Using the force constant in equations

[marcus]

Yuk, this is hard work. and because the project is eccentric if not bizarre it is mostly in social isolation (but thanks to those who have dropped in so far!)

I am checking to see about these natural units----like ordinary Planck but with |F|=1
instead of the more usual |G|=1

First, they do seem to work better than conventional Planck and this is confirmed by what I notice in Quantum Gravity articles. Increasingly I notice a kappa ("gravitational constant") which is 8piG. And which can be set to equal one to further simplify the equations. The Newtonian G is yielding a little---not yet a secondary constant but not as predominant.

the moment one sets
|F|= |c|=|hbar|=|k|=|e|=1
one has a fairly universal set of units and it is interesting to see what some familiar quantities come out to be.

In part I am just interested in rough sizes, in part I want to know basic constants like Hubble parameter, proton mass, cosmological constant, in these terms because oddly enough I've found it is sometimes actually convenient to work with data that way

rough sizes:

pound E8
year E50
handbreadth E33
pace (32 inch) E34
halfmile E37
lightyear E50
food Calorie E-5
lab calorie E-8
quartervolt E-28
green photon energy 10E-28
average Earth surface temp E-29
2/3 mph E-9
67 mph E-7
cold air speed of sound E-6
one "gee" acceleration E-50
weight of 50 kg sack of cement E-40
power of 160 watt bulb E-49

some constants (approx.):

reciprocal proton mass 2.6E18
electron mass 2.1E-22
Hubble time 1.6E60
Lambda 0.85 E-120
rho-Lambda 0.85 E-120
rho-crit (critical density) 1.16 E-120
more exact Earth year 1.1676 E50
more exact lightyear 1.1676 E50
avg Earth orbit speed E-4
earth mass 1.38 E33
earth radius 7.86 E40
sun mass 4.6 E38
solar surface temp 2.0E-28
CMB temperature 9.6E-32
earth surface pressure 1.4E-106
earth surface gravity 0.88E-50
fuel energy released by one O₂ 17E-28
density of water 1.225 E8/E99

the time scale is important enough to treat separately:
1/222 of a minute E42
4.5 minutes E45
(to have a named power-of-ten for a convenient time interval, imagine counting out loud rapidly, at the rate of 222 counts a minute, each count is E42 natural time units. A thousand counts is 4 and 1/2 minutes. It just happens that one year is roughly E8 counts, or E50 natural.)

Named powers of ten can help assimilate and remember quantities expressed in natural units. The way I remember Earth surface air pressure is to think of the weight of a sack of cement (E-40) on a sq. handwidth area (E66) which gives me an idea of the pressure E-106, and it is 1.4 of those.
I remember the density of water as 1.225 E8/E99, that is somewhat more than one "pound" per
pint-size cubic handbreadth volume, which simplifies to 1.225E-91.
Richard suggested that a distaste for extreme numbers be called "googlephobia"---isn't google
somebody's name for 10¹⁰⁰?
I think of the Earth radius as 7860 halfmiles (a halfmile being E37)
instead of 7.86E40. another way of coping with googlephobia, or of
bridging between humanscale and natural.

 

[marcus]

special numbers

some special numbers go with these units, most are pure math numbers and would be factors in the equation no matter what system of units, but here they sometimes jump out a little more clearly.

80/pi this tells the evap time of a BH. cube the mass and multiply by 80/pi

pi²/15 tells the per-volume radiant energy density at some temp. quart the temp (raise to fourth) and multiply by pi²/15

pi²/60 tells the brightness at some temp (power radiated per unit area). quart the temp (raise to fourth) and multiply by pi²/60

3zeta(4)/zeta(3) = 2.701 tells the average photon energy at some temp.
just multiply the temperature by 2.701. Since sun temp is 2E-28, the average sunlight photon has energy 5.402E-28---anyway that's the idea.

1 tells the bekenhawking temperature of a BH. just take 1 over the mass.

1/4pi tells the Schw. radius of a BH. just take that times the mass.

1/4pi tells the area of the BH. take that times the square of the mass.

