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Sorry if this was answered before, but is FAR acceptable to use or is everybody using the stock aero physics?

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Ok. I've got a really chonky ssto called the "Maiden", but it needs an extended runway, as it is so heavy. Is it ok, if I take off on the fields around the KSC? Thanks!

Also, I'm going for the heavy rain division- 333 tons

Know what? Scew that. It can't even take off at 140 m/s with an extended runway

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17 hours ago, έķ νίĻĻάίή said:

Ok. I've got a really chonky ssto called the "Maiden", but it needs an extended runway, as it is so heavy. Is it ok, if I take off on the fields around the KSC?

I'm more concerned about the "SSTO" part, since Rule #7 states that you must stay below 35 km ASL at all times. You should be okay as long as you stay within the atmosphere. 

 

Also, just how long does it take for your plane to reach takeoff speed? Personally, I would suggest:

  1. Applying the brakes
  2. Starting engines at full throttle.
  3. Waiting a bit (I don't know how long for you) for the engines to warm up.
    1. Jet engines take a while to reach maximum thrust output.
  4. Releasing the brakes for a faster takeoff speed.

 

18 hours ago, έķ νίĻĻάίή said:

Also, I'm going for the heavy rain division- 333 tons

333 tons? My Poseidon SSTO weighs only 184. If you can take off, fly around Kerbin, and land in one piece in less than 80 minutes, then I say go for it.

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4 minutes ago, Raptor kerman said:

For the record, the Kerbin circumnavigation record by plane is ~37 minutes.

Tell @nelso that.

 

  • And this isn't even his best time.

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22 minutes ago, Mars-Bound Hokie said:

Tell @nelso that.

 

  • And this isn't even his best time.

Realised my mistake, asked the mods to remove the post. The post was based on the old kerbin (air breathing) circumnavigation challenge where the record was at around 37 minutes. Bear in mind, I saw this thread and didn't read the posts. 

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Okay, I'm mounting a serious attempt at this challenge. I have spent an incredible amount of time making 1600 m.s^-1 ULR aircraft. But I haven't done circumnavigations in a LONG TIME.

Question: what is the distance to travel the full circumference at the equator? 

 

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jwMQP8u.png

 

Which means that, in order to pass the challenge, you have to go around the planet in less than 80 minutes:

EbFYe5M.png

 

Now, that does not mean that you're fine if you cruise at 786 m/s. This average time also includes the time it takes to take off, get up to speed, and land.

  • And yes, your total entry time includes takeoff and landing (up to a complete stop)

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8 minutes ago, Mars-Bound Hokie said:

jwMQP8u.png

 

Which means that, in order to pass the challenge, you have to go around the planet in less than 80 minutes:

EbFYe5M.png

 

Now, that does not mean that you're fine if you cruise at 786 m/s. This average time also includes the time it takes to take off, get up to speed, and land.

  • And yes, your total entry time includes takeoff and landing (up to a complete stop)

cheers, just fleshed out basic operational requirements, and I'm looking at a a cruise speed of 2100 m.s^-1  among other things. While I have previous high-alt, high-speed experience, this is completely new for me. I'm wondering if I've bitten off more than I can chew.....

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8 minutes ago, Raptor kerman said:

Question: what is the distance to travel the full circumference at the equator? 

It depends from that runway you take. If you take off from 09 (bearing 090 - as AFAIK it's the intended bearing on the challenge rules) and keep going, at 1400ms (my previous entry), KSP says you travelled about 5.860 Km.

I understand that this is due Kerbin spinning under you once you reach some altitude: the more time you spend flying, more time Kerbin have to spin under you, "pushing" the destination a bit further.

My previous entry took ~50 Min, or ~3000 secs, to circumnavigate Kerbin. Well, If I travelled 5.860Km in 3000 secs, I had an average speed of 1,89333333 Km/Sec. What's impossible, because my top speed was 1,4 Km per sec! :D 

As in the previous post, Kerbin has ~3.770 kms on the Equator. Since the day on Kerbin is 6 hours, Kerbin "runs" 3.770 Km each 6 hours, or ~628.33 Km each hour, or ~174m/s.  Problem is.... At 174m/s, on 3.000 secs the distance covered is about 522 Kms, 25% of the difference between what KSP says I traveled and the circumference of Kerbin: 5860 - 3770 = 2090.

