Someone check my anomaly detection thinking?

[Foxster]

OK, what I'm trying to do is find the 'best' way to detect anomalies (monoliths, Easter eggs etc) on any body.

I need someone to check my logic (as I am mathematically challenged).

Here's what I have so far...

1. Use a RoveMate probe core. These have a limited Field of View (10°) but the detection rate is 100%, which means you only have to scan an area once to be guaranteed to detect anything there.

2. Calculate how high above the body's centre you need to be so that a 10° FoV exactly fills the KerbNet screen i.e  body-radius / tan(5°)    (radius of each body is available in the Wiki and 5° is half the 10° FoV) e.g. for Kerbin:

3. Use Alt-F12 cheat menu option 'Set Orbit' to set the Semi Major Axis to this, with an inclination of 90°.

4. Open KerbNet (right click RoveMate) and set FoV to the max of 10°, set KerbNet refresh to 3.5s, set warp to whatever is good. 

I have tried it for a few bodies and it works so far, with anomalies (?s) popping right up.

Here are some bodies and SMA (altitude is SMA minus radius of body):

Kerbin - 6858km (altitude above surface=6258km)

Eeloo - 2400km (alt=2190km)

Duna - 3657km (alt=3337km) (watch out for Ike!)

Does this all make sense?

(PS - This is actually also not a bad way of mapping and tagging biomes if you use the Cycle Display Mode button and then adjust the FoV to zoom in on KerbNet)

Edited April 23, 2017 by Foxster

[Numerlor]

I always take probe core that has detection above 15 and high FOV then spam that refresh button

Edited April 23, 2017 by Numerlor

[Foxster]

25 minutes ago, Numerlor said:

I always take prove core that has detection above 15 and high FOV then spam that refresh button

Yup, pretty much what I have done before. I wanted to try to figure out something better. 

[LordFerret]

An interesting exercise for sure.

And then there's SCANsat.

[StahnAileron]

Actually, shouldn't the radius be at a right angle to the edges of the FOV rather then the centerline? Unless the FoV can see through terrain to some extent, as your graphic shows, you would technically be blocked from seeing the maximum amount of the sphere. (The LoS pierces the circle; it should be tangential, no?)

I've done similar calculations for RemoteTech. (e.g. Figuring out the altitude I would need to be at around the poles to get line-of-sight to equatorial satellites at various altitudes. It was for drone planes; I once lost an unmanned cargo shuttle because I had it de-orbit at a pole without full planetary coverage.) Of course this all depends on the "penetrating" power of the signals.

Edited April 23, 2017 by StahnAileron

[Archgeek]

On 4/23/2017 at 1:21 PM, StahnAileron said:

Actually, shouldn't the radius be at a right angle to the edges of the FOV rather then the centerline? Unless the FoV can see through terrain to some extent, as your graphic shows, you would technically be blocked from seeing the maximum amount of the sphere. (The LoS pierces the circle; it should be tangential, no?)

I've done similar calculations for RemoteTech. (e.g. Figuring out the altitude I would need to be at around the poles to get line-of-sight to equatorial satellites at various altitudes. It was for drone planes; I once lost an unmanned cargo shuttle because I had it de-orbit at a pole without full planetary coverage.) Of course this all depends on the "penetrating" power of the signals.

Nope, the idea is that we want to see the whole hemisphere at once with our little 10° field o' view by varying our semi-major axis - or the length of that centreline.  We want that right angle right where it is, so we can use trig.  Also, the other angle will always be 85°, because triangles gotta add up to 180.

Don't rely on the graphic too much -- that half FoV angle's well over 5° ('looks around 12 to me), and the lines are very not to scale.  Also, you'll find the degree of penetration is amusingly proportional to the thickness of lines, as the line o' sight is kind of centred in the line describing the planet surface, instead of attop it.

[ExtremeSquared]

The term StahnAileron is trying to use is "tangent line" and he's correct. You can't see an entire hemisphere at once. Depending on how KSP models occlusion, that is.

The correct calculated SMA should be
SMA = 600 / sin(5 degrees)
At such low angles, making this trig mistake results in a ~0.3% error based on my quick calculations.

Edited April 24, 2017 by ExtremeSquared

[Foxster]

20 minutes ago, ExtremeSquared said:

At such low angles, making this trig mistake results in a ~0.3% error based on my quick calculations.

Oh, I think I can live with that . It's not like the KerbNet display is exactly the Hubble. 

[ExtremeSquared]

The fact that sin(theta) / tan(theta) approaches 1 as theta approaches zero ends up being somewhat horrifying in real life.
Tracking down huge discrepancies in calculations is usually way easier than a result being consistently off by 0.3%.

[StahnAileron]

36 minutes ago, ExtremeSquared said:

The fact that sin(theta) / tan(theta) approaches 1 as theta approaches zero ends up being somewhat horrifying in real life.
Tracking down huge discrepancies in calculations is usually way easier than a result being consistently off by 0.3%.

And astronomy/astrophysics is one case where you can come across that deviation often.

Thanks for clarifying my statement. I initially made the same mistake when I was doing RT calculations. I caught myself at the time because I was tutoring math at the time, some of it being basic trigonometry. I realized looking at JUST the FoV vs planetary radius would cause me to intersect through the planet (or exclude part of an orbit). With RT, it's more obvious because the highest FoV (at the time) was 45 degrees. Suffice to say my calculations needed total accuracy or else I risked having blackout zones at the most inopportune times. (Which is why I wound doing the basic calcs in the first place as my initial post mentions.) I floated the thought of making a radio tower at both poles. The numbers made it unfeasible.

[Archgeek]

Oh heck, you're quite right.  I just figured some crazed identity was being used to make tangent a shortcut of some sort, did the math with the given numbers in my head and figured it seemed about right (.3% is close enough to fool my estimation circuits, that for sure).  At least I was right about the 85°, in spite of it not mattering really.

[Callmedave]

i would like to add my suggestion too.

 

Starting off in a perfect 90° inclined orbit is good so that you cover the poles, but then you end up scanning areas you have already scanned.

 

after 'a few' orbits change your inclination +/- 10° and repeat after a few orbits.If you use scan sat it is easy to track when you have covered the areas of the planet near the poles.

 

[Foxster]

Yea, I've seen that issue with an exact 90° orbit a while ago when playing with the ScanSat mod. You just don't move on to the next "stripe" quickly. 

I have tended to pick an orbit a couple of degrees either side of 90 but I'm sure there is a clever way to calculate that best angle for a body.