3 tells the critical density of the universe. just multiply 3 by the square of the Hubble parameter

6 tells the density of a round planet. divide 6 by the square of the radian time in low orbit.

9 or thereabouts is the heat capacity of a molecule of water

29 is the molecular weight of air. It is handy to know.
(atomic and molecular weights generally are)

Oh, they tell us that the density of the universe is at or very close to the critical value. So 3 also tells the actual density of the universe.

1/137 (more exactly 1/137.036...) is the coulomb constant. it tells the force between two charges separated by a distance. just multiply the charges by 1/137 and divide by the square of the distance.

1/137 also tells the force between parallel currents (measured on a test segment with length equal half the separation). just multiply the currents by 1/137

(1/137)² tells the energy needed to ionize a hydrogen atom. multiply the rest energy of an electron (2.1E-22) by it and you get a quantity of energy called the Hartree----which is twice the ionization energy (so you still need to divide by two)

in each case i am assuming that the calculation is done in natural units terms, so that I don't have to specify the units each time I say something.

there's lots more but maybe this is enough for now

 

[marcus]

finally a way to remember the solar constant

Back in post #71 I listed some rough sizes, including these force and power benchmarks.

weight of 50 kg sack of cement E-40
power of a 160 watt lightbulb E-49

I am thinking of the force E-40 as a "sack" force benchmark
and imagine a 50 kg weight on a pulley descending at speed E-9
(which is 2/3 mph, or a billionth 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 E-49. To see that,
you just have to multiply the force E-40 by the speed E-9
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 sun---the power per unit area delivered by direct unattenuated sunlight----is 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 E-49. So a bulb of power spread over a square pace is
E-49/E68 = E-117
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 pace-wide square.

In natural units, 5.7E-117 is what the handbook value of the solar constant actually turns out to be. but I don't find that so easy to remember. So I visualize it as 5.7 bulbs per sq. pace.

A pace is just one of my steps----around 32 inches----so 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 5.7E-117
===================

In the "Force" system of natural units, the unit of power is of course E49 bulbs (because bulb was defined as E-49) 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 Bang-scale. 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 one----like with speed too.

 

[marcus]

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/T²
you just divide 6 by the square of the radian time, so the three densities are
6/(49E90) = 1.224E-91
6/(16E90) = 3.75E-91
and 6/(9E90) = 6.66E-91

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 earth-like 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)

 

[marcus]

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:

1.4E-117
1.9E-117
and 2.5E-117

the question is, what is the night-time 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/pi².
However, pi-square 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/pi²

8.511 E-117
11.55 E-117
15.20 E-117

and press the squareroot button twice to get the temps

0.960 E-29
1.037 E-29
1.110 E-29

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 E-29 and 1.000E-29 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 F-degrees, 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 18-some Fahrenheit-degrees. 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 E-117 in the infrared.

 

[marcus]

two more benchmarks
the temperature in the energyproducing core of the sun is 5E-25

(remember that solar surface temp is 2E-28
and avg Earth surface temp is E-29
the way to remember solar surface is that E-28 is an eQ
and green photons have energy 10eQ, so solar surface is going to be
around E-28, and the temp there happens to be 2E-28.
so from core out to surface, temp goes down by factor of 2500)

and middle D on piano is (1/2)E-39, have to go, back later
that means the D in the soprano/alto range, on the fourth line
of the treble clef, is E-39
the natural unit frequency is E39 times higher than that note
the sopranos in our chorus sing---maybe 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 Bagel-shaped troopships. I hope this will scandalize Kea.

 

[marcus]

------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 ten-million-pound troopship wishes to achieve a speed of E-3 (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 E-3 = E12 momentum units. This requires discharging a photon pulse with E12 energy units, which consumes 1000 cats.

 

[marcus]

---------FAST FOOD EXERCISE------
According to Defense Analyst Daniel Pinkwater, the Earth is in danger of invasion by the Fat Men: a space-faring 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 sports-jackets, 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.

