Just posting this because I couldn't find it when I searched and needed to know because my CV wouldn't take off and rather than spam thrusters I needed a method. If you need to calculate the downward thrust in Newtons that's required order to take off vertically (i.e. number and/or strength of thrusters) you use the following formula: Planetary G * 9.81 * Ship weight in Kg. E.g. (1.03 G planet * 9.81) * 4.03kt => 10.1043 * 4,030,000 => 40.7MN Thrust required That's an absolute minimum for any vertical speed and it's insanely slow. To calculate your maximum weight limit on a planet you twiddle the numbers around: If you have 80MN Thrust available on a 1.03 G planet, your total mass can be 7.917 kt (80MN / 10.1043)

I've had this formula on a spreadsheet now for ages. It's always been quite helpful - but especially now that volume / mass is a thing.

Even easier is to approximate g by 10, which has the added benefit of building in a little extra thrust.

Very useful post @pachangas , I was just thinking about sharing similar info, and yay, perfect spot to do so As @geostar1024 said, factor of 10 is readily remembered. *No offense intended to anyone in the following bits. Posting simply in case it proves useful For SV/HV which use kN (kiloNewton) & t (metric ton, 1,000 kg) =>> 10 kN per 1 t For CVs, MN (million Newtons) & kt (1000 tons or 1,000,000 kg) =>> 10 MN per 1 kt { 10,000 kN (10 MN) per 1,000 t (1 kt) } *** Worth knowing that for a nominal 1 g planet in Empyrion where re-entry means ~40 m/s and it's only ~900 meters to the ground you'll need an additional 1 m/s^2 thrust to decelerate from 40 m/s to zero m/s in 800 meters. If you want to be safe and go from 40 to 0 in 400 meters (mountains) then you'll need 2, or a Total of ~12 m/s^2 For me the advantage of thinking this way is the ease of quickly estimating how much Mass a given ship can lift. For CVs it's handy that Large Steel Blocks mass 1 t each, so a stack = 1 kt. Lrg Hardened Steel runs 2 t per blk so 2 kt per stack. So if I'm looking at a CV in the workshop that masses 3 kt and has 200 MN lift I can quickly mumble, " so 30 MN for the ship leaves 170 MN or 17 kt for salvage... but if I'm doing space salvage I should use 12 so.. 120 gives 10 kt and then 72 for 6 and another half so 16 and a half minus 3 so about 14 kt of space salvage if I don't want to go splat returning to the planet." Last bit is rough estimating for higher gee planets. Since 1 g ~ 9.8 m/s^2, which we've been rounding to 10 m/s^2, can just continue using a 'times ten'. So a planet with 1.38 g needs roughly 14 m/s^2 thrust to just lift off. So for a SV/HV instead of 10 kN per 1 t, use 14 kN per 1 t And adding another 1 or 2 m/s^2 to the 14 m/s^2 for reentry will result in the same 'stopping distances'.

Oh boy this is highschool physics. I (everyone) should remember... Someone please make it sticky....Thank you.

Thank you! Guess one has played too much E:GS when one builds 'just according to feeling' and usualy ends up having around 30-40m/s2 thrust I dont care about numbers (doing math) when building a ship, but I do know the values I want it to have when it's done, the rest 'goes natural'.... This said, I recomend 30-40m/s2 thrust (bottomside, to lift off) at the very least - to carry cargo as much/heavy as the ship itself and still be able to lift off a 1.5-2g planet. If you want comfort and 'secured' cargo in a 1g unverse that is. And if I may advertise my 'Creative Studio', which offers different planets, with gravity values 0.7 | 1.5 | 2 | 2.5 | 3 | 4 | 5 to build and test your vehicles. In that scenario, lava is for hardcore creators, as it offers 5g - and an atmospheric density of 5kg/m3 ('Akua' has about 1kg).

I encountered a planet that had gravity and/or atmo density so high the only way off was to build a vessel with only a single thruster pushing against gravity. Math is always handy.

Yeah I've been using your studio to test out various things. Very useful. For fun I took my latest CV project to the lava planet and as expected it sank like a rock. Just running some basic caculations: 82.2 kt on 5g requires 4031.91 thrust to lift and maintain height. Not sure how many people play on 5g planets, but I sort of chalked it up as a trade-off. It's a bit difficult to create something very stylized and have it function in absolutely every environment. Also, an interesting thing about atmospheric density is that on the default yaml, it states the range is 0.9 - 1.2. To be on the safe side I usually test in 2.0. This setting doesn't seem to effect thrust so much as power draw - w/o surplus power, 2.0+ is going to yellow / redline your output.

