The Moose In Your House

But It doesn't have an antenna, but it might be the science module.

That's strange, I have a Mun base and it connects to the space center without any relays. Even when the base had no probes/crew inside.

Here's a segment of the 'KSP Plague' with a crew capacity of 30 and enough fuel and monoprop for most small missions. The plan is to have 3 of these hooked up to a central fuel core. Finally.. Got imgur to work. Seems legit,

Also if you are having trouble with tsiolkovsky rocket equation read this. (Information from Wikipeda) Newton's second law of motion relates external forces (Fi{\displaystyle F_{i}\,}) to the change in linear momentum of the whole system (including rocket and exhaust) as follows: ∑Fi=limΔt→0P2−P1Δt{\displaystyle \sum F_{i}=\lim _{\Delta t\to 0}{\frac {P_{2}-P_{1}}{\Delta t}}} where P1{\displaystyle P_{1}\,} is the momentum of the rocket at time t=0: P1=(m+Δm)V{\displaystyle P_{1}=\left({m+\Delta m}\right)V} and P2{\displaystyle P_{2}\,} is the momentum of the rocket and exhausted mass at time t=Δt{\displaystyle t=\Delta t\,}: P2=m(V+ΔV)+ΔmVe{\displaystyle P_{2}=m\left(V+\Delta V\right)+\Delta mV_{e}} and where, with respect to the observer: V{\displaystyle V\,} is the velocity of the rocket at time t=0 V+ΔV{\displaystyle V+\Delta V\,} is the velocity of the rocket at time t=Δt{\displaystyle t=\Delta t\,} Ve{\displaystyle V_{e}\,} is the velocity of the mass added to the exhaust (and lost by the rocket) during time Δt{\displaystyle \Delta t\,} m+Δm{\displaystyle m+\Delta m\,} is the mass of the rocket at time t=0 m{\displaystyle m\,} is the mass of the rocket at time t=Δt{\displaystyle t=\Delta t\,} The velocity of the exhaust Ve{\displaystyle V_{e}} in the observer frame is related to the velocity of the exhaust in the rocket frame ve{\displaystyle v_{e}} by (since exhaust velocity is in the negative direction) Ve=V−ve{\displaystyle V_{e}=V-v_{e}} Solving yields: P2−P1=mΔV−veΔm{\displaystyle P_{2}-P_{1}=m\Delta V-v_{e}\Delta m\,} and, using dm=−Δm{\displaystyle dm=-\Delta m}, since ejecting a positive Δm{\displaystyle \Delta m} results in a decrease in mass, ∑Fi=mdVdt+vedmdt{\displaystyle \sum F_{i}=m{\frac {dV}{dt}}+v_{e}{\frac {dm}{dt}}} If there are no external forces then ∑Fi=0{\displaystyle \sum F_{i}=0} (conservation of linear momentum) and mdVdt=−vedmdt{\displaystyle m{\frac {dV}{dt}}=-v_{e}{\frac {dm}{dt}}} Assuming ve{\displaystyle v_{e}\,} is constant, this may be integrated to yield: ΔV =veln⁡m0m1{\displaystyle \Delta V\ =v_{e}\ln {\frac {m_{0}}{m_{1}}}} or equivalently m1=m0e−ΔV /ve{\displaystyle m_{1}=m_{0}e^{-\Delta V\ /v_{e}}} or m0=m1eΔV /ve{\displaystyle m_{0}=m_{1}e^{\Delta V\ /v_{e}}} or m0−m1=m1(eΔV /ve−1){\displaystyle m_{0}-m_{1}=m_{1}(e^{\Delta V\ /v_{e}}-1)} where m0{\displaystyle m_{0}} is the initial total mass including propellant, m1{\displaystyle m_{1}} the final total mass, and ve{\displaystyle v_{e}} the velocity of the rocket exhaust with respect to the rocket (the specific impulse, or, if measured in time, that multiplied by gravity-on-Earth acceleration). The value m0−m1{\displaystyle m_{0}-m_{1}} is the total mass of propellant expended, and hence: Mf=1−m1m0=1−e−ΔV /ve{\displaystyle M_{f}=1-{\frac {m_{1}}{m_{0}}}=1-e^{-\Delta V\ /v_{\text{e}}}} where Mf{\displaystyle M_{f}} is the propellant mass fraction (the part of the initial total mass that is spent as working mass). ΔV {\displaystyle \Delta V\ } (delta v) is the integration over time of the magnitude of the acceleration produced by using the rocket engine (what would be the actual acceleration if external forces were absent). In free space, for the case of acceleration in the direction of the velocity, this is the increase of the speed. In the case of an acceleration in opposite direction (deceleration) it is the decrease of the speed. Of course gravity and drag also accelerate the vehicle, and they can add or subtract to the change in velocity experienced by the vehicle. Hence delta-v is not usually the actual change in speed or velocity of the vehicle. If special relativity is taken into account, the following equation can be derived for a relativistic rocket,[3] with Δv{\displaystyle \Delta v} again standing for the rocket's final velocity (after burning off all its fuel and being