Updated Vehicle Handbook in the works

Geir- Last minute thought on the Vehicle Handbook update. Have you scoured all of the rest of the printed material from all Mongoose printed books for any vehicle rules that weren't in the previous VH? The same way ship construction rules are scattered in a dozen different books. TL-16 stuff in Behind the Claw. Sub-bridges in Element Cruisers. etc. All of this should be in High Guard. Just like all of the components for vehicles should be in the VH.

Edit: This would actually be a good use of the community if anyone knows where such things are to help Geir out in locating them all.
 
Do heavy gravity worlds actually exist without attracting an atmosphere? Actual question. Sorry.
According to the rules, there can be size 5 (.45 G) worlds with no atmosphere, and size 6 (.7 G) with trace atmospheres which is the same thing for this kind of thing. So, no heavy gravity but either of these would be more than enough to make the ship leave a nice crater. Even so, a thrust 1 ship is fine.
 
There isn't as much out there on vehicles. JTAS 16 has some, most of which (not just my own) are being implemented. But if there's something useful out there, by all means let me know.

But, you know, there may be things out there that should just be smothered in their crib. Not that I have an example in mind, but not all ideas are great.
 
Just less than terminal velocity is still not a "soft" landing. As you slow down using a combination of air resistance and your M-drive, the slower you go, the less air resistance that you have and the more M-drive you need to overcome it. At terminal velocity, it would be in balance, but you are still speeding towards the ground. Someone would have to do the math to know for sure, but I do not see how your statement makes any sense. If I do an orbital jump out of a ship to a 0.75G planet, I am still dead when I hit the ground from massive blunt force trauma. What is the terminal velocity for 0.75G and an Earth-like atmospheric density? Use the same brick in both scenarios.
It makes sense if you know basic physics. You don't hit the ground at terminal velocity, you decelerate with your 0.25g engine.

You can even calculate at what altitude you need to switch it on.

Watch Space X land a booster... it falls at terminal velocity with a couple of small fins maintaining stability. At the last moment it turns the engine back on and soft lands. With a 0.25g engine you would have to switch it on at a greater altitude. For a typical UK house brick the altitude is ~310m.
 
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It makes sense if you know basic physics. You don't hit the ground at terminal velocity, you decelerate with your 0.25g engine.

You can even calculate at what altitude you need to switch it on.

Watch Space X land a booster... it falls at terminal velocity with a couple of small fins maintaining stability. At the last moment it turns the engine back on and soft lands. With a 0.25g engine you woud have to switch it on at a greater altitude.
That does not even remotely make sense 'basic' physics or no. The engines on the Falcon & Super-heavy are both capable of lifting the entire mass of the booster at far more than 1G acceleration. Neither does 'terminal velocity' help much on worlds with thinner-than-standard atmosphere; impacting the ground at 310 m/s is gonna make a crater. 'Turning on the engine at the last moment' works for Falcon because it can exert enough thrust in the last few seconds to completely negate that 310 m/s PLUS the ~10 m/s^2 of acceleration due to gravity; an engine which grants only 2.5 m/s^2 is NOT sufficient. Turning such an engine on earlier 'to burn longer' also has exactly no effect; increase in speed is reduced to only 7.5 m/s^2 -- but anything already at terminal velocity is (due to drag) not gaining speed anyway.
 
The reason it wouldn't work is that when you switch the engine on and slow the brick down it loses air resistance, with less air resistance its terminal velocity actually goes up ;) well sort of.

So you would slow down, then start to speed back up again, the question is how much do you slow down before you speed up...

It isn't actually a simple problem at all because you have two variables changing continuously.

When in free fall the brick is at terminal velocity so the weight force and the air resistance forces balance.

Switching on the 0.25g engine now causes a resultant force... but as this resultant force slows the brick the air resistance is reduced.

A new terminal velocity will be achieved when the new air resistance + 0.25g thrust are equal to the weight.

The initial terminal velocity of the brick is ~40m/s, how fast will in now be falling towards the ground...
 
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The reason it wouldn't work is that when you switch the engine on and slow the brick down it loses air resistance, with less air resistance its terminal velocity actually goes up ;)

So you would slow down, then start to speed back up again, the question is how much do you slow down before you speed up...
No, it wouldn't. It would find equilibrium at a velocity that can be calculated using 0.75G instead of 1.0G
It isn't actually a simple problem at all because you have two variables changing continuously.
The object's Terminal Velocity never changes unless the acceleration due to gravity, the object's cross section changes, or the object's mass changes. An object's terminal velocity is determined by this formula.

Terminal velocity = √(2mg/ρACd)

Terminal velocity depends on the mass, cross-sectional area, and drag coefficient of the object, as well as the density of the fluid through which the object is falling and gravitational acceleration
 
No, it wouldn't. It would find equilibrium at a velocity that can be calculated using 0.75G instead of 1.0G
It's not that simple. it would decelerate until the new equilibrium is established.
The object's Terminal Velocity never changes unless the acceleration due to gravity, the object's cross section changes, or the object's mass changes. An object's terminal velocity is determined by this formula.
You missed out air resistance...
drag coefficient and fluid density or air resistance for the sake of simplicity.
 
