. The power output of that descending weight is E49. To see that, you just have to multiply the force E40 by the speed E9 and you get the power. of course if you are generating electricity there will be some loss because of inefficiencies. but basically this force exerted at that speed delivers that much power. and I'm going to call that level of power a BULB of power. this is a drastic solution to the problem of remembering the brightness of sunlight. the solar constant at this distance from the sunthe power per unit area delivered by direct unattenuated sunlightis 5.7 BULBS PER SQUARE PACE. In natural unit terms, a pace (81 cm) is E34 and a square pace is E68 and a bulb of power is E49. So a bulb of power spread over a square pace is E49/E68 = E117 I am saying that the brightness of sunlight is 5.7 times that. It is like about SIX of those 160 watt litebulbs set in a pacewide square. In natural units, 5.7E117 is what the handbook value of the solar constant actually turns out to be. but I dont find that so easy to remember. So I visualize it as 5.7 bulbs per sq. pace. A pace is just one of my stepsaround 32 inchesso I can easily pace out a square that size on the flagstones in the garden. It is an easy area for me to visualize. and the litebulbs are easy to visualize. so I have a visual handle on this 5.7E117 =================== In the "Force" system of natural units, the unit of power is of course E49 bulbs (because bulb was defined as E49) and it is the power delivered by the unit Force pushing at the speed of light. this is a lot of power and if you count as fast as you probably can outloud, say 222 counts a minute, then WITH EVERY COUNT UNIT POWER DELIVERS ENOUGH ENERGY TO CREATE 2000 SUNS. We discussed this, it is enough power to create a galaxy in something on the order of 100 days. or if you wanted to produce such a power by annihilating stars and converting their whole mass into energy then you would have to annihilate about 2000 stars like the sun with every count. As with conventional Planck units, these natural units are fundamentally Big Bangscale. the temperature, the density, the pressure, ...and so on...are mostly at the level of big bang conditions. I guess that could be seen as reassuring. You can be sure ahead of time that you are not going to encounter any temperature less than zero or greater than one. the physical scales tend to be bounded between zero and onelike with speed too. 



#74
Jan2105, 05:49 PM

Astronomy
Sci Advisor
PF Gold
P: 22,800

An explorer once visited three planets and went into low orbit around each in order to take pictures with his digital camera to put on his website. He finds each planet more delightful than the one before it and, while skimming around the third, he breaks down and calls you on the cellphone.
Hello, says the explorer, on each planet it took a different length of time to travel one radian of the low orbit. In natural time units it took 7E45 4E45 and 3E45 on planets A, B, and C respectively. what are the densities of the three planets? the formula for the density is D = 6/T^{2} you just divide 6 by the square of the radian time, so the three densities are 6/(49E90) = 1.224E91 6/(16E90) = 3.75E91 and 6/(9E90) = 6.66E91 the first, you tell the explorer, is virtually the same the density of water. planet B, on the other hand, is slightly over 3 times the density of water and is therefore comparable to many of the solar system's satellites including the earth's moon planet C, however, is 5.4 times the density of water, quite close to earth itself, which is 5.5! Indeed, says the explorer, that is within experimental error. I believe I am just passing over Sausalito. BTW E45 natural time units was listed a couple of posts back as lasting 4.5 minutes, so the radiantime for low orbit in the earthlike case, namely 3E45, becomes 3 x 4.5 = 13.5 minutes. that is for a hedgetop skimming orbit neglecting air not practical, of course, but raising it above the atmosphere does not make the orbit all that much slower. so it is a pretty good estimate. (multiply 13.5 minutes by 2pi to get the period) 