reduced to a rest mass of m1{\displaystyle m_{1}}) in the inertial frame of reference where the rocket started at rest (with the rest mass including fuel being m0{\displaystyle m_{0}} initially), and c{\displaystyle c} standing for the speed of light in a vacuum: m0m1=[1+Δvc1−Δvc]c2ve{\displaystyle {\frac {m_{0}}{m_{1}}}=\left[{\frac {1+{\frac {\Delta v}{c}}}{1-{\frac {\Delta v}{c}}}}\right]^{\frac {c}{2v_{e}}}} Writing m0m1{\displaystyle {\frac {m_{0}}{m_{1}}}} as R{\displaystyle R}, a little algebra allows this equation to be rearranged as Δvc=R2vec−1R2vec+1{\displaystyle {\frac {\Delta v}{c}}={\frac {R^{\frac {2v_{e}}{c}}-1}{R^{\frac {2v_{e}}{c}}+1}}} Then, using the identity R2vec=exp⁡[2vecln⁡R]{\displaystyle R^{\frac {2v_{e}}{c}}=\exp \left[{\frac {2v_{e}}{c}}\ln R\right]} (here "exp" denotes the exponential function; see also Natural logarithm as well as the "power" identity at Logarithmic identities) and the identity tanh⁡x=e2x−1e2x+1{\displaystyle \tanh x={\frac {e^{2x}-1}{e^{2x}+1}}} (see Hyperbolic function), this is equivalent to Δv=c⋅tanh⁡(vecln⁡m0m1){\displaystyle \Delta v=c\cdot \tanh \left({\frac {v_{e}}{c}}\ln {\frac {m_{0}}{m_{1}}}\right)} Other derivations[edit] Impulse-based[edit] The equation can also be derived from the basic integral of acceleration in the form of force (thrust) over mass. By representing the delta-v equation as the following: Δv=∫t0t1|T|m0−Δmt dt{\displaystyle \Delta v=\int _{t0}^{t1}{\frac {|T|}{{m_{0}}-\Delta {m}{t}}}~dt} where T is thrust, m0{\displaystyle m_{0}} is the initial (wet) mass and Δm{\displaystyle \Delta m} is the initial mass minus the final (dry) mass, and realising that the integral of a resultant force over time is total impulse, assuming thrust is the only force involved, ∫t0t1F dt=J{\displaystyle \int _{t0}^{t1}F~dt=J} The integral is found to be: J ln⁡(m0)−ln⁡(m1)Δm{\displaystyle J~{\frac {\ln({m_{0}})-\ln({m_{1}})}{\Delta m}}} Realising that impulse over the change in mass is equivalent to force over propellant mass flow rate (p), which is itself equivalent to exhaust velocity, JΔm=Fp=Vexh{\displaystyle {\frac {J}{\Delta m}}={\frac {F}{p}}=V_{exh}} the integral can be equated to Δv=Vexh ln⁡(m0m1){\displaystyle \Delta v=V_{exh}~\ln \left({\frac {m_{0}}{m_{1}}}\right)} Acceleration-based[edit] Imagine a rocket at rest in space with no forces exerted on it (Newton's First Law of Motion). However, as soon as its engine is started (clock set to 0) the rocket is expelling gas mass at a constant mass flow rate M (kg/s) and at exhaust velocity relative to the rocket ve (m/s). This creates a constant force propelling the rocket that is equal to M × ve. The mass of fuel the rocket initially has on board is equal to m0 - mf. It will therefore take a time that is equal to (m0 - mf)/M to burn all this fuel. Now, the rocket is subject to a constant force (M × ve), but at the same time its total weight is decreasing steadily because it's expelling gas. According to Newton's Second Law of Motion, this can have only one consequence; its acceleration is increasing steadily. To obtain the acceleration, the propelling force has to be divided by the rocket's total mass. So, the level of acceleration at any moment (t) after ignition and until the fuel runs out is given by; Mvem0−(Mt).{\displaystyle ~{\frac {Mv_{e}}{m_{0}-(Mt)}}.} Since the time it takes to burn the fuel is (m0 - mf)/M the acceleration reaches its maximum of Mvemf{\displaystyle ~{\frac {Mv_{e}}{m_{f}}}} the moment the last fuel is expelled. Since the exhaust velocity is related to the specific impulse in unit time as Isp=veg0,{\displaystyle I_{\rm {sp}}={\frac {v_{\text{e}}}{g_{0}}},}[4] where g0 is the standard gravity, the corresponding maximum g-force is MIspmf.{\displaystyle ~{\frac {MI_{\rm {sp}}}{m_{f}}}.} Since speed is the definite integration of acceleration, and the integration has to start at ignition and end the moment the last propellant leaves the rocket, the following definite integral yields the speed at the moment the fuel runs out; ∫0m0−mfMMvem0−(Mt) dt= −veln⁡(mf)+veln⁡(m0)= veln⁡(m0mf){\displaystyle ~\int _{0}^{\frac {m_{0}-m_{f}}{M}}{\frac {Mv_{e}}{m_{0}-(Mt)}}~dt=~-v_{e}\ln(m_{f})+v_{e}\ln(m_{0})=~v_{e}\ln \left({\frac {m_{0}}{m_{f}}}\right)}

Because it says the max power of the engine on the part.

