How to of the Day

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• How to Play the Hidden Basketball Game on Facebook Messenger
• How to Derive the Faraday Tensor
• How to Modify a Nerf N Strike Elite Stryfe

How to Play the Hidden Basketball Game on Facebook Messenger Posted: 10 May 2016 05:00 PM PDT In March 2016, Facebook released an update to its Messenger app containing a hidden basketball game. This minigame proved to be a hit, racking up over 300 million plays and counting since its launch.[1] Wanna get in on the action? See Step 1 to begin. EditSteps Update the Messenger app to its latest version. Facebook rolled out the initial update on March 18, 2016. To update the app on iOS, visit the App Store. To update the app on Android, visit the Play Store. Open the app and select a conversation. Find the basketball emoji and send it to a friend. On Android, tap the emojis icon beside the text field, select the third tab, and scroll halfway down to find the orange basketball. On iOS, tap the world icon at the lower left of the keyboard. Then hit or scroll to the activities emoji tab (indicated by a grey soccer ball) and find the basketball emoji at the beginning of the second row. Tap the basketball in the conversation to start the game. Flick the basketball with your finger to shoot. Your goal is to make as many balls in the basket as you can, before missing. The ball will change positions as the game progresses. At 10 points, the game will up the difficulty by moving the backboard side-to-side. At 20 points, it will move faster. Once you reach 30 points, it'll start moving diagonally.[2] The game ends when you miss. Your highest score will be saved and recorded in the conversation window so your friend can challenge your score. EditTips The game was released to coincide with March Madness, a college basketball tournament organized by the National Collegiate Athletic Association (NCAA) and held every year in the United States. EditRelated wikiHows Get Facebook Messenger on Your Phone Use Facebook Messenger for Android Chat Using Facebook Messenger App on iOS EditSources and Citations http://thenextweb.com/facebook/2016/03/17/messenger-basketball/ http://bgr.com/2016/03/18/how-to-play-hidden-facebook-messenger-basketball-game/ Cite error: <ref> tags exist, but no <references/> tag was found

How to Derive the Faraday Tensor Posted: 10 May 2016 09:00 AM PDT While Maxwell's equations demonstrate the connections between the electric field ${\displaystyle {\mathbf {E} }}$ and the magnetic field ${\displaystyle {\mathbf {B} },}$ in special relativity, they are really two aspects of the same force – electromagnetism. It is therefore a necessity to derive a mathematical object that describes both these fields in a useful fashion. We start from the Lorentz force and basic principles of special relativity to arrive at a mathematical formulation of the electromagnetic field and its associated Lorentz transformation. EditSteps EditDeriving the Faraday Tensor Begin with the Lorentz force. The Lorentz force is the result of observations in the 19th century that describe the way electric and magnetic fields exert forces on charged particles. While it may seem innocuous at first, the relation is actually a relativistic one, if formulated as such. Below, we write the force in terms of change of momentum. ${\displaystyle {\frac {{\mathrm {d} }{\mathbf {p} }}{{\mathrm {d} }t}}=q({\mathbf {E} }+{\mathbf {v} }\times {\mathbf {B} })}$ A central tenet of special relativity is that the conservation laws in Newtonian mechanics also apply to the upgraded 4-vectors. This implies that the above relation holds for 4-momentum ${\displaystyle p^{\mu }}$ and 4-velocity ${\displaystyle v^{\mu }.