 

[marcus]

Baez posted some facts about Titan on SPR today including at what altitude the atmosphere was coldest----in natural units terms it was
surface temp 3.33E-30
lowest temp 2.48E-30 (at altitude 6.2E38)

let's look at the indicated gradient 0.85E-30/6.2E38 =1.4E-69
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 be-----this can be compared with the measured gradient.

http://www.esa.int/SPECIALS/Cassini-Huygens/SEMMF2HHZTD_0.html

-------GRAVITY ON TITAN-----
Titan surface gravity is 0.12E-50.
(I write it that way because I think of acceleration E-50 as about one "gee")
-----TEMPERATURE GRADIENT ON TITAN-----
in natural units terms the lapse-rate (rate of change in temp with altitude) is just 2/7* x wt of molecule

the weight of a nitrogen molecule = 0.12E-50 x 28/(2.6E18) = 1.29E-68
and 2/7 of that is 0.37E-68 = 3.7E-69

So the theoretical lapse-rate (with no methane assumed) is 3.7E-69
but what was actually observed was 1.4E-69

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 E-50 natural acceleration units-----official Earth surface gravity is 0.88E-50.
So* titan is 0.12E-50

Wind is the result of convection which happens when the temperature gradient (cooling off with altitude) exceeds the threshold gradient called "lapse-rate".

The lapse rate in moist air is less than the dry-air lapse rate, because having moisture (able to condense to form clouds and give up its energy) serves as a "fuel" for convection---a 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 1.4E-69, think of E-32 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 E-32/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.

 

[marcus]

Numbers that mean home

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 likewise---so 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: 5.7E-117 spells life and it is a longterm number. It's the brightness of sunlight expressed in natural units----as 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

E-29 is global avg. surface temp
E-50 is a "gee"
E-91 is the density of water
E-106 is a "bar"----standard air pressure

but 5.7E-117 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 5.7E-117 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 0.9E-50
and the normal density of water is more like 1.2E-91
and the usual sealevel pressure norm is 1.4E-106
but I don't care too much about a few percent. I think I could be adapt
(as long as the chemistry was good) somewhere with
gravity E-50 and pressure E-106. But I think I might get depressed,
or sick of the place, if illumination wasnt right.

 

[marcus]

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.

E-29 is global avg. surface temp
E-50 is a "gee"
E-106 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 E-50 but more like 0.9E-50,
the usual sealevel pressure norm more like 1.4E-106,
the planet mass 1.4E33,
although estimates vary the average temperature is close to E-29,
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 5.7E-117 expressed in natural units. And as to some things like the proton mass and the density of water we have no choice:
1.2E-91 is the density of water at standard temperature and pressure (E-29 and E-106).
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 exercises---a kind of natural units Yoga---which is finding out about this round number planet: the speed of sound, the threshold 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.

 

[marcus]

summary of inputs
easy numbers:
E-29 is global avg. surface temp
E-50 is a "gee"
E-106 is a "bar"----standard air pressure
E33 is the planet's mass
E50 is the year

hard numbers:
5.7E-117 is the brightness of sunlight.
1.2E-91 is the density of water.
2.6E18 protons make one mass unit.
-------------------

so to begin, let's calculate the planet radius.

surface gravity = E-50 = GM/R² = (1/8pi)E33/R²

R² = (1/8pi)E83

R = 6.308E40 = 6.3E40 rounded.

to humanize this recall E37 is half a mile. the radius is 6308 halfmiles.

 

[marcus]

summary of inputs
easy numbers:
E-29 is global avg. surface temp
E-50 is a "gee"
E-106 is a "bar"----standard air pressure
E33 is the planet's mass
E50 is the year
E-4 is the planet's orbit speed around its sun

hard numbers:
5.7E-117 is the brightness of sunlight.
1.2E-91 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
speed² = (7/5) kT/(mass of molecule)

In our case the average mass of the air molecule is 29/(2.6E18) and temp is E-29

speed² = (7/5) E-29 (2.6E18)/29 = 1.255E-12
speed = 1.12E-6
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.