What do you mean by "numbers"?

Every time I create a SSTO and ignore the science stuff the SSTO keeps pulling up, and then stalls and smashes into the ground.

I have just the thing. (I didn't make the video)

Kein Deutsch?

A Fighter inspired from light bombers designed in the 30’s and 40’s era during WWII. (Action groups) 1 toggle bomb bay 2 deploy bomb (Download available here:https://kerbalx.com/TheDumbKerbal/NB-1-Realistic-WWII-Bomber) (Hint: the propellers move) Evidence that the propellers move

I'm interested into what you guys have come up for your craft designs. Go ahead and post your awesome creation here.

Can they survive in shallow reefs?

But can they life in shallow reefs?

Oh, but can they live in the ocean?

Can you do a suicide burn on any planet/moon?

If they can live with no oxygen. Could they survive in space?

"Real" suicide burns?

is http://forum.kerbalspaceprogram.com/index.php?/profile/16426-sal_vager/ keeping his forum account?

"Wedging Kracken into the Mohole" "Training trainees" "Ignoring Dres"

Verboten für nicht meine Muttersprache sprechen. (Translation: Banned for translating.)

Wann ist die Making History DLC Verfügung?

Parece legítimo.

Banned for spreading "false rumors".

The Fairchild Republic A-10 Thunderbolt II is a single-seat, twin turbofan engine, straight wing jet aircraft developed by Fairchild-Republic for the United States Air Force (USAF). Commonly referred to by the nicknames Warthog or Hog, its official name comes from the Republic P-47 Thunderbolt, a World War II fighter that was effective at attacking ground targets. The A-10 was designed for close air support (CAS) of friendly ground troops, engaging armored vehicles and tanks, and providing quick-action support against enemy ground forces. It entered service in 1976 and is the only production-built aircraft that has served in the USAF that was designed solely for CAS. Its secondary mission is to provide forward air controller â airborne (FAC-A) support, by directing other aircraft in attacks on ground targets. Aircraft used primarily in this role are designated OA-10. The A-10 was intended to improve on the performance of the A-1 Skyraider and its poor firepower. The A-10 was designed around the 30 mm GAU-8 Avenger rotary cannon. Its airframe was designed for durability, with measures such as 1,200 pounds (540 kg) of titanium armor to protect the cockpit and aircraft systems, enabling it to absorb a significant amount of damage and continue flying. Its short takeoff and landing capability permits operation from airstrips close to the front lines, and its simple design enables maintenance with minimal facilities. The A-10 served in the Gulf War (Operation Desert Storm), the American intervention against Iraq’s invasion of Kuwait, where the A-10 distinguished itself. The A-10 also participated in other conflicts such as Operation Urgent Fury in Grenada, the Balkans, Afghanistan, Iraq, and against ISIL in the Middle East. (Download available here:https://kerbalx.com/TheFatKerbal/A-10-Thunderbolt-II)

The Fairchild Republic A-10 Thunderbolt II is a single-seat, twin turbofan engine, straight wing jet aircraft developed by Fairchild-Republic for the United States Air Force (USAF). Commonly referred to by the nicknames Warthog or Hog, its official name comes from the Republic P-47 Thunderbolt, a World War II fighter that was effective at attacking ground targets. The A-10 was designed for close air support (CAS) of friendly ground troops, engaging armored vehicles and tanks, and providing quick-action support against enemy ground forces. It entered service in 1976 and is the only production-built aircraft that has served in the USAF that was designed solely for CAS. Its secondary mission is to provide forward air controller â airborne (FAC-A) support, by directing other aircraft in attacks on ground targets. Aircraft used primarily in this role are designated OA-10. The A-10 was intended to improve on the performance of the A-1 Skyraider and its poor firepower. The A-10 was designed around the 30 mm GAU-8 Avenger rotary cannon. Its airframe was designed for durability, with measures such as 1,200 pounds (540 kg) of titanium armor to protect the cockpit and aircraft systems, enabling it to absorb a significant amount of damage and continue flying. Its short takeoff and landing capability permits operation from airstrips close to the front lines, and its simple design enables maintenance with minimal facilities. The A-10 served in the Gulf War (Operation Desert Storm), the American intervention against Iraq’s invasion of Kuwait, where the A-10 distinguished itself. The A-10 also participated in other conflicts such as Operation Urgent Fury in Grenada, the Balkans, Afghanistan, Iraq, and against ISIL in the Middle East. (Download available here:https://kerbalx.com/TheFatKerbal/A-10-Thunderbolt-II)