}$ Meanwhile, charge ${\displaystyle q}$ is an invariant. Recall the relationship between power, force, and velocity. Because power is defined as work per unit time, and magnetic fields do no work, the Lorentz force can be written as ${\displaystyle {\mathbf {F} }=q{\mathbf {E} }.}$ The usefulness of this relation will be seen later. ${\displaystyle {\begin{aligned}P={\frac {{\mathrm {d} }E}{{\mathrm {d} }t}}&={\mathbf {F} }\cdot {\mathbf {v} }\\&=q{\mathbf {E} }\cdot {\mathbf {v} }\end{aligned}}}$ Do not be confused by ${\displaystyle E}$ in this context, which stands for energy, not electric field. Recall the relationship between coordinate time ${\displaystyle t}$ and proper time ${\displaystyle \tau }$. The Lorentz force, while true, is not very useful in its current state. The reason why this is the case is because coordinate time is not invariant in Minkowski space. We need to reformulate the Lorentz force in terms of proper time, for proper time is invariant. ${\displaystyle {\mathrm {d} }t=\gamma {\mathrm {d} }\tau ,\gamma ={\frac {1}{\sqrt {1-{\frac {v^{2}}{c^{2}}}}}}}$ When derivatives are taken with respect to these variables, the relation is ${\displaystyle {\frac {\mathrm {d} }{{\mathrm {d} }t}}={\frac {1}{\gamma }}{\frac {\mathrm {d} }{{\mathrm {d} }\tau }}.}$ Therefore, in order to convert to proper time, we must multiply by ${\displaystyle \gamma .}$ Rewrite power and the Lorentz force with respect to proper time. The result is simply an extra ${\displaystyle \gamma }$ factor on the right side. ${\displaystyle {\frac {{\mathrm {d} }E}{{\mathrm {d} }\tau }}=\gamma q({\mathbf {E} }\cdot {\mathbf {v} })}$ ${\displaystyle {\frac {{\mathrm {d} }{\mathbf {p} }}{{\mathrm {d} }\tau }}=\gamma q({\mathbf {E} }+{\mathbf {v} }\times {\mathbf {B} })}$ Write the Lorentz force in manifestly covariant form. This form is similar in appearance to a matrix equation, in which a matrix acting on a vector outputs another vector. We can rewrite it like this because the above two equations describe everything we need to know about the matrix. Recognize the 4-momentum and 4-velocity in component form below. ${\displaystyle {\frac {\mathrm {d} }{{\mathrm {d} }\tau }}{\begin{pmatrix}E/c\\p_{x}\\p_{y}\\p_{z}\end{pmatrix}}=\gamma q{\begin{pmatrix}F^{0}{}_{0}&F^{0}{}_{1}&F^{0}{}_{2}&F^{0}{}_{3}\\F^{1}{}_{0}&F^{1}{}_{1}&F^{1}{}_{2}&F^{1}{}_{3}\\F^{2}{}_{0}&F^{2}{}_{1}&F^{2}{}_{2}&F^{2}{}_{3}\\F^{3}{}_{0}&F^{3}{}_{1}&F^{3}{}_{2}&F^{3}{}_{3}\end{pmatrix}}{\begin{pmatrix}1\\v_{x}\\v_{y}\\v_{z}\end{pmatrix}}}$ The matrix above is the Faraday tensor ${\displaystyle F^{\mu }{}_{\nu },}$ written out in its component form. (Don't worry about the placement of the indices for now.) From here, it is clear that we need to find these components such that they satisfy ${\displaystyle {\frac {{\mathrm {d} }E}{{\mathrm {d} }\tau }}=\gamma q({\mathbf {E} }\cdot {\mathbf {v} })}$ and ${\displaystyle {\frac {{\mathrm {d} }{\mathbf {p} }}{{\mathrm {d} }\tau }}=\gamma q({\mathbf {E} }+{\mathbf {v} }\times {\mathbf {B} }).