 

[marcus]

summary of inputs
easy numbers:
E-29 is global avg. surface temp
E-50 is a "gee"
E-106 is a "bar"----standard air pressure
E33 is the planet's mass
E50 is the year
E-4 is the planet's orbit speed around its sun

hard numbers:
5.7E-117 is the brightness of sunlight.
1.2E-91 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 v³ = E50 (E-4)v³ = E50 E-12 = 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 won't 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 wouldn't be a bad guess)

 

[marcus]

Over in general astronomy there is Little Orbits thread
https://www.physicsforums.com/showthread.php?p=447657#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 24E-91, 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
T² = 3³/8E-91
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
T² = 3 S³/density
all we needed to do was put 24E-91 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
T² = 3 4³/24E-91 = 4³/8E-91 = 8E91

 

[marcus]

just now out in the garden was impressed by the vivid colors
blue sky, green leaves, red camelia (we have this big tree-almost, which is covered)
and recalled that I knew the energies and frequencies and wavelengths of these colors
green energy is E-27
which means the frequency is also E-27 (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 0.8E-27, which makes its frequency the same 0.8E-27
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 1.25E-27 also with frequency 1.25E-27, 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 E-39
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 4-part harmony is about
and maybe it doesn't matter but her D which is E-39 natural
is related to the green in sunlight and outside in the garden, which is E-27

by merely a trillion, by merely a factor of E12

when the sopranos sing it is a transfusion of light
or sometimes fire

 

[marcus]

round# planet inputs
easy numbers:
E-29 is global avg. surface temp
E-50 is a "gee"
E-106 is a "bar"----standard air pressure
E33 is the planet's mass
E50 is the year
E-4 is the planet's orbit speed around its sun

hard numbers:
5.7E-117 is the brightness of sunlight.
1.2E-91 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 E-40 natural.

How many nucleons are in my body?

 

[marcus]

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
2E-40 of the natural constant force

so how many baryons am I made of? how many protons and neutrons?
well gee is E-50

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.

 

[marcus]

round# planet inputs
easy numbers:
E-29 is global avg. surface temp
E-50 is a "gee"
E-106 is a "bar"----standard air pressure
E33 is the planet's mass
E50 is the year
E-4 is the planet's orbit speed around its sun

hard numbers:
5.7E-117 is the brightness of sunlight.
1.2E-91 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 rate----the 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 threshold for convection.

If I am on the round number planet, what is that threshold 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 E-50

threshhold cooling rate for convection = (2/7)(29/2.6)E-68 = 3.2E-68

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 E-32 of the natural unit. So a halfdegree per halfmile drop in temperature is E-32/E37 = E-69. so what the calculation showed is that on the round number planet, which is prettymuch like earth, a drop of 32E-69, 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 3.2E-68. And implicit in the clouds of moisture lofted by the rising air.

 

[marcus]

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 E-29 since that is typical of the planet

the weight of an airmolecule in this planet's 'gee' gravity E-50 is
gee times 29/(2.6E18), which is 11.15E-68

the pressure drop scale D = temp/air molecule weight = E-29/(11.15E-68) = 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 well-pump that we learned about in middleschool)

 

[marcus]

...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 well-pump that we learned about in middleschool)

the relevant arithmetic fact 1/8 is 12.5 percent, you know: 8 and 0.125 are reciprocals

and the density of water, in natural units, is 1.225E-91
or to put it more humanly, 1.225 pound/pint.

In the round number ocean, pressure rises by one atmoshere for every 8 paces you go down.

why? on this planet one standard atmosphere is E-106
and gravity is E-50.
I want a depth D such that
D x E-50 x 1.225E-91 = E-106
D x E-50 = 8E-16
D = 8E34 = 8 paces

a eucalyptus across the street is 40 paces tall (I paced it off as that yesterday---to a point where its elevation was half a rightangle) and 40 = 5 x 8, so today I pictured being in clear water up to the top of the eucalyptus, which I guess was some kind of giant seaweed they have there. I was tankdiving at a depth of 40E34 and the pressure was 5 atmospheres. Amazing conditions! Water so clear I could see the top of the eucalyptus.