#75
Jan2205, 05:48 PM

Astronomy
Sci Advisor
PF Gold
P: 22,800

Another time the explorer cruises by the nightsides of each of 3 planets in order to gauge the infrared heat brightness from each. He wants to know how warm or cold they are. He finds that the heatglow brightness of the three is as follows:
1.4E117 1.9E117 and 2.5E117 the question is, what is the nighttime surface temperature on each planet? Simply put, you just multiply each number by 6 and take the fourth root (press square root twice) although officially what you multiply each number by is 60/pi^{2}. However, pisquare is almost the same as ten, and 60/10 is six, so it's almost the same either way. So let's multiply each planet's heatglow by 60/pi^{2} 8.511 E117 11.55 E117 15.20 E117 and press the squareroot button twice to get the temps 0.960 E29 1.037 E29 1.110 E29 The first is below freezing, the second is room temperature, and the third is the perfect temperature for a hot tub! To help with interpreting these temperatures, remember all those we usually experience are close to E29 and 1.000E29 is our basic reference 49 Fahrenheit. Going up from 1.000 to 1.110 is equivalent to going up 110 halfFahrenheit steps, that is 55 Fdegrees, which if you add it to 49 gets you 104 Fahrenheit. On the other hand, going from 1.000 up to 1.037 is equivalent to 37 of those steps which is 18some Fahrenheitdegrees. Adding that which to 49 gets you 67 Fahrenheit, and what could be a more comfortable than that? So the moral is: go for the planet that glows 1.9 E117 in the infrared. 



#76
Jan2705, 02:24 PM

Astronomy
Sci Advisor
PF Gold
P: 22,800

two more benchmarks
the temperature in the energyproducing core of the sun is 5E25 (remember that solar surface temp is 2E28 and avg earth surface temp is E29 the way to remember solar surface is that E28 is an eQ and green photons have energy 10eQ, so solar surface is going to be around E28, and the temp there happens to be 2E28. so from core out to surface, temp goes down by factor of 2500) and middle D on piano is (1/2)E39, have to go, back later that means the D in the soprano/alto range, on the fourth line of the treble clef, is E39 the natural unit frequency is E39 times higher than that note the sopranos in our chorus singmaybe I could get it in falsetto. THE CAT ENGINES OF ORNISH I suspect Kea of liking cats, so I have devised a story about the wicked space pirates of Planet Ornish and how they propel their giant Bagelshaped troopships. I hope this will scandalize Kea. 



#77
Jan2705, 03:17 PM

Astronomy
Sci Advisor
PF Gold
P: 22,800

THE CAT ENGINE OF THE SHIPS OF ORNISH
The ships of Ornish are driven by Cat Motors which consume cats as fuel by converting each cat entirely into energy. The captain of a tenmillionpound troopship wishes to achieve a speed of E3 (one thousandth of the speed of light, about 300 km/second) in order to depart a plundered system. Once in the clear he will enter warp and rendezvous with the rest of the pirate band. The initial change of velocity is accomplished by the ship's efficient photon drive. How many standard 10 pound cats must be converted? answer "pound" is just a handle on E8 mass units so clearly the mass of each cat is one billion mass units (10 pounds is 10 E8 units). so since c = 1 each cat yields one billion natural energy units. (the mass is the same number as the energy in natural units) The ship mass is 10 million pounds (E15), so the desired momentum change is E15 x E3 = E12 momentum units. This requires discharging a photon pulse with E12 energy units, which consumes 1000 cats. 



#78
Jan2705, 03:33 PM

Astronomy
Sci Advisor
PF Gold
P: 22,800

FAST FOOD EXERCISE
According to Defense Analyst Daniel Pinkwater, the earth is in danger of invasion by the Fat Men: a spacefaring race which plunders other planets for their fast food. Their planet, planet Ornish, is deficient in basic resources like french fries, mayonaise, potato pancakes, Colonel Sanders fried chicken, and sour cream, which has forced them into a life of nomadic piracy. The Fat Men spacewar uniforms consist of loud plaid sportsjackets, green dacron slacks, loafers, and eyeglasses with heavy black plastic frames. They travel by the millions in a fleet of troopships shaped like enormous bagels. Professor Pinkwater fears that, before long, hordes of Fat Men will descend on earth and ravage our fast food outlets. The defense analyst has calculated that one raid by the Fat Men could deplete the earth of a million pounds of its mayonaise. How much energy does this represent? answer If you just go into the kitchen and look on a jar of Best Foods Real Mayonaise it will say that a 13 gram serving has 90 food Calories. One of our (E8 natural mass unit) pounds is 434 grams so it contains almost exactly 3000 Calories. Also to a reasonably close approximation, the natural unit of energy is 100 thousand food Calories. So a million pounds of mayonaise, with its 3 billion Calories, represents 30 thousand natural energy units. 