}$ Solve the matrix equation for ${\displaystyle F^{\mu }{}_{\nu }}$ by direct comparison. It is easy to do this one equation at a time. ${\displaystyle {\frac {\mathrm {d} }{{\mathrm {d} }\tau }}(E/c)=\gamma q(F^{0}{}_{0}+F^{0}{}_{1}v_{x}+F^{0}{}_{2}v_{y}+F^{0}{}_{3}v_{z})={\frac {\gamma q}{c}}({\mathbf {E} }\cdot {\mathbf {v} })}$ Here, the answer is trivial. ${\displaystyle F^{0}{}_{0}=0,F^{0}{}_{1}=E_{x}/c,F^{0}{}_{2}=E_{y}/c,F^{0}{}_{3}=E_{z}/c}$ ${\displaystyle {\frac {{\mathrm {d} }p_{x}}{{\mathrm {d} }\tau }}=\gamma q(F^{1}{}_{0}+F^{1}{}_{1}v_{x}+F^{1}{}_{2}v_{y}+F^{1}{}_{3}v_{z})}$ Here, the answer is slightly less obvious, because we need to incorporate the ${\displaystyle {\mathbf {B} }}$ field as well. Since this is the ${\displaystyle x}$ component of the force, we have to look for fields that generate forces in that direction. We know ${\displaystyle {\mathbf {E} }}$ fields generate forces parallel to them, while a moving charged particle in a ${\displaystyle {\mathbf {B} }}$ field generates a force in the direction orthogonal to both ${\displaystyle {\mathbf {v} }}$ and ${\displaystyle {\mathbf {B} }.}$ Of course, a particle moving in the ${\displaystyle x}$ direction cannot possibly generate a force in that same direction, given how ${\displaystyle {\mathbf {B} }}$ fields interact with them, so that term is 0. Therefore, ${\displaystyle F^{1}{}_{0}=E_{x}/c,F^{1}{}_{1}=0,F^{1}{}_{2}=B_{z},F^{1}{}_{3}=-B_{y}.}$ We can proceed to derive the last two rows of the tensor in the same manner. The important part is the antisymmetry exhibited in the lower-right 3x3 partition of the tensor, which stems from the cross product in the Lorentz force. In doing so, the diagonal elements of the tensor get sent to 0. The last two rows are as follows. ${\displaystyle F^{2}{}_{0}=E_{y}/c,F^{2}{}_{1}=-B_{z},F^{2}{}_{2}=0,F^{2}{}_{3}=B_{x}}$ ${\displaystyle F^{3}{}_{0}=E_{z}/c,F^{3}{}_{1}=B_{y},F^{3}{}_{2}=-B_{x},F^{3}{}_{3}=0}$ Arrive at the Faraday tensor. This tensor, also called the electromagnetic tensor, describes the electromagnetic field in spacetime. Two fields, previously thought of as separate, shown to be interconnected via Maxwell's equations, are finally united by special relativity into a single mathematical object. The tensor shown below is in mixed-variant form because of how we have derived it from the Lorentz force. ${\displaystyle F^{\mu }{}_{\nu }={\begin{pmatrix}0&E_{x}/c&E_{y}/c&E_{z}/c\\E_{x}/c&0&B_{z}&-B_{y}\\E_{y}/c&-B_{z}&0&B_{x}\\E_{z}/c&B_{y}&-B_{x}&0\end{pmatrix}}}$ EditLorentz Transformations of Electromagnetic Fields Begin with the covariant forms of the Lorentz force, 4-momentum, and 4-velocity. Index notation allows these quantities to be described more compactly and in a coordinate-independent manner. ${\displaystyle {\frac {{\mathrm {d} }p^{\alpha }}{{\mathrm {d} }\tau }}=qF^{\alpha }{}_{\beta }v^{\beta }}$ ${\displaystyle p^{\alpha \prime }=\Lambda ^{\alpha \prime }{}_{\beta }p^{\beta }}$ ${\displaystyle v^{\beta \prime }=\Lambda ^{\beta \prime }{}_{\delta }v^{\delta }}$ Above, ${\displaystyle \Lambda }$ is the Lorentz transformation tensor. For a boost in the ${\displaystyle x}$ direction, it can be written as below. ${\displaystyle \Lambda ^{-1},}$ of course, has positive ${\displaystyle \gamma \beta }$ on the off-diagonal. ${\displaystyle \Lambda ={\begin{pmatrix}\gamma &-\gamma \beta &0&0\\-\gamma \beta &\gamma &0&0\\0&0&1&0\\0&0&0&1\end{pmatrix}}}$ Write the Lorentz force as measured in the boosted frame. The power of writing the above quantities in the covariant form stems from the fact that the Lorentz transformation is a linear transformation. ${\displaystyle {\frac {{\mathrm {d} }p^{\alpha \prime }}{{\mathrm {d} }\tau }}=qF^{\alpha \prime }{}_{\beta \prime }v^{\beta \prime }}$ Write the boosted Lorentz force in terms of quantities measured in the coordinate frame. Then left-multiply each side by the inverse Lorentz tensor ${\displaystyle \Lambda ^{-1}.