 

[marcus]

A couple of posts back I mentioned that the force I exert on the ground is 2E-40.
Just went for a walk up the hill behind campus. Still open undeveloped land? it is fairly steep and you can do a 200 pace change of altitude (500 feet if you like feet). along some steep firetrails. you go up 200E34 = 2E36.

I am curious about my "wattage" on this climb. In natural units there is a natural unit of power which is pretty huge and E-49 of that is like a 160 watt lightbulb. So E-49 power can be visualized, and there is a nice bit of serendipity that E45 time units is 4.5 minutes.

It just happens that I can do this climb of 2E36 in a time of 4E45 (four of those 4.5 minute periods). So I calculate the rate I do lifting work as
the force of my weight x height/time

2E-40 x 2E36 /4E45 = E-49

not surprising, knew I could probably crank 160 watts on a stationary bicycle.have to go, back later

 

[marcus]

inputs for the round number planet
easy numbers:
E-29 is global avg. surface temp
E-50 is a "gee"
E-106 is a "bar"----standard air pressure
E33 is the planet's mass
E50 is the year
E-4 is the planet's orbit speed around its sun

hard numbers:
5.7E-117 is the brightness of sunlight.
1.2E-91 is the density of water.
2.6E18 protons make one mass unit.
-------------------

well I've been using myself as a guineapig and it seems to me that this is a nicely proportioned set of natural units. one can calculate all sorts of human and earthly things with considerable ease, or so it seems to me, once one gets the hang of it.

I would like to know something. Suppose someone reading this thread wants to test-drive these units and see how they work for them. What sorts of things would you need to know to get started?

And what would it occur to someone to calculate?

So if anything occurs to you as you read, constructive suggestions of that sort would be welcome.
here are a few more convenient humanscale handles on the units, which I mentioned earlier and have been using
------
time E45 = 4.5 minutes
length E34 = pace (32 inches, 81 cm)
length E37 = halfmile
force E-40 = 480 Newton, like usual weight of 50 kg.
energy E-5 = roughly one food Calorie
power E-49 = 160 watt bulb
mass E8 = about one pound
voltage E-28 = one quarter volt
angular frequency E-39 = D on treble staff.

 

[marcus]

I tried the classic problem of the airplane flying over the N geomagnetic pole, which asks what is the voltage difference between the wingtips.

Maybe I made a mistake. i got that the voltage difference (for a 100 foot wingspan plane) was very small, like 0.6 of a conventional volt.

worked in natural units it went this way
Speed E-6
Wingspan 4E35 (this is 40 paces, about 100 ft)
Vertical component of geomagn. field 6E-58

multiplying these together gives 24E-29 for the potential difference betw wingtips of the aircraft.

this is only a couple of quartervolt (it comes to 2.4E-28, so 2.4 which is
like 0.6 conventional volts.) I don't know or can't remember if that is about right.

the natural unit of charge used is the electron charge
the form of the Lorentz force equation adopted is
F = q( E + beta X B)
this means that the natural units of electric and magnetic fields are the same unit---can be thought of as voltage/distance or force/charge

we should have an exercise about lightning.
he're link with some background
http://www.weatherwise.org/qr/qry.lightningpower.html
http://www.space.com/scienceastronomy/lightning_backgrounder.html

 

[marcus]

time E45 = 4.5 minutes
length E34 = pace (32 inches, 81 cm)
length E37 = halfmile
force E-40 = 480 Newton, like usual weight of 50 kg.
energy E-5 = roughly one food Calorie
power E-49 = 160 watt bulb
mass E8 = about one pound
voltage E-28 = one quarter volt
angular frequency E-39 = D on treble staff
magnetic field E-57 = gauss
magnetic field E-53 = tesla

another unexpected coincidence that makes natural units potentially easier to use is the fact that a tesla is very close to E-53 (force units per electron charge)
here is how close, in case you are curious.
1 Tesla = 0.9974 E-53 natural = 1.00E-53
1 gauss = 0.9974 E-57 natural = 1.00E-57
if one rounds to two decimal accuracy the relevant factor is just one!