Furthermore, Forum says that the Kerbin's atmosphere rotates together the ground, so that 174m/s should not influence your travelling.

But so, why KSP says I travelled 5.860 Km? :P If someone has an explanation, I'm hearing! :)

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I'm not counting this as an entry ATM, but to show that FAR does not affect the heating of the plane's parts at high speeds. It just alters how the game looks at the plane as a whole, not as individual parts.

Anyhow, this is what I did. I did lose the nose antenna (does help with the aero in FAR) so it's disqualified as is.

vcO8RCh.png
JWgtn6x.png
wEjckpy.png

Not too shabby for a single Whiplash powered Jet plane. :)

Edited by GDJ

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15 hours ago, Mars-Bound Hokie said:

I'm more concerned about the "SSTO" part, since Rule #7 states that you must stay below 35 km ASL at all times. You should be okay as long as you stay within the atmosphere. 

The battle plan - Since the Maiden class SSTOs carry a lot of oxidiser, I’m just going to use the rapiers in closed cycle, meaning that I’ll have MOAR POWER under 35km where they usually lose power due to lack of air

Edited by έķ νίĻĻάίή

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On 1/21/2020 at 1:28 AM, GDJ said:

I'm not counting this as an entry ATM, but to show that FAR does not affect the heating of the plane's parts at high speeds. It just alters how the game looks at the plane as a whole, not as individual parts.

I'm toying with FAR now and then, and I like the new challenges it imposes - as the CoL changing on supersonic speeds. I agree that it makes aircraft designs more realistic - some stunts just doesn't pays anymore (or even works at all) with FAR.

However... I think that FAR should be applied on specific Challenges just for it.

-- POST EDIT --

About the rule of the craft being capable of landing.... I ask that the rule doesn't imposes that the landing must be done on the same run. By the time I get my high score, I end up without fuel to the journey home!

Edited by Lisias
post edit

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On 1/20/2020 at 11:24 PM, Lisias said:

It depends from that runway you take. If you take off from 09 (bearing 090 - as AFAIK it's the intended bearing on the challenge rules) and keep going, at 1400ms (my previous entry), KSP says you travelled about 5.860 Km.

I understand that this is due Kerbin spinning under you once you reach some altitude: the more time you spend flying, more time Kerbin have to spin under you, "pushing" the destination a bit further.

My previous entry took ~50 Min, or ~3000 secs, to circumnavigate Kerbin. Well, If I travelled 5.860Km in 3000 secs, I had an average speed of 1,89333333 Km/Sec. What's impossible, because my top speed was 1,4 Km per sec! :D 

As in the previous post, Kerbin has ~3.770 kms on the Equator. Since the day on Kerbin is 6 hours, Kerbin "runs" 3.770 Km each 6 hours, or ~628.33 Km each hour, or ~174m/s.  Problem is.... At 174m/s, on 3.000 secs the distance covered is about 522 Kms, 25% of the difference between what KSP says I traveled and the circumference of Kerbin: 5860 - 3770 = 2090.

Furthermore, Forum says that the Kerbin's atmosphere rotates together the ground, so that 174m/s should not influence your travelling.

But so, why KSP says I travelled 5.860 Km? :P If someone has an explanation, I'm hearing! :)

remember your not at sea level. Adding your sea level altitude component to the radius of kerbin and calculating circumference from there might be of slight help.

Edited by Raptor kerman

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45 minutes ago, Raptor kerman said:

remember your not at sea level. Adding your sea level altitude component to the radius of kerbin and calculating circumference from there might be of slight help.

But then this is not "Ground Distance Covered"!! It's "linear distance" or whatever, the ground below me doesn't "stretches" as I fly higher!

Anyway, let's check: C = 2πr ; r = C/2π;

C(Kerbin) = ~3.770Km; r(Kerbin) = ~600Km

C(Me) = 5860Km; r(Me) = ~932.648 Km

But I had flown at ~21Km high (average), or r(Kerbin)+21; or r(me) = 621Km.

So...

C(me) = 2π * 620 = ~3901.946Km , being this number that "linear distance" thingy I mentioned. It's merely a 131Km more than travelling on the ground, not 2090!

Edited by Lisias
Tyop! Surprised?

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