}$ ${\displaystyle (\Lambda ^{-1})^{\mu }{}_{\alpha \prime }{\frac {\mathrm {d} }{{\mathrm {d} }\tau }}\Lambda ^{\alpha \prime }{}_{\beta }p^{\beta }=q(\Lambda ^{-1})^{\mu }{}_{\alpha \prime }F^{\alpha \prime }{}_{\beta \prime }\Lambda ^{\beta \prime }{}_{\delta }v^{\delta }}$ Factor in the inverse Lorentz tensor. Because the Lorentz tensor can be treated as a constant, it can be inserted inside the derivative operator. Observe that ${\displaystyle (\Lambda ^{-1})^{\alpha }{}_{\beta \prime }\Lambda ^{\beta \prime }{}_{\sigma }=\delta ^{\alpha }{}_{\sigma },}$ where ${\displaystyle \delta }$ is the Kronecker delta (not to be confused by the index below, which only represents numbers). ${\displaystyle {\begin{aligned}{\frac {\mathrm {d} }{{\mathrm {d} }\tau }}(\Lambda ^{-1})^{\mu }{}_{\alpha \prime }\Lambda ^{\alpha \prime }{}_{\beta }p^{\beta }&=q(\Lambda ^{-1})^{\mu }{}_{\alpha \prime }F^{\alpha \prime }{}_{\beta \prime }\Lambda ^{\beta \prime }{}_{\delta }v^{\delta }\\{\frac {\mathrm {d} }{{\mathrm {d} }\tau }}\delta ^{\mu }{}_{\beta }p^{\beta }&=q(\Lambda ^{-1})^{\mu }{}_{\alpha \prime }F^{\alpha \prime }{}_{\beta \prime }\Lambda ^{\beta \prime }{}_{\delta }v^{\delta }\end{aligned}}}$ When the Kronecker delta acts on a vector, the same vector is outputted. The only difference is that here, the ${\displaystyle \beta }$ index is contracted. ${\displaystyle {\frac {{\mathrm {d} }p^{\mu }}{{\mathrm {d} }\tau }}=q(\Lambda ^{-1})^{\mu }{}_{\alpha \prime }F^{\alpha \prime }{}_{\beta \prime }\Lambda ^{\beta \prime }{}_{\delta }v^{\delta }}$ Obtain the boosted Faraday tensor. Notice that on the right-hand side, ${\displaystyle (\Lambda ^{-1})^{\mu }{}_{\alpha \prime }F^{\alpha \prime }{}_{\beta \prime }\Lambda ^{\beta \prime }{}_{\delta }}$ describes the Faraday tensor in the coordinate frame ${\displaystyle F^{\mu }{}_{\lambda },}$ so that ${\displaystyle {\frac {{\mathrm {d} }p^{\mu }}{{\mathrm {d} }\tau }}=qF^{\mu }{}_{\lambda }v^{\lambda }}$ (where we originally started). Therefore, ${\displaystyle (\Lambda ^{-1})^{\mu }{}_{\alpha \prime }F^{\alpha \prime }{}_{\beta \prime }\Lambda ^{\beta \prime }{}_{\delta }=F^{\mu }{}_{\delta }.}$ However, this tells us how to boost from the moving frame to the coordinate frame. To perform the inverse operation, simply switch the Lorentz tensors by left-multiplying by ${\displaystyle \Lambda }$ and right-multiplying by ${\displaystyle \Lambda ^{-1}.}$ The below equation gives us the relation that we want. ${\displaystyle F^{\alpha \prime }{}_{\beta \prime }=\Lambda ^{\alpha \prime }{}_{\alpha }F^{\alpha }{}_{\beta }(\Lambda ^{-1})^{\beta }{}_{\beta \prime }}$ Evaluate the Faraday tensor in the boosted frame. Below, we boost in the ${\displaystyle +x}$ direction. Remember that in the process of evaluating, all the diagonal elements of the tensor must be 0. ${\displaystyle {\begin{pmatrix}0&{\frac {E_{x}}{c}}&\gamma ({\frac {E_{y}}{c}}-\beta B_{z})&\gamma ({\frac {E_{z}}{c}}+\beta B_{y})\\{\frac {E_{x}}{c}}&0&\gamma (-\beta {\frac {E_{y}}{c}}+B_{z})&\gamma (-\beta {\frac {E_{z}}{c}}-B_{y})\\\gamma ({\frac {E_{y}}{c}}-\beta B_{z})&\gamma (\beta {\frac {E_{y}}{c}}-B_{z})&0&B_{x}\\\gamma ({\frac {E_{z}}{c}}+\beta B_{y})&\gamma (\beta {\frac {E_{z}}{c}}+B_{y})&-B_{x}&0\end{pmatrix}}}$ Obtain the Lorentz transformations for the ${\displaystyle {\mathbf {E} }}$ and ${\displaystyle {\mathbf {B} }}$ fields. There are two things of note here. First, from the above tensor, we see that the components of both fields parallel to the direction of motion remain unchanged. Second, and more importantly, the transformations for components perpendicular to the direction of motion show that a field that is zero in one reference frame may not be in another. In general, this will be the case (especially with electromagnetic waves, which cannot exist without mutual induction), so special relativity tells us that these two fields are really just two aspects of the same electromagnetic field. Electric fields (note that we have multiplied by ${\displaystyle c}$ to both sides) ${\displaystyle E_{x}^{\prime }=E_{x}}$ ${\displaystyle E_{y}^{\prime }=\gamma (E_{y}-\beta cB_{z})}$ ${\displaystyle E_{z}^{\prime }=\gamma (E_{z}+\beta cB_{y})}$ Magnetic fields ${\displaystyle cB_{x}^{\prime }=cB_{x}}$ ${\displaystyle cB_{y}^{\prime }=\gamma (cB_{y}+\beta E_{z})}$ ${\displaystyle cB_{z}^{\prime }=\gamma (cB_{z}-\beta E_{y})}$ EditTips The Lorentz force as written in Part 2 uses the mixed-variant form of the Faraday tensor. To obtain the more familiar contravariant form, rewrite the Lorentz force as ${\displaystyle {\frac {{\mathrm {d} }p^{\mu }}{{\mathrm {d} }\tau }}=qF^{\mu \nu }v_{\nu }.}$ Here, ${\displaystyle v_{\nu }}$ is the covariant 4-velocity, which differs from the contravariant 4-velocity ${\displaystyle v^{\nu }}$ by negating the spatial components, where we are using the timelike Minkowski metric with signature ${\displaystyle (+---)}$ on the main diagonal. Therefore, to compensate, columns 2-4 of the Faraday tensor must negate as well, leading to the contravariant Faraday tensor below. ${\displaystyle F^{\mu \nu }={\begin{pmatrix}0&-E_{x}/c&-E_{y}/c&-E_{z}/c\\E_{x}/c&0&-B_{z}&B_{y}\\E_{y}/c&B_{z}&0&-B_{x}\\E_{z}/c&-B_{y}&B_{x}&0\end{pmatrix}}}$ Another way to derive the Faraday tensor is to start from its definition, which is written in terms of potentials. The definition makes the antisymmetry of the tensor clear, because when ${\displaystyle \mu =\nu ,F^{\mu \nu }=0.}$ Below, ${\displaystyle \partial ^{\mu }=\left({\frac {1}{c}}{\frac {\partial }{\partial t}},-\nabla \right)}$ is the contravariant 4-gradient and ${\displaystyle A^{\mu }=\left({\frac {\phi }{c}},{\mathbf {A} }\right)}$ is the 4-potential. ${\displaystyle F^{\mu \nu }=\partial ^{\mu }A^{\nu }-\partial ^{\nu }A^{\mu }}$ Recall that ${\displaystyle {\mathbf {E} }=-\nabla \phi -{\frac {\partial {\mathbf {A} }}{\partial t}}}$ and ${\displaystyle {\mathbf {B} }=\nabla \times {\mathbf {A} }.}$ Then, we can immediately derive the components of the tensor through direct substitution. Below, we derive the ${\displaystyle {\mathbf {E} }}$ field components, and recognize that antisymmetry automatically gives us the transpose elements. Latin indices are used for spatial components, as per convention. ${\displaystyle {\begin{aligned}F^{0i}&=\partial ^{0}A^{i}-\partial ^{i}A^{0}\\&={\frac {1}{c}}{\frac {\partial A^{i}}{\partial t}}-(-\nabla )\left({\frac {\phi }{c}}\right)\\&=-{\frac {1}{c}}\left(-{\frac {\partial A^{i}}{\partial t}}-\nabla \phi \right)\\&=-{\frac {E^{i}}{c}}\end{aligned}}}$ The ${\displaystyle {\mathbf {B} }}$ field components are slightly more complicated, but