I haven't been bothering to show precision in this thread since we rarely if ever need it, but it is always available
natural energy unit = 3.9018E8 joule
natural charge = electron charge = 1.602176E-19 coulomb
1 conventional volt = 4.1062E-28 natural voltage units
1 meter = 1.2342E34 natural length units.
Enough digits! the upshot is that a magnetic field that registers as 1 Tesla on a metric gauge will read 1.00E-53 on the natural scale.

And my handbook's value of 0.58 gauss for the Earth magnetic field at the north pole converts directly to 0.58E-57 natural.

 

[marcus]

we should have an exercise about lightning.
he're link with some background
http://www.weatherwise.org/qr/qry.lightningpower.html
http://www.space.com/scienceastronomy/lightning_backgrounder.html

Every minute we have 6000 flashes of lightning (worldwide).
this is because the Fat Men of Ornish are angry with us for not providing enough sour cream with the potato pancakes or enough other junk food in which they delight.

Each bolt of lightning releases some 1 to 25 natural units of energy, according to background info.

The captain of an Ornish battle cruiser wishes to hurl a lightning bolt at Trenton New Jersey to express his dissatisfaction with their offering of Tartar sauce with the fried scallops. Scallops require enough Tartar sauce.

He sets the voltage to 1.4E-20 and the current to 1.4E-19.
What duration of flash should he choose if he wishes exactly 20 units of energy to be delivered upon the helpless city?

answer. the pulse of current should last E40 time units
E40 x 1.4E-20 x 1.4E-19 = 20

-------------
How many lightning bolts occur in E45 time units?

recall that E45 = 4.5 conventional minutes. Multiply by 6000.
27,000 bolts
-------------
If the average energy per flash is 10 natural units, the Ornish ships must be expending energy on the Earth at the rate of 270,000 units per E45 interval.
The Ornish ships are powered by Cat Engine which converts a standard 10 pound cat (E9 mass units) entirely to energy in accordance with the usual emceesquare proportion. Over what length of time is one Catsworth of energy expended on lightning?

answer: a catsworth is E9, a billion units. divide E9 by 270,000 to learn the number of E45 intervals required to consume one cat.
3700 of these intervals. multiply by 4.5 if you want to know minutes.

 

[marcus]

inputs for the round number planet, and other benchmarks
easy numbers:
E-29 is global avg. surface temp
E-50 is a "gee"
E-106 is a "bar"----standard air pressure
E33 is the planet's mass
E50 is the year
E-4 is the planet's orbit speed around its sun

hard numbers:
5.7E-117 is the brightness of sunlight.
1.2E-91 is the density of water.
2.6E18 protons make one mass unit.
-------------------

convenient handles on the units:
time E45 = 4.5 minutes
length E34 = pace (32 inches, 81 cm)
length E37 = halfmile
force E-40 = 480 Newton, like usual weight of 50 kg.
energy E-5 = roughly one food Calorie
power E-49 = 160 watt bulb
mass E8 = about one pound
voltage E-28 = one quarter volt
angular frequency E-39 = D on treble staff
magnetic field E-57 = gauss (earth's field is about half a gauss)

----------
what about cyclotron frequency of the proton, in a field of some strength B?

say it's the geomagnetic field somewhere on Earth where it's between 1/2 and 1/3 gauss. plenty of places like that! (only gets strong like 0.6 gauss near the poles)

let's say it is (1/2.6) gauss

now a proton will spiral around in that field at an angular frequency (radians per unit time) called "cyclotron frequency". the stronger the field the higher the frequency and this is a nice weak field so the frequency should be low. maybe even audible!

cyclotron frequency = qB/m where q=1, and m= 1/(2.6E18) and B = (1/2.6)E-57

cyclotron freq. = (1/2.6)E-57 x 2.6E18 = E-39

that is the D on the treble staff (one line from the top) a soprano note.

definitely audible, maybe some radio noise frequencies come from spiraling particles.

 

[marcus]

measuring a 1 Tesla field with a stirrup gauge

classic gauge is like a stirrup, you lower it down into the magn. field B and run a current in the crossbar of the stirrup, and measure the force it gets pulled.

the crossbar of the stirrup can have several parallel conductors, it can be a sector of a coil in other words, but for simplicity just think of one conductor with a lot of amps

E-24 natural current unit is 0.6 amps, so let's say we put E-23 current thru the stirrup (6 amps). And suppose the crossbar is handbreath long, or E33

and the field is one Tesla, which is E-53 natural

then what is the pull?

E-23 x E33 x E-53 = E-43

that is half a Newton. So if you put E-23 current across a tesla field, a conductor which is E33 long will experience half Newton force.

 

[marcus]

another unexpected thing. it turns out that at standard conditions of temperature and pressure (E-29 and E-106) the mass of air is
about 1 pound per cubic pace. I calculate it's 1.115 E8 mass units.

the number of molecules in (E34)³ volume

PV/T = n
E-106 x E102/E-29 = E25

the mass of E25 molecules, each 29/(2.6E18).
E25 x 29/(2.6E18) = 1.115E8, to humanize it, call it 1.1 "pound"

--------------
It was a crisp Fall day, the temperature was E-29.
The hills of rural Vermont had turned bright colors.
A dog and a goat wished to take a ride in a balloon, so they went
to the goat's barn and got out his hot-air balloon.

the mass of gear and passengers is 400 pounds----that is 4E10 natural.

the goat asked the dog, who was a physicist,
how much they would have to heat the air to get lift-off.

What is the volume of the balloon? said the dog.
8000 cubic paces, replied the goat, naturally that is 8E105.

Well, said the dog, who enjoyed off-the-cuff order-of-magnitude calculation, the mass of air at ambient conditions is about 8000 pounds.
We have to heat it by about 5 percent, to lighten it by 5 percent, which is the weight of us and our equipage.

Excellent said the goat, as he opened the propane valve and pressed the igniter. We will raise the temperature in the bag from 1.00E-29 to 1.05E-29. Being a Vermonter, I call that 25 Fahrenheit degrees----shouldn't take too long!

 

[marcus]

Lo Fat was a pirate in the South China Sea who practiced piracy in an environmentally sustainable manner. He and his men ate organic vegan food and their ship was powered by oars.

Lo Fat had a crew of blond well-muscled young Republican captives to row the vessel. He motivated them by giving lectures on dismantling social security and the graduated income tax. The pirate vessel cut swiftly through the waves, searching for merchant prey.

After many years of successful piracy, Lo Fat noticed that the air temperature was more often than not 1.10E-29---which is human body temperature and eventually makes a man long for nice chilly air.

So he decided to equip the captain's cabin on his ship with air conditioning.
You may recall picturing a power of E-49 natural as a 160 watt lightbulb, in which case you know what it means for the pirate deciding on a 5E-49 airconditioner model for his cabin.

Now because of his committment to sustainable piracy, and thorough-going rejection of fossil fuels, the AC unit had to be solar powered.

And because of 10 percent efficiency, the PV panel had to get 5E-48 of sunlight!

Now the problem is how big a panel does the comfort-seeking pirate require?

 

[marcus]

remember that the solar constant (sunlight power per unit area) is
5.7E-117
in the South China Sea let us suppose that 5E-117 is available during the hours of peak demand. comparing this to 5E-48 we see that the area must be E69. My goodness that is ten square paces (a square pace area is (E34)²)

 

[marcus]

... at standard conditions of temperature and pressure (E-29 and E-106) the mass of air is
about 1 pound per cubic pace. I calculate it's 1.115 E8 mass units.

the number of molecules in (E34)³ volume

PV/T = n
E-106 x E102/E-29 = E25

the mass of E25 molecules, each 29/(2.6E18).
E25 x 29/(2.6E18) = 1.115E8, to humanize it, call it 1.1 "pound"

--------------
It was a crisp Fall day, the temperature was E-29.
The hills of rural Vermont had turned bright colors.
A dog and a goat wished to take a ride in a balloon, so they went
to the goat's barn and got out his hot-air balloon.

the mass of gear and passengers is 400 pounds----that is 4E10 natural.

the goat asked the dog, who was a physicist,
how much they would have to heat the air to get lift-off.

