Plotting the Solar System's planets' distances logarithmically

[DaveC426913]

TL;DR Summary

distances of planets, converting AU to inches and plotting logarithmically

I'm designing a growth chart for my granddaughter with the planets on it. I wanted them to be proportional* as far as distance goes - with the sun at the floor and Pluto at, say, six feet, but obviously that's impossible since all the inner planets would be in the first inch or two from the floor.

So the next best thing is to do it logarithmically. Unfortunately, I never studied logs.

Instead of putting the sun at 0", I'll put it at 19" (her height at birth)
and I'll set 40AU at, say, 72"** from the floor.
So that's a span of 53" with an offset of 19".

Now I want to spread the planets logarithmically from zero to 53.

These are the distances in AUs, rounded to about 2 decimals.

Body	Distance (AU)	Height (Inches)
S	0	0
M	.4
V	.7
E	1
M	1.5
AB	3
J	5.2
S	9.6
U	19
N	30
P	40	53

Not sure how to calculate logs for these. I can probably do some of it in Excel, but I'm not sure how to set the fractional log so that 40AU=53"* Is it overkill? She's too young, of course, but ideally she'll be using this into her tweens. How old were you before you realized that the planets are not all close-by and evenly spaced? It's the kind of entrenched illusion that results in long lists of "Today I Learned". IMO, there's no harm in starting children off with accurate images.

** How likely is it she'll exceed six feet before she gives up on her childhood growth chart? Pretty unlikely. But one's reach should exceed one's grasp.

 

[fresh_42]

Here is another nice link to get an impression:
"If the Moon were only a pixel"
made by a (former?) member of us.

 

[Ibix]

I think you just calculate $x=\log r$ for each orbit radius $r$, call the height $h$, write $h=mx+c$, plug in a couple of radii that you want to correspond to specific heights (Mercury = height today, and your Pluto figure, perhaps), and solve simultaneously for $m$ and $c$. Then generate your heights.

 

[fresh_42]

I get a base $b=\sqrt[53]{40} \approx 1.0720808709628515077933844785535...$ and calc.exe in Windows can compute $\log_b 40 = 53$.

 

[DaveC426913]

Here is another nice link to get an impression:
"If the Moon were only a pixel"
made by a (former?) member of us.

Yeah. I'm not doing the planets to-scale, just their orbits.

(Someday, when I have a plot of land more than 120 feet long I'll make a model to scale. I wish I had done it on my nerd friend's 14 acre farm.)

 

[Keith_McClary]

How likely is it she'll exceed six feet before she gives up on her childhood growth chart?

Or that they'll find another planet.

 

[DaveC426913]

calc.exe in Windows can compute $\log_b 40 = 53$.

I don't know how to do that with Calulator.

 

[Ibix]

I don't know how to do that with Calulator.

More problematically, $\log 1=0$ and $\log 0.4<0$, in any base. That's why I suggested my approach.

 

[DaveC426913]

These are the numbers I got, but I don't know if I did it right.

Body	Distance (AU)	log d	Height (Inches) ( 53*log d/1.60 )
S	0	n/a	?
M	.4	-.398	-13
V	.7	-.155	-0.25
E	1	0	0
M	1.5	.176	5.4
AB	3	.477	16
J	5.2	.716	24
S	9.6	.982	32.5
U	19	1.28	42
N	30	1.48	49
P	40	1.60	53

They look OK inasmuch as they sort of make a flat curve, as they should.

But Earth (1AU) is at the zero mark, not the Sun. The log of a fraction is a negative number. I guess I'm not doing it right.

(I also guess this may be a futile effort. I might as well just add the planets where ever, just per aesthetics. Although it would have been nice to be able to declare it as an accurate diagram.)

 

[DaveC426913]

I think you just calculate $x=\log r$ for each orbit radius $r$, call the height $h$, write $h=mx+c$, plug in a couple of radii that you want to correspond to specific heights (Mercury = height today, and your Pluto figure, perhaps), and solve simultaneously for $m$ and $c$. Then generate your heights.