#79
Jan2805, 05:27 PM

Astronomy
Sci Advisor
PF Gold
P: 22,800

Baez posted some facts about Titan on SPR today including at what altitude the atmosphere was coldestin natural units terms it was
surface temp 3.33E30 lowest temp 2.48E30 (at altitude 6.2E38) let's look at the indicated gradient 0.85E30/6.2E38 =1.4E69 this was for a mainly nitrogen atmosphere which was as much as 7 percent methane at some places, but for simplicity I want to calculate as if it is "dry" nitrogen with no methane "vapor" and see what the theoretical lapse rate would bethis can be compared with the measured gradient. http://www.esa.int/SPECIALS/Cassini...F2HHZTD_0.html GRAVITY ON TITAN Titan surface gravity is 0.12E50. (I write it that way because I think of acceleration E50 as about one "gee") TEMPERATURE GRADIENT ON TITAN in natural units terms the lapserate (rate of change in temp with altitude) is just 2/7* x wt of molecule the weight of a nitrogen molecule = 0.12E50 x 28/(2.6E18) = 1.29E68 and 2/7 of that is 0.37E68 = 3.7E69 So the theoretical lapserate (with no methane assumed) is 3.7E69 but what was actually observed was 1.4E69 The presence of a few percent methane could explain why the observed gradient was less. notes That ESA link says Titan radius is 0.4 earth's, and Titan mass is 1/45 of earth. this means surface gravity is around 13.8 percent of earth As a crutch I remember an order of magnitude "gee" is E50 natural acceleration unitsofficial earth surface gravity is 0.88E50. So* titan is 0.12E50 Wind is the result of convection which happens when the temperature gradient (cooling off with altitude) exceeds the threshold gradient called "lapserate". The lapse rate in moist air is less than the dryair lapse rate, because having moisture (able to condense to form clouds and give up its energy) serves as a "fuel" for convectiona reservoir of energy which makes convection easier and able to happen with more gradual temperature gradients. As a way of interpreting the observed gradient of 1.4E69, think of E32 on the temp scale as half a Fahrenheit degree and E37 on the distance scale is half a mile so this is saying 1.4 E32/E37 which is 1.4 fahrenhalfsteps per halfmile.* it is a very slight gradient, because of the weak gravity, by earth standards. so convection happens easily and there is a lot of wind action, and the convection shifts heat upwards and evens out the gradient. 



#80
Jan3005, 02:53 PM

Astronomy
Sci Advisor
PF Gold
P: 22,800

I think Quantum Gravity research is probably going to succeed and get a quantum theory of spacetime (and of its shape, which we call gravity)
and I expect physics will be rebuilt on this new spacetime (just like since 1680 it has been built mostly on newton's absolute flat spacetime and the 1905 minor variation of that) and today as usual I looked at a QG paper and as usual it was using nonmetric QG units where c and hbar are set equal one (or something else convenient) and G likewiseso it is some variant of the planck units (which we've had around since 1899 or so). As physics gets rebuilt on a quantum spacetime basis, I expect people will gradually get used to gauging things in natural units: scales intrinsic to quantum spacetime. So after reading I went out in the garden, bright with insects flying around and some stuff opening, a few leaves coming out to get the sun, and I thought: 5.7E117 spells life and it is a longterm number. It's the brightness of sunlight expressed in natural unitsas long as sunlight is this level, as long as there is a rich chemistry, as long as the rivers run... some crows were talking, a redtail hawk cried, some other birds were calling too well I am going for a walk up the hill with my wife, for exercise etc. be back later these longlasting numbers that spell earth, spell home, most are not too interesting E29 is global avg. surface temp E50 is a "gee" E91 is the density of water E106 is a "bar"standard air pressure but 5.7E117 is the brightness of sunlight, not too strong or too weak, that is driving all this life, the crows cawing, trees budding, wind stirring the branches, and all that, so if I happened to be riding around with some aliens in their saucer I think I would would ask to be put off at a planet with steady 5.7E117 sunlight. it is the number that more than any other characterizes home btw in natural units what I gave for gee and bar etc were only approximate the official gee is more like 0.9E50 and the normal density of water is more like 1.2E91 and the usual sealevel pressure norm is 1.4E106 but I dont care too much about a few percent. I think I could be adapt (as long as the chemistry was good) somewhere with gravity E50 and pressure E106. But I think I might get depressed, or sick of the place, if illumination wasnt right. 



#81
Jan3105, 11:25 AM

Astronomy
Sci Advisor
PF Gold
P: 22,800

Freshman physics on the round number planet.
I'm thinking of a planet with the same general chemistry as on earth, air has the same composition, soil and vegetation similar etc. but where some basic features like the surface gravity are round numbers. E29 is global avg. surface temp E50 is a "gee" E106 is a "bar"standard air pressure E33 is the planet's mass E50 is the year (Note that on earth the official gee is not exactly E50 but more like 0.9E50, the usual sealevel pressure norm more like 1.4E106, the planet mass 1.4E33, although estimates vary the average temperature is close to E29, and the year is 1.167E50 instead of exactly E50. So this round number planet is not an exact match to Terra but akin to it.) One thing we will not compromise about is the brightness of sunlight. It is 5.7E117 expressed in natural units. And as to some things like the proton mass and the density of water we have no choice: 1.2E91 is the density of water at standard temperature and pressure (E29 and E106). 2.6E18 protons make one mass unit, so if we need a figure for the mass of a water or air molecule we can say 18/(2.6E18) and (29/2.6E18). I'm thinking of a series of exercisesa kind of natural units Yogawhich is finding out about this round number planet: the speed of sound, the threshhold of convection in the air, the minimal orbit time, the planet's radius and density, the escape velocity, the rate pressure falls with altitude, the equilibrium temperature in direct sunlight. these might be good exercises to do for several reasons. 



#82
Jan3105, 03:24 PM

Astronomy
Sci Advisor
PF Gold
P: 22,800

summary of inputs
easy numbers: E29 is global avg. surface temp E50 is a "gee" E106 is a "bar"standard air pressure E33 is the planet's mass E50 is the year hard numbers: 5.7E117 is the brightness of sunlight. 1.2E91 is the density of water. 2.6E18 protons make one mass unit.  so to begin, let's calculate the planet radius. surface gravity = E50 = GM/R^{2} = (1/8pi)E33/R^{2} R^{2} = (1/8pi)E83 R = 6.308E40 = 6.3E40 rounded. to humanize this recall E37 is half a mile. the radius is 6308 halfmiles. 



#83
Jan3105, 04:18 PM

Astronomy
Sci Advisor
PF Gold
P: 22,800

summary of inputs
easy numbers: E29 is global avg. surface temp E50 is a "gee" E106 is a "bar"standard air pressure E33 is the planet's mass E50 is the year E4 is the planet's orbit speed around its sun hard numbers: 5.7E117 is the brightness of sunlight. 1.2E91 is the density of water. 2.6E18 protons make one mass unit.  Let's calculate the speed of sound at the planet's surface. the speed of sound formula for biatomic gas at temp T speed^{2} = (7/5) kT/(mass of molecule) In our case the average mass of the air molecule is 29/(2.6E18) and temp is E29 speed^{2} = (7/5) E29 (2.6E18)/29 = 1.255E12 speed = 1.12E6 just over a millionth of the speed of light BTW I added the planet's orbit speed to the list of easy numbers. It matches the earth's orbit speed almost exactly and will allow us to calculate the mass of the planet's sun. 



#84
Jan3105, 05:40 PM

Astronomy
Sci Advisor
PF Gold
P: 22,800

summary of inputs
easy numbers: E29 is global avg. surface temp E50 is a "gee" E106 is a "bar"standard air pressure E33 is the planet's mass E50 is the year E4 is the planet's orbit speed around its sun hard numbers: 5.7E117 is the brightness of sunlight. 1.2E91 is the density of water. 2.6E18 protons make one mass unit.  I want to use just these inputs and demonstrate how various things follow from them. mass of the sun, in natural units the mass M is related to the year period P and the orbit speed v by M/4 = P v^{3} = E50 (E4)v^{3} = E50 E12 = E38 M = 4E38 remember the planet mass is E33, so this is 400,000 times that. (rather like how earth and sun masses are related, you can always try comparing between what we get for the round number planet and the handbook data for Earth and it usually wont be far off. in this case jupiter is a bit less than 1/1000 of sun and earth is roughly 1/300 of jupiter and 400,000 wouldnt be a bad guess) 



#85
Jan3105, 07:16 PM

Astronomy
Sci Advisor
PF Gold
P: 22,800

Over in general astronomy there is Little Orbits thread
http://physicsforums.com/showthread....657#post447657 which among other things dealt with calculating the circular orbit period of two identical gold balls separated by 3 times the radius, center to center. A toy gravitating system consisting of two equal balls with a gap between them equal to their common radius. That should be a snap for us with these units. In interpreting the orbit time we are going to calculate remember that the time interval E45 is about 4.5 minutes. The density of gold is 24E91, which is to say 24 pounds per pint (24 E8/E99). I had to look it up in the CRC handbook. If T is the radian time (that is P/2pi) then T^{2} = 3^{3}/8E91 T= 5.8E45 = about 26 minutes Multiply by 2pi to get 164 minutes for the period. For those who like more symbols let S be the ratio of the separation to the radius, which in this case is 3 (sep is thrice radius) and then T^{2} = 3 S^{3}/density all we needed to do was put 24E91 in for the density. As another illustration, if the separation is 4 radii, so that the gap is equal to the diameter of one of the balls, then T^{2} = 3 4^{3}/24E91 = 4^{3}/8E91 = 8E91 



#86
Feb105, 03:28 PM

Astronomy
Sci Advisor
PF Gold
P: 22,800

just now out in the garden was impressed by the vivid colors
blue sky, green leaves, red camelia (we have this big treealmost, which is covered) and recalled that I knew the energies and frequencies and wavelengths of these colors green energy is E27 which means the frequency is also E27 (radians per natural unit of time) and the angular wavelength is E27 (natural length units) so it is so easy: you know this one number, for the energy of a green photon, and it works for the frequency and wavelength too. and the extemes of the visible tend to be about 25 percent higher and lower than green, which is right in the middle so red energy is 0.8E27, which makes its frequency the same 0.8E27 and its wavelength 1.25E27 so that is the camelias (we have white and pink ones too but I am just thinking of the very red ones) and the blue of the sky, its photon energy is 1.25E27 also with frequency 1.25E27, and of course the wavelength is shorter than green, namely the reciprocal number 0.8E27 So I am remembering last night, our chorus rehearsal, where I was in the bass section sitting right next to the sopranos and right at my elbow was Dacia or Dasha whose high range is wonderful and she was singing plenty of frequency E39 which is the D on the fourthline of the trebleclef and we of course match each other singing by whole numbers and precise fractions because that is what 4part harmony is about and maybe it doesnt matter but her D which is E39 natural is related to the green in sunlight and outside in the garden, which is E27 by merely a trillion, by merely a factor of E12 when the sopranos sing it is a transfusion of light or sometimes fire 



#87
Feb105, 04:43 PM

Astronomy
Sci Advisor
PF Gold
P: 22,800

round# planet inputs
easy numbers: E29 is global avg. surface temp E50 is a "gee" E106 is a "bar"standard air pressure E33 is the planet's mass E50 is the year E4 is the planet's orbit speed around its sun hard numbers: 5.7E117 is the brightness of sunlight. 1.2E91 is the density of water. 2.6E18 protons make one mass unit.  I want to use just these inputs and demonstrate how various things follow from them. Say that here is the round number planet and a can judge force on the soles of my feet. I go out in the garden and just stand there feeling the sunlight and the force of my weight on the ground, and watching what is going on. On each foot I feel a force of E40 natural. How many nucleons are in my body? 



#88
Feb105, 04:52 PM

Astronomy
Sci Advisor
PF Gold
P: 22,800

these natural units are based on the universal force constant which is the force that appears in the Einst. eqn. relating energy density in space to curvature of space
the force of my own weight is something I know. I have to cope with this force when I get out of bed in the morning, and it helps me compact the trash so I can get a bit more in the can on pickup day. this force is 2E40 of the natural constant force so how many baryons am I made of? how many protons and neutrons? well gee is E50 so my mass is 2E10 natural mass units (that is force/gee) and baryons are 2.6E18 to each mass unit it says in the list of 3 hard numbers. So I am made of 5.2E28 baryons. 



