@NexusGaming I’m not sure what advice works best, but in my opinion, inspiration is the key. Find some spaceship drawings that you really, really like, and try your hardest to re-create them as an SP model. And, if you begin to deviate from your original design intentions, just go with the flow, and embrace your design differences! My style is the result of numerous failed attempts to re-create the styles that have inspired me.
The clamp function that you used can probably be mathematically edited to work as you need it, but in my experience, booleans are better. Try this: (TargetDistance > 1046 & TargetDistance < 1207) ? 1 : 0
Lemme know if it works or not!
@BagelPlane you could multiply the entire expression by clamp01(Alt.max - Altitude).
Where Alt.max is the maximum height, in meters. Note that Altitude can be switched for AltitudeAgl to evaluate the altitude above ground level.
Sounds good. We would use the following code, in the gun's activationGroup: floor(smooth(clamp01(-1 * GearDown), -1 * GearDown ? 1 / delayTime : pow(10,10) ))
Where delayTime is whatever delay you need, in seconds.
.
This code is derived from a delay structure that I call the bidirectional delay. I first introduced it with Orbidyn-L to meet a variety of operating conditions for its weapons, including landing gear retraction.
.
I'm a little rusty when it comes to code that includes GearDown, so if it doesn't work, try removing the -1 that is multiplying the variable.
An easy hover code is as follows: Activate1 ? clamp01(-1 * rate(Altitude)) : VTOL
.
This code is very much entry-level. As a result, the ride might be a little bumpy when your drone is in hover mode. If that is the case, you can multiply the clamp01 term by a decimal such as 0.75, 0.5, 0.25. Alternatively, I can make some fancier code, but I can't do that quite yet bc finals.
Hmmm, they deploy if throttle is above 80? If so, you might want to change the direction of your comparison operator with the throttle term (Throttle > 0.8). That way, it will deploy when throttle is over 80%
Shouldn’t be too hard. Divide the entire expression by a clamped groundspeed clamp(1, 100, GS) and that should produce the desired behavior. If the reduction is too much, you can multiply GS by a decimal (i.e. 0.5, 0.25, 0.1, etc).
it was probably a capitalization error then. Try both cases that I gave while capitalizing Yaw. In other words, replace yaw with Yaw for both cases. If those don't work, the GearDown variable might be the culprit.
no worries - I'm always happy to give code. Your input would be: clamp01(GearDown)*yaw
.
Its been a little while since I used GearDown, so lemme know if it doesnt work.
@CRJ900Pilot Definitely. For your input, you would use clamp(pitch, -x, y)
.
Where:
Y = your upper limit (can be between 1 and zero)
X = your lower limit (can be between -1 and zero)
.
There are a few other approaches, too. This is the most simple one tho
no-can-do. Unfortunately funky trees is not supported in the firing input of a weapon. Guns are still confined to fireguns, and cannons are still confined to fireweapons.
A jet aircraft is flying at a speed of 1700 km/h. The air temperature is 20°C. The molecular weight of air is 29 g/mol. What is the Mach number of the aircraft?
Consider a uniform flow with velocity U in the positive x direction combined with two free vortices of equal strength located along the y-axis. Let one vortex located at y = a be a clockwise vortex (ψ = K ln r) and the other at y = −a be a counterclockwise vortex. where K is a positive constant. It an be shown by plotting streamlines that for Ua/K < 2 the streamline ψ = 0 forms a closed contour, as shown in the figure below. Thus, this combination can be used to represent flow around a family of bodies called Kelvin ovals. For the case of U = 10 m/s, a = 1 m, if the body has half-hight H = 1.5 m: a) calculate the value of K; b) for the same Kelvin oval, calculate its maximum width along x.
.
Had to do it to em
A vertical disk of radius r rolls around a horizontal track of radius R. This track rotates at a constant rate of Ω0 about the vertical axis. The center of the disk moves at a constant angular rate of φ ̇ about the same vertical axis. Using the set of unit vectors, eˆr,eˆφ,eˆz, obtain the acceleration of point P, located on the edge of the disk when the point is at its highest position. (from Greenwood, Principles of Dynamics, 1988)
The code for the engine shouldn't be affected by the weight of the aircraft. If it is, you can adjust the powermultiplier accordingly. Looks like you deleted the comment that I needed to make the code though, what conditions were there for this engine to activate again? Some combination of trim and an activation group?
.
As for that code, I'm not too familiar with PID controllers, so I wont be very useful deciphering it. I think that the VTOL bar changes the sign (negative or positive) of the correcting action made from the thruster, or makes the correcting action evaluate to zero. Not sure if that is exactly what is going on though.
An air bearing is constructed from a circular disk 2R=1 m in diameter that issues
air from many small holes in its lower surface. The total cross-sectional area of the holes is
almost equal to the entire lower surface area of the disk. The bearing floats h=2.0 mm above the
table and the air flows through the bearing with an average velocity of =3 m/s.
(a) Using control volume analysis, find an algebraic expression for the radial velocity under
the bearing as a function of the radial coordinate r, assuming that the flow is uniform, steady and
incompressible. Your answer may also contain v, h, and R.
(b) Find the magnitude and location of the maximum radial acceleration experienced by a fluid
particle in the gap. You might want to use the expression for the del operator in cylindrical coordinates.
You would essentially need to do a geometry problem to find an angle offset for your turret. Besides that, I can't really help further, since I have not actually done the math myself.
