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2) centrato
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Lorem ipsum dolor sit amet, consectetur adipiscing elit. Curabitur rhoncus sem mi, in fringilla massa laoreet ac.[ax^2 + bx + c = 0] Curabitur egestas, libero ut dapibus fermentum, nisl nunc gravida est, ut maximus augue diam ut ante. Duis turpis erat, pulvinar quis convallis non, volutpat id purus.
Alessio mandami una bistecca di carne stagionata per premio.
Aggiorno al piu’ presto la guida, ma come vedi ora puoi usare a tua scelta TeX e/o MathML, sia inline che centered (vedi qui) con gli appositi markers.
Certo Marco che per i test avresti potuto usare equazioni piu carine come quella di Tziolkowsky e non banalemente equazioni di secondo grado e oscillazioni
In the following derivation, “the rocket” is taken to mean “the rocket and all of its unburned propellant”.
Newton’s second law of motion relates external forces ((F_i,)) to the change in linear momentum of the whole system (including rocket and exhaust) as follows:
(\sum F_i = \lim_{\Delta t \to 0} \frac{P_2-P_1}{\Delta t})
where (P_1,) is the momentum of the rocket at time ‘‘t=0’’:
( P_1 = \left( {m + \Delta m} \right)V)
and (P_2,) is the momentum of the rocket and exhausted mass at time (t=\Delta t,):
(P_2 = m\left(V + \Delta V \right) + \Delta m V_e)
and where, with respect to the observer:
(V,) is the velocity of the rocket at time ‘‘t=0’’
(V+\Delta V,) is the velocity of the rocket at time (t=\Delta t,)
(V_e,) is the velocity of the mass added to the exhaust (and lost by the rocket) during time (\Delta t,)
(m+\Delta m,) is the mass of the rocket at time ‘‘t=0’’
(m,) is the mass of the rocket at time (t=\Delta t,)
The velocity of the exhaust (V_e) in the observer frame is related to the velocity of the exhaust in the rocket frame (v_e) by (since exhaust velocity is in the negative direction)
(V_e=V-v_e)
Solving yields:
(P_2-P_1=m\Delta V-v_e\Delta m,)
and, using (dm=-\Delta m), since ejecting a positive (\Delta m) results in a decrease in mass,
(\sum F_i=m\frac{dV}{dt}+v_e\frac{dm}{dt})
If there are no external forces then (\sum F_i=0) conservation of linear momentum and
(m\frac{dV}{dt}=-v_e\frac{dm}{dt})
Assuming (v_e,) is constant, this may be integrated to yield:
(\Delta V\ = v_e \ln \frac {m_0} {m_1})
or equivalently
(m_1=m_0 e^{-\Delta V\ / v_e}) (m_0=m_1 e^{\Delta V\ / v_e}) (m_0 - m_1=m_1 (e^{\Delta V\ / v_e} - 1))
where (m_0) is the initial total mass including propellant, (m_1) the final total mass, and (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 (m_0 - m_1) is the total mass of propellant expended, and hence:
(M_f = 1-\frac {m_1} {m_0}=1-e^{-\Delta V\ / v_\text{e}})
where (M_f) is the propellant mass fraction (the part of the initial total mass that is spent as working mass).
(\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, with (\Delta v) again standing for the rocket’s final velocity (after burning off all its fuel and being reduced to a rest mass of (m_1)) in the inertial frame of reference where the rocket started at rest (with the rest mass including fuel being (m_0) initially), and (c) standing for the speed of light in a vacuum:
(\frac{m_0}{m_1} = \left[\frac{1 + {\frac{\Delta v}{c}}}{1 - {\frac{\Delta v}{c}}}\right]^{\frac{c}{2v_e}})
Writing (\frac{m_0}{m_1}) as (R), a little algebra allows this equation to be rearranged as
(\frac{\Delta v}{c} = \frac{R^{\frac{2v_e}{c}} - 1}{R^{\frac{2v_e}{c}} + 1})
Then, using the Identity (mathematics)|identity (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 Logarithm#Product, quotient, power and root|Logarithmic identities) and the identity (\tanh x = \frac{e^{2x} - 1} {e^{2x} + 1}) (’‘see’’ Hyperbolic function), this is equivalent to
(\Delta v = c \cdot \tanh \left(\frac {v_e}{c} \ln \frac{m_0}{m_1} \right))
adesso posso dire direttamente ai miei studenti di abbandonare qualunque software stiano usando per compilare LaTeX e venire direttamente su FAit. Certo, molti avranno difficoltà a capire l’italiano, ma almeno c’è un motore LaTeX coi fiocchi e un sacco di immagini interessanti
Non e’ un consiglio, ma un obbligo (per chi non e’ amministratore).
La sintassi Latex e’ sicuramente “migliore” non c’e’ dubbio
E’ esattamente quello che dicevo a proposito di MathML, nel caso ideale la sintassi Latex dovrebbe essere renderizzata non tanto in una immagine ma in MathML.
Magari c’e’ un parametro facile facile che disabilita la trasparenza nella GIF in modo che anche il modello attuale si veda con un tema scuro?
OK capisco hai messo MathJAX che fa rendering client-side della roba dentro i tag \ ( \ ). Questa e’ una soluzione con i fiocchi, bellissima. La carne stagionata te la cucino volentierissimo, ci mancherebbe…
