Jump drive loses input power

Sure, I'm happy to explain.

Take the capacity and the rate and divide them: 3MWh / 32MW. Since a megawatt (MW) is inherently "per second" (1 watt is defined as 1 joule per second), a megawatt-hour (MWh) is how many megawatts can be provided in an hour. Since the dividend is smaller than the divisor, we know that the resulting number is going to be less than 1. Take 3/32, you get 0.09375. This is the proportion of an hour that 32 MW would take to charge 3 MWh of storage. Multiply that by 60 to get minutes, and you get 5.625 minutes. Multiply 0.625 by 60 to get the seconds, and you have 5 minutes, 37.5 seconds.

Yes, the jump drive states that it will take 7 minutes to charge. This I don't dispute. The problem is that the jump drive loses input power along the charge path and thus the required power is that much greater than the published storage capacity.

The only possible way that a jump drive can take 7 minutes to charge, with a capacity of 3 MWh and an input of 32 MW, is that a portion of the charge is being lost, just like batteries. If we assume that it's the same as a battery and that it loses 20% of its input power, we can take the inverse of 20% and multiply the time by it to check the math.

1/(1-0.2) => 1/0.8 = 1.25

Multiply 5.625 (the figure obtained earlier by dividing 3/32) by 1.25: 5.625 * 1.25 = 7.03125

Multiply 0.03125 by 60 to get seconds, and you get 7 minutes, 1.875 seconds. Almost exactly what's shown in game, and plenty close due to rounding.

You can very easily prove this, even!

• Create a new game in Space Engineers in a survival world (not creative, because batteries have infinite charge in creative).
• Press Alt-F10 and check Enable Creative Mode Tools
• Place down a jump drive on the ground.
• Confirm via terminal that it starts with 0Wh Stored Power
• Place a large grid battery near it (not attached) for verification purposes. Confirm that the battery holds 900 kWh of power via terminal. You can remove this battery afterward as it's just to verify that the battery holds 900 kWh.
• Place a battery on the jump drive so they're on the same grid.
• Observe that the jump drive's lights change from red to yellow as it begins charging.
• Open the terminal of that grid and watch Stored Power increase.

If the jump drive really does lose 20% power, then the battery should run out of power when the Jump Drive is charged to around 790 kWh of power. In reality, this test ran out of power with around 725 kWh of power in the jump drive, but that's very close to 20% loss.

You can also do this with uranium and a reactor! If you place down a jump drive and a Large Reactor, try putting 32 kg of Uranium into the reactor. Since 1 kg of uranium is 1 MWh of power, that much should charge a jump drive fully in 7 minutes! It won't though.

My reason for submitting this bug is that the amount of charge shown does not accurately reflect the amount of charge or fuel needed to be input to actually charge the jump drive. Yes, time to charge is correct, but unless you have infinite fuel then the actual amount of fuel consumed is 25% more than the power stored, due to the 20% loss.

Hello, Merii,

wow... I have no words :) Asking for simple formula, getting full math/physics lecture in return.

I tried it the way you described and do the math along with you. All is right the way you have written it.

Put this issue into our internal system.

Kind Regards

Keen Software House: QA Department

Thanks! I wasn't sure what formula you meant so I gave you everything I had on it, I figured too much info was better than not enough.