Yeah, I see what you're saying, but I'm seein' it through a fog of 4 decades since math.

 

[Janus]

If I take the natural log of the orbital radii (as measured in millions of km), I get
Mercury = 4.05
Venus = 4.60
Earth = 5
Mars = 5.4
Jupiter = 6.66
Saturn = 7.27
Uranus = 7.96
Neptune = 8.4
Pluto = 8.6

If I plug those values into a spreadsheet and plot them on a log scale, I get:

With the y-axis in millions of Km

If we set Pluto at Six ft (72 in) then the log scale should be
Mercury = 37.6 in.
Venus = 38.5 in.
Earth = 41.9 in
Mars = 45.2 in
Juptiter = 55.8 in
Saturn = 60.9 in
Uranus = 66.6 in
Neptune = 70.3 in
Pluto = 72 in
Which roughly lines up with the numbers along the right side of the chart denoting ft above the floor. (I manually added these to the image, so they are not perfectly spaced.)

 

[DaveC426913]

Which roughly lines up with the numbers along the right side of the chart denoting ft above the floor. (I manually added these to the image, so they are not perfectly spaced.)

Heh. That was how I was thinking about solving it: brute-forcing it by plotting it on a chart, then measuring it in pixels. :)

(It'll be even more linear when I add the AB/Ceres.)

Thanks!

 

[mfb]

A logarithmic scale cannot include the Sun. 0 is infinitely far away. While you could take the radius of the Sun it would still be very far away from the rest.
Let's take Janus' logarithms: They range from 4.05 to 8.6. You want 8.6 to correspond to 72 inch. If you put Mercury at 19 inch then you get a simple linear relation. A range of 4.55 is now 53 inch, that's almost 10 inch per 1 different in the logarithm. Simply multiplying by 10 would put Pluto at 86 inch, however, so we need to subtract 14. That means distances are now ln(d)-14 where d is in million km and the distance is in inch.
Pluto -> 72 by construction.
Mercury -> 27.5
Surface of the Sun (d=0.6) -> 19 below the floor

 

[DaveC426913]

A logarithmic scale cannot include the Sun. 0 is infinitely far away.

Ah. Right. Like Antarctica on a flat map. If it were logarithmic and accurate, you should literally be able to see a smeared-out polar bear or penguin in the bottom millimetre of the map.

 

[Ibix]

Ah. Right. Like Antarctica on a flat map. If it were logarithmic and accurate, you should literally be able to see a smeared-out polar bear or penguin in the bottom millimetre of the map.

But the Sun has a finite radius. Its surface can be included, and you just cut the chart off at some point below that.

 

[pasmith]

A logarithmic scale cannot include the Sun. 0 is infinitely far away.

One possibility, which avoids this problem, is to set $H(r) = \frac{H_0}{\log 2}\log\left(1 + \frac{r}{r_0}\right)$ where $r_0$ is the distance of Pluto from the Sun and $H_0$ is the height where you want Pluto to appear.

But you can use $H(r) = H_0 f(r/r_0)$ where $f$ is any strictly increasing function with $f(0) = 0$, $f(1) = 1$, and $f'$ is strictly decreasing (so that the inner planets are placed further apart and the outer planets closer together than would be the case for a linear scaling).

 

[DaveC426913]

Dang. This really is going to amount to nothing but a labour of love.

Especially considering I hope to render the planets to-scale size-wise (though not the the same scale as the orbital radius) - any semblance of the effect will be lost.

I think maybe I might be better to just hint qualitatively at the geometric scale of orbital radii, and not try to be accurate.

Hm. Even that's not going to work so well...

 

[Keith_McClary]

I hope to render the planets to-scale size-wise

I made them scaled by the log of radius in a Java applet (now deprecated).