there is a pattern here. The cross product ${\displaystyle {\mathbf {B} }=\nabla \times {\mathbf {A} }}$ can be written as ${\displaystyle B^{i}=\varepsilon ^{i}{}_{jk}\partial ^{j}A^{k},}$ where ${\displaystyle \varepsilon ^{i}{}_{jk}}$ is the Levi-Civita symbol. This antisymmetric symbol is defined as ${\displaystyle \varepsilon ^{1}{}_{23}=1,}$ any swapping of two numbers negates the symbol, and any two numbers that are the same sends it to 0. Therefore, after choosing ${\displaystyle i,}$ only two components of ${\displaystyle \varepsilon ^{i}{}_{jk}}$ are non-negative; for ${\displaystyle i=1,\varepsilon ^{1}{}_{23}=1}$ and ${\displaystyle \varepsilon ^{1}{}_{32}=-1.}$ Knowing that the spatial components of the contrapositive 4-gradient are negative, the ${\displaystyle {\mathbf {B} }}$ field components can therefore be written as ${\displaystyle B^{i}=-\varepsilon ^{i}{}_{jk}F^{jk}.}$ ${\displaystyle {\begin{aligned}F^{12}&=\partial ^{1}A^{2}-\partial ^{2}A^{1}\\&=-\partial ^{x}A^{y}-(-\partial ^{y}A^{x})\\&=-(\partial ^{x}A^{y}-\partial ^{y}A^{x})\\&=-B^{z}\end{aligned}}}$ ${\displaystyle {\begin{aligned}F^{13}&=\partial ^{1}A^{3}-\partial ^{3}A^{1}\\&=-\partial ^{x}A^{z}-(-\partial ^{z}A^{x})\\&=\partial ^{z}A^{x}-\partial ^{x}A^{z}\\&=B^{y}\end{aligned}}}$ ${\displaystyle {\begin{aligned}F^{23}&=\partial ^{2}A^{3}-\partial ^{3}A^{2}\\&=-\partial ^{y}A^{z}-(-\partial ^{z}A^{y})\\&=-(\partial ^{y}A^{z}-\partial ^{z}A^{y})\\&=-B^{x}\end{aligned}}}$ Once again, antisymmetry of the Faraday tensor automatically gives us the transpose elements of the ${\displaystyle {\mathbf {B} }}$ field components. Check to see that this derivation gives us the correct components of ${\displaystyle F^{\mu \nu }.}$

How to Modify a Nerf N Strike Elite Stryfe Posted: 10 May 2016 01:00 AM PDT Looking to step up your Nerf game? Here is a tutorial on how to do all of the basic lock-removals and modifications for the Nerf N-Strike Elite Stryfe. EditSteps EditDisassembling the Stryfe Get to know your gun. Learn the internals of the Stryfe; this will help just in case you seem to have misplaced a piece or the blaster is no longer functional. Begin by un-screwing all of the screws on the outside of the blaster, except for the one located on the bottom-left most side of the blaster. Carefully separate the top half of the shell from the bottom half and set it aside for later. You are now ready to modify your Nerf N-Strike Elite Stryfe. EditReleasing Trigger Locks (Stage 1) Start by removing the small "L" shaped piece at the top of the magazine well, it also has a spring. You may throw this piece away as now to serves no purpose. Unscrew the middle most big silver screw located in the trigger assembly. Make sure to keep that screw for later. Gently raise the arm that it held down and rotate it out of the way. Unscrew the bottom most screw from the trigger assembly and remove the trigger itself. One again, make sure you keep that screw with the trigger. Unscrew the two screws sitting diagonally from one an other located on a retaining plate to the left of the screw you undid in step three. Again, keep the screw. Remove the retaining plate entirely and set it aside. Lift up the orange tab which is currently sticking into the magazine well of the blaster, and set it aside for later. Remove the remaining orange piece, which also has a spring attached. You may throw the piece away. Replace the orange tab which was removed earlier and replace