Another practical example of my 'goal'... Lets say I have a ship of 100kt weight, which holds 50 CC with each providing 80k SU. I want this ship to leave 3g. (3 * 10) * (50 * 80'000 + 100'000) = 123'000'000 30 * 4'100'000 = 1230 kn // 1.230 Mn So, to achieve this, I need 1.3 Mega Newton?

Is there a specific max weight a cargo container can hold? In my estimation, the volume is separated from the mass - two very different things. For my own simple spreadsheet calculation, I take the base weight of the ship in kilotons, run if through the forumla which incorporates planet gravity, and keep increasing it to find the exact MN needed for the weight of the ship w/o cargo and with a full cargo. Sometimes its easier to go back and add additional thrusters to compensate, sometimes not. The max cargo weight is basically an approximation because different stuff takes up different volume, which may require additional storage units, which changes the total weight, etc,. 1.3 MN sounds off though (unless my eyes are failing me and I'm tired), because even at a base weight of 100kt, the minimum thrust needed to just maintain position in atmosphere at 3g would be 2943MN. Of course this is just vertical thrust - you could always tack on rear thrust too but that's kind of 'cheating'.

Well I was going by: * 1 Liter Water = 1 Kg * 8000 Storage Units = 8000 Liter = 8000 Kg / 8T But yeah, 1.3Mn seemed rather low, hence my question. Cant do proper math with the wrong numbers. 'Good' to know that it doesnt work that way

For cargo extensions/containers, the max SU for a cargo 'grid' is 320k SU, which comes out to 39 cargo extensions. I don't believe we have anything in the game that is 1kg and 1SU so this is why you have to approximate using their current system. So that 320k SU can have vastly different weights depending on what exactly you're storing. Sometimes to ballpark I just toss a bunch of combat steel stacks in, because eventually what you'll come up to is a MN that just gets exponentially more difficult to achieve (depending on build of course). For reference, 32ok SU will hold about 5334 large combat steel blocks, which comes to 21.3kt. Putting XL CV thrusters in (the largest volume item in the game IIRC) allows 160, and that comes out to 69.3kt. Obviously a huge difference in weight. So just a basic estimate then: 2x 320k SU that will hold 21.3kt each + 100kt dry weight of ship needs 4196.718 MN thrust to operate in 3g. Sorry if this isn't terribly helpful, but the volume system makes it difficult to actually predict how much weight something will end up because everything takes up different volume in cargo. Best thing to do would be plan for a set amount of total cargo you'd like to hold for a given max gravity, then work backwards from there, leaving some wiggle room (3-5kt).

The best way to test this is to just fill you containers with the heaviest thing you can find then check the stats page for the total mass. The stats page changes the ships mass based on cargo now. That gives you your maximum mass which you can then apply to the calculation.

I went back and checked the 9.0 patch notes and it states liters was changed to SU, but even that doesn't quite make sense given that liter is for volume and kg is for mass, and the density of the liter would need to be known to do a proper conversion. So even if they're going with a 1:1 ratio of 1 liter = 1kg (which is wrong), you won't be able to exactly fill that out in cargo extensions. If you're going for 2 theoretically 'maxed out' bays, I'd probably shoot for between 30-45kt each, bringing the total weight up to 190kt (using 45kt at the upper limit), which means you'll need 5591 MN to operate in 3g. So quite a bit of thrust. In standard gravity that means you could haul 364kt of stuff, which is kind of a crazy amount. Sorry, this was just bugging me!

I just use wolfram-alpha for this sort of stuff. Example: https://www.wolframalpha.com/input/?i=force+m=1t,+a=10m/s^2

Figured the CV Project has aleady 8000Mn while beeing 60'ish kt (instead of the previous named 100kt, but then again, it's not done yet either). But I'm still worried about the Cargo weight, as I plan/need to add another (at least) 8 containers of preferable 200-300k SU or more. -> That's another 2.5 mil SU to the exiting (rounded up) 4mil SU. Regarding the topic, the Question is: How can we integrate SU into a forumla to figure the amount of thrust we need (eg: mass)?

I use 5kg/L for iron ore, and 10kg/L as an average for components containing denser materials like sathium and zascosium. In your case, 6.5 ML (MSU) would be 32 ktons of iron ore, or 64 ktons of advanced components. Use 20 kg/L if you really want to be absolutely certain you can lift a full load of extremely dense material (though I don't think any in-game materials actually have a density of 20 kg/L at this time).