It's not that simple. it would decelerate until the new equilibrium is established.
If you are using your M-Drive the whole time, it is a fixed number, so no.
You missed out air resistance...
No, I didn't. We are comparing things within the same atmosphere. If you are landing on a planet, you can't change the air's density. You have to deal with that as a constant.
 
Perhaps I should explain this using simpler terms.

Having an M/0 Drive just changes the acceleration due to gravity from 9.8m/s/s on Earth to 7.35m/s/s on Earth. Since 1G is Earth Gravity.

2 Identical Objects
Cube Shape
Coefficient of Drag 1.05
Mass of 200 metric tons
Cross Section of 125m2
Density of Fluid 1.204kg/m3

1G
Acceleration due to Gravity 9.81m/s/s
Terminal Velocity 567.3kph

0.75G
Acceleration due to Gravity 7.3575m/s/s
Terminal Velocity 491.3kph

In both instances they make craters and everyone is dead.
 
If you are using your M-Drive the whole time, it is a fixed number, so no.

No, I didn't. We are comparing things within the same atmosphere. If you are landing on a planet, you can't change the air's density. You have to deal with that as a constant.
Air resistance varies with speed, that's what causes terminal velocity...

It's all moot anyway, m-drives include lifters now.
 
Perhaps I should explain this using simpler terms.

Having an M/0 Drive just changes the acceleration due to gravity from 9.8m/s/s on Earth to 7.35m/s/s on Earth. Since 1G is Earth Gravity.

2 Identical Objects
Cube Shape
Coefficient of Drag 1.05
Mass of 200 metric tons
Cross Section of 125m2
Density of Fluid 1.204kg/m3

1G
Acceleration due to Gravity 9.81m/s/s
Terminal Velocity 567.3kph

0.75G
Acceleration due to Gravity 7.3575m/s/s
Terminal Velocity 491.3kph

In both instances they make craters and everyone is dead.
So an 1-G drive only slows a craft (landing on Terra) down via air-resistance; when air resistance is negligible, so is the rate of slowing. On a Thin / Very Thin / Vacuum world with a 1-G gravity field, both of the ships above will have incredibly higher impact velocities.

Including 'Contra Gravity Lifters' as a hand wave in all M-drives is execrable. It really should not be difficult to describe different TL capabilities of the different 'Gravitic' devices. As it is TL-30 'Contra Grav' is exactly the same as TL-9 'Contra Grav'; and it is all hidden away from the folks who want to design a vehicle or ship.
 
So an 1-G drive only slows a craft (landing on Terra) down via air-resistance; when air resistance is negligible, so is the rate of slowing. On a Thin / Very Thin / Vacuum world with a 1-G gravity field, both of the ships above will have incredibly higher impact velocities.

Including 'Contra Gravity Lifters' as a hand wave in all M-drives is execrable. It really should not be difficult to describe different TL capabilities of the different 'Gravitic' devices. As it is TL-30 'Contra Grav' is exactly the same as TL-9 'Contra Grav'; and it is all hidden away from the folks who want to design a vehicle or ship.
I figure having a M-1 Drive is nearly pointless for ships designed to land on a planet, but fine for ships that never land on a planetary surface. Anything less than M-2 and the number of planets that you can land on becomes a lot more limited. I dislike lifters because local gravity doesn't matter. Lifters negate local gravity. Period. Doesn't matter if the local gravity is 0.1G or 10G, your lifters work the same. That makes no sense. If that were actually true, then I can make "lifters" that will give Me 10Gs of thrust on a 1G world.
 
I don't see why a ship designed to land on planets, even with a 1G drive, wouldn't have contra grav lifters in it as part of the drive machinery.
 
I figure having a M-1 Drive is nearly pointless for ships designed to land on a planet, but fine for ships that never land on a planetary surface. Anything less than M-2 and the number of planets that you can land on becomes a lot more limited. I dislike lifters because local gravity doesn't matter. Lifters negate local gravity. Period. Doesn't matter if the local gravity is 0.1G or 10G, your lifters work the same. That makes no sense. If that were actually true, then I can make "lifters" that will give Me 10Gs of thrust on a 1G world.
Personally I see 'lifters' starting out as:
1} limited to a small percentage of the shielded mass,
2} up to a maximum effect of x kgs of mass negated,
3} requiring high power per negated kg.
All of these improve as the TL of the lifter rises.

For extra-crunch give some lifters an inverse of third (or higher) -power-of-distance relationship with the local gravity field, or a 'ground effect' where they are more effective -- so we can build 'hover bikes' without having supersonic stratosphere racers.

Early lifters evenly cover the entire lower surface of the (spherical) shielded region; higher TL allows the lifter to generate non-spherical regions, to be evenly distributed over a smaller percentage of the bottom surface, to be placed at the center of mass instead of evenly distributed, or be located at any arbitrary point inside the shielded region with no need of emitters on the bottom surface.

Inertial dampers independently improve by TL in similar ways; maximum capacity, maximum off-normal angle, etc.

Going this route, a Man-1 Trader could possibly visit a 1-G world -- but the performance of the lifters might limit what can be safely delivered or lifted to orbit.
 
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