Jak Nazryth
Mongoose
This came up in another thread and I think the Mongoose guys can clarify.
When installing higher tech drives and power plants, you can reduce the required size but increase the cost.
High Guard MgT2, page 48 and 49
Since cost of a drive is based on tonnage, is the final price based on the default size or the new modified size?
Example
A 200 ton ship wants a Jump 2 and Thrust 2 and is TL 14 design.
You want to take take 2 advantages for each dive by reducing the size.
You can reduce each size by 20% but you pay and additional 25% more for the drives.
Jump Drive (2) = 5% of the hull size, so a Jump 2 Drive takes 15 tons. (10+5 per the rules page 15)
Maneuver Drives thrust 2 = 2% of the hull size, so a thrust 2 M-Drive takes 4 tons.
By taking size reduction twice (page 48) your J-Drive now takes up 12 tons
By taking size reduction twice (page 49) your M-Drive now takes up 3.2 tons
Question. In both cases this reduction is a 25% cost increase, but since the price of a drive is based on tonnage, do you calculate the price of the drives on the orignal size BEFORE the size reduction or AFTER the size reduction?
Here is the difference
J-Drive based on original size = 37.5 MCr (with 25% price increase) J-Drive based on modified size = 30 MCr ( with 25% price increase)
You can do the math for the M-Drives.
this will also affect power plants or anything else that the price is based on tonnage, when you can reduce size based on advantages.
So which is correct, base price on original tonnage or modified tonnage?
Thanks
When installing higher tech drives and power plants, you can reduce the required size but increase the cost.
High Guard MgT2, page 48 and 49
Since cost of a drive is based on tonnage, is the final price based on the default size or the new modified size?
Example
A 200 ton ship wants a Jump 2 and Thrust 2 and is TL 14 design.
You want to take take 2 advantages for each dive by reducing the size.
You can reduce each size by 20% but you pay and additional 25% more for the drives.
Jump Drive (2) = 5% of the hull size, so a Jump 2 Drive takes 15 tons. (10+5 per the rules page 15)
Maneuver Drives thrust 2 = 2% of the hull size, so a thrust 2 M-Drive takes 4 tons.
By taking size reduction twice (page 48) your J-Drive now takes up 12 tons
By taking size reduction twice (page 49) your M-Drive now takes up 3.2 tons
Question. In both cases this reduction is a 25% cost increase, but since the price of a drive is based on tonnage, do you calculate the price of the drives on the orignal size BEFORE the size reduction or AFTER the size reduction?
Here is the difference
J-Drive based on original size = 37.5 MCr (with 25% price increase) J-Drive based on modified size = 30 MCr ( with 25% price increase)
You can do the math for the M-Drives.
this will also affect power plants or anything else that the price is based on tonnage, when you can reduce size based on advantages.
So which is correct, base price on original tonnage or modified tonnage?
Thanks