it in the blaster. Push the orange tab right until is presses the little black button below it all the way down, then use a small piece of electrical tape to hold it in position. You use a smaller second piece in you feel one is not enough. When putting the electrical tape in the blaster, do not cover the screw holes or the pin wholes shown in the photo above. Replace the retaining plate in the blaster and screw it back in tightly. This will also help the electrical tape to stay put. EditReassembling the Trigger Assembly Place the trigger back into place and screw it back in. Make sure not to screw it too tightly or the trigger will be too tight. Swing the arm from step 2 of "Releasing Trigger Locks" back into position and screw it back in. EditReleasing Trigger Locks (Stage 2) Unscrew the silver screw located to the right of the rev trigger and take out the orange plate. Remove the entire rev trigger and make sure to keep the spring with it. Remove the orange tab. It also has a spring attached to it. You may throw it away. Place the rev trigger back into position and make sure the spring is pushed past the small wall between the area with the rev trigger and the area with the orange tab you removed. Make sure to push the rev trigger all the way down before proceeding, or else the button will not be pressed down. You may add a small piece of electrical tape onto the little metal box which holds the button, just in case. Re-screw the orange plate from step one. EditModifying the Voltage Find, in the top right of the blaster internals, the small chip called a Thermistor which shuts the blaster down if there is too much energy running through it. On it is what looks like a yellow omega (Ω) symbol. Gently lift the omega shaped piece until it's standing upright. Twist it slowly and gently until the 2 wires are touching. Make sure not to over-twist the omega shaped piece or attempt to re-twist it because it will break and require soldering. You may press the omega shaped piece down slightly but make sure not to push it down to far and too hard or the wires will snap. EditReassembling Your Blaster Put the other half of the shell on the half you were working on. Screw everything back in. EditTips Having both a precision screwdriver and a normal screw driver is very important because some of the screws are smaller or bigger than others. After un-screwing the screws on the outside of the blaster keep them all together in a cup or plastic bag, make sure not to lose them. Despite becoming obsolete in the blaster your currently modifying, you should keep all of the spring which you remove, incase you need it in a blaster later on. EditWarnings Remember, when un-screwing the blaster, do not un-screw the screw on the bottom-left most side of the blaster, located on the orange area because it is not necessary to open the blaster. When opening the blaster pieces may get lose or springs might pop out, so make sure you have everything in a cup or a plastic bag before beginning to modify. When modifying the voltage of the blaster, pay great attention to the polarity of the batteries when you place them in, because it will completely break the blaster. When modifying your voltage, make sure to not over-twist the omega shaped piece or attempt to re-twist it because it will break and require soldering. EditSources and Citations Voltage Modification Video Releasing Trigger Locks Video

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