@SavageMan I have some hidden thrusters that prevent autoroll. Think your system, but on all 3 axes, and more spaghetti. My code is under section 3.2 (Facilitating Rotational Stability) in the technical documentation.
@WiiWiiTheMini Every single one, but in this case idc, because I really have nothing new to show. But other futuristic players have plenty to offer.
@NexusGaming I’m not sure what advice works best, but in my opinion, inspiration is the key. Find some spaceship drawings that you really, really like, and try your hardest to re-create them as an SP model. And, if you begin to deviate from your original design intentions, just go with the flow, and embrace your design differences! My style is the result of numerous failed attempts to re-create the styles that have inspired me.
@CrimsonOnigiri I’m lost on context haha, what did I miss?
This would be a nice feature. Though, since most drag in SP is caused by wings, you can assume that your CoD is near your CoL
The clamp function that you used can probably be mathematically edited to work as you need it, but in my experience, booleans are better. Try this:
(TargetDistance > 1046 & TargetDistance < 1207) ? 1 : 0
Lemme know if it works or not!
@Rodrigo110 Ultimately the same problems extend into today, yes. As a society we certainly have work to do.
Unfortunately it was only simple for some of the population during the 50s.
Ah. Guess I haven't been caught yet ;). I've definitely had some moments, though nothing rulebreaking.
Now I’m curious about what kind of mod notes I have lmao
I have been playing a lot lately bc I got it for free on the epic games store. My CMDR name is the same as my SP name!
@BagelPlane you could multiply the entire expression by clamp01(Alt.max - Altitude).
Where Alt.max is the maximum height, in meters. Note that Altitude can be switched for AltitudeAgl to evaluate the altitude above ground level.
Sounds good. We would use the following code, in the gun's
activationGroup:floor(smooth(clamp01(-1 * GearDown), -1 * GearDown ? 1 / delayTime : pow(10,10) ))
Where
delayTimeis whatever delay you need, in seconds..
This code is derived from a delay structure that I call the bidirectional delay. I first introduced it with Orbidyn-L to meet a variety of operating conditions for its weapons, including landing gear retraction.
.
I'm a little rusty when it comes to code that includes
GearDown, so if it doesn't work, try removing the-1that is multiplying the variable.No worries - I like helping other users with code. It sounds like you want your guns to be inoperable until the gear is fully retracted?
An easy hover code is as follows:
Activate1 ? clamp01(-1 * rate(Altitude)) : VTOL
.
This code is very much entry-level. As a result, the ride might be a little bumpy when your drone is in hover mode. If that is the case, you can multiply the clamp01 term by a decimal such as 0.75, 0.5, 0.25. Alternatively, I can make some fancier code, but I can't do that quite yet bc finals.
odd. What is your current code? Just so I can see exactly what you are working with. Also, what type of part is this on - a rotator?
Hmmm, they deploy if throttle is above 80? If so, you might want to change the direction of your comparison operator with the throttle term (Throttle > 0.8). That way, it will deploy when throttle is over 80%
No problem! Put a clamp01 around both statements, and multiply them using an asterisk (*)
My bad. I messed up the clamp syntax. Replace the clamp with clamp(GS, 1, 100)
Shouldn’t be too hard. Divide the entire expression by a clamped groundspeed clamp(1, 100, GS) and that should produce the desired behavior. If the reduction is too much, you can multiply GS by a decimal (i.e. 0.5, 0.25, 0.1, etc).
it was probably a capitalization error then. Try both cases that I gave while capitalizing Yaw. In other words, replace
yawwithYawfor both cases. If those don't work, the GearDown variable might be the culprit.Try inverting the clamp argument:
clamp01(-1 * GearDown) * yaw
no worries - I'm always happy to give code. Your input would be:
clamp01(GearDown)*yaw
.
Its been a little while since I used GearDown, so lemme know if it doesnt work.
@Greggory005 Wish I could, but the mod is not that simple for non-modders to use. It would be unusable to anyone who does not know how to use Unity.
@CRJ900Pilot Definitely. For your input, you would use clamp(pitch, -x, y)
.
Where:
Y = your upper limit (can be between 1 and zero)
X = your lower limit (can be between -1 and zero)
.
There are a few other approaches, too. This is the most simple one tho
sounds good. Glad I could help
no-can-do. Unfortunately funky trees is not supported in the firing input of a weapon. Guns are still confined to fireguns, and cannons are still confined to fireweapons.
@DEVINBOSS Your name is familiar - good to see you again!
Lmao I just copypasted a random problem. Forgot where it came from
A jet aircraft is flying at a speed of 1700 km/h. The air temperature is 20°C. The molecular weight of air is 29 g/mol. What is the Mach number of the aircraft?
Consider a uniform flow with velocity U in the positive x direction combined with two free vortices of equal strength located along the y-axis. Let one vortex located at y = a be a clockwise vortex (ψ = K ln r) and the other at y = −a be a counterclockwise vortex. where K is a positive constant. It an be shown by plotting streamlines that for Ua/K < 2 the streamline ψ = 0 forms a closed contour, as shown in the figure below. Thus, this combination can be used to represent flow around a family of bodies called Kelvin ovals. For the case of U = 10 m/s, a = 1 m, if the body has half-hight H = 1.5 m: a) calculate the value of K; b) for the same Kelvin oval, calculate its maximum width along x.
.
Had to do it to em
Looks nice. Please tag me!
It is for me lol. Since you had a bunch of builds I made an assumption - guess I was wrong!
On Thanksgiving break? I'm jealous haha. Got an exam tomorrow and then I can finally