#89
Feb105, 06:14 PM

Astronomy
Sci Advisor
PF Gold
P: 22,800

round# planet inputs
easy numbers: E29 is global avg. surface temp E50 is a "gee" E106 is a "bar"standard air pressure E33 is the planet's mass E50 is the year E4 is the planet's orbit speed around its sun hard numbers: 5.7E117 is the brightness of sunlight. 1.2E91 is the density of water. 2.6E18 protons make one mass unit.  I want to use just these inputs and demonstrate how various things follow from them. right outside the front gate there are tall eucalyptus, and a lot of the time we hear the wind. So I often think of the lapse ratethe temperature gradient necessary for convection. there is a deadend barrier here and beyond it the ravine with the creek. for much of the year we hear the creek making a quiet roaring and that also is driven by convection, since it lifts water to make rain uphill from us. so these constant voices tell me that somewhere the gradient has become steeper than the threshhold for convection. If I am on the round number planet, what is that threshhold gradient? It is going to be some temperature drop per pace or per mile, in human terms, some number of degrees cooling off per measure of height. but energy per distance is force, so it is basically a certain force which we will measure in relation to the universal force constant. the formula is simply the WEIGHT of an average air molecule, divided by (7/2)k. but k = 1, so we just have to divide by 7/2. the average air molecule mass is 29/(2.6E18) in natural terms. And gee on this planet is E50 threshhold cooling rate for convection = (2/7)(29/2.6)E68 = 3.2E68 so this is the number which I want to hear in the wind, if I live on the round number planet. to humanize it, a halfmile is E37 length units, and a halfFahrenheit step is E32 of the natural unit. So a halfdegree per halfmile drop in temperature is E32/E37 = E69. so what the calculation showed is that on the round number planet, which is prettymuch like earth, a drop of 32E69, which is 32 of those temp steps per halfmile (16 Fahrenheit if you have to think F) is a kind of limit on the rate the air can cool with height. If it cools off faster then convection, and mixing, will set in. Interestingly, Richard advised against humanizing numbers too often. maybe he would say to keep the number that is implicit in the wind a simple 3.2E68. And implicit in the clouds of moisture lofted by the rising air. 



#90
Feb105, 11:17 PM

Astronomy
Sci Advisor
PF Gold
P: 22,800

Some of the eucalyptus are 4E35 tall. that is 40E34, or 40 paces, in more familiar terms 100 feet. They tower.
we are all familiar with the drop in pressure as you gain altitude. what is the scale of this. it is exponential, so there must be some distance D such that the pressure falls off with height h as exp(h/D). that is, if you want the pressure to be less by 5 percent you should ascend by 5 percent of D. we are on the round number planet, this pressure drop distance is part of getting acquainted with the planet. what is it? the average temperature in the air column we are looking at might as well be E29 since that is typical of the planet the weight of an airmolecule in this planet's 'gee' gravity E50 is gee times 29/(2.6E18), which is 11.15E68 the pressure drop scale D = temp/air molecule weight = E29/(11.15E68) = 9.0E37, do you remember E37 as a halfmile? the distance D is 9 of those (or 4.5 miles if you prefer miles to halfmiles a whole lot) so D = 9 halfmiles. that means if you want the pressure to decline by 1/9 of what it is now, you should go up the mountain 1 halfmile (one ninth of D). What physics facts about round number planet might you want to know that you think we might calculate from the given data? Suppose we go scuba diving on this planet. how deep do you go to get an increased pressure of one atmosphere? (it's the limit of a suction wellpump that we learned about in middleschool) 


Register to reply 
Related Discussions  
[SOLVED] Application of the equations of motion with constant acceleration  Introductory Physics Homework  3  
Using force equations and kinematic equations  Introductory Physics Homework  29  
Constant Force  Introductory Physics Homework  5  
Quick question about Rydberg Constant Equations  Introductory Physics Homework  5  
constant force?  Introductory Physics Homework  4 