What is the volume of the balloon? said the dog.
8000 cubic paces, replied the goat, naturally that is 8E105.

Well, said the dog, who enjoyed off-the-cuff order-of-magnitude calculation, the mass of air at ambient conditions is about 8000 pounds.
We have to heat it by about 5 percent, to lighten it by 5 percent, which is the weight of us and our equipage.

Excellent said the goat, as he opened the propane valve and pressed the igniter. We will raise the temperature in the bag from 1.00E-29 to 1.05E-29. Being a Vermonter, I call that 25 Fahrenheit degrees----shouldn't take too long!

an obvious followup concerns how much propane they are going to burn in the initial heating, to raise the temperature in the 8000 cubic pace bag by 0.05E-29---or 5E-31
we just calculated that the number of molecules in a cubic pace is E25
so we are talking about some 8000E25---or 8E28---molecules.

In natural units, heat capacities are generally easy because k=1, so for most metals it is around 3k per atom, for water 9k per molecule,...I will drop the k since it is one... for monatomic gasses at constant pressure 5/2
for biatomic gasses 7/2, per molecule always.

so the energy to heat the air in that balloon is about
(7/2) x 8E28 x 5E-31 = 140E-3 = 0.14 energy unit.

we just multiplied the delta-tee by the number of molecules by the heat capacity per molecule. In cases like this there is nothing to look up.

How much propane does the goat require, to get lift-off?

 

[marcus]

In answer to this inquiry, the dog recited a poem:

"whether it's to burn or eat
the Oh-Two count will tell the heat
on the average each Oh-Two
releases 17 eekyoo."

Now a propane is C3H8, and it uses 5 Oh-Twooz when it burns,
so it releases 85 eekyoo which is 85E-28 natural energy unit.
Moreover a propane is 36+8 = 44 baryons, and therefore a pound (mass E8) is 2.6E18 x E8 baryons. Accordingly the dog calculated that a pound is (2.6E26)/44 propanes and must therefore supply 85E-28 x 2.6E26 x (1/44) = 0.05 energy units.

To get off the ground, said the dog, we will need 0.14 unit of heat.
Behold! said he, we will burn 3 pounds of propane for lift-off!

That is fine, said the goat, who had experience in these matters: after that we will not need so much, because we will mostly just be keeping the air in the bag warm. The propane tank will be ample for our trip.

 

[marcus]

The captain of a million-pound scout vessel of the Ornish fleet is searching for planets rich in junk food for his men to plunder. As the ship comes out of warp, he discovers that it is bearing directly down on Atlantic City New Jersey at a speed of E-4.
With no time to spare he must order a photon pulse to cancel the ships momentum.

How many standard 10 pound cats will be consumed?

----
answer: a pound is E8 so the ship mass is E14 and the speed (30 km/second) is E-4, so the momentum to be canceled is E10 natural momentum units.
The light pulse with this momentum, directed at Atlantic City so as to avoid collision, delivers E10 natural energy units.

This will of course vaporize the famed vacation spot and some of the surrounded ocean. Of interest to the captain, however, is how many cats need to be converted to supply the energy.

the mass of a 10-pound cat is E9 and therefore, when converted in accordance to the emcee-square rule, the cat will yield E9 units of energy. Therefore 10 cats are needed from the ships fuel reserves to accomplish this maneuver.

 

[marcus]

It might be of interest to judge the effects of the scout ship's maneuver on the environs of Atlantic City. The boardwalk and casino Mecca is located on a narrow sandbar with water on both sides. For simplicity let us assume that the pulse of light misses the ciity and is entirely absorbed by the adjacent ocean.

the handbook figure for the latent heat of vaporization of water (2260 joules per gram) translates into one natural unit of energy vaporizing 399 pounds of water, let us say 400 pounds for round numbers. If we allow for some of the energy to go into preliminary heating then we can estimate that one unit is sufficient to vaporize 300 plus pounds.

We may estimate that ten billion units from the Ornish ship, on being absorbed, then vaporizes 3E12 pounds of water. It is clear why junk food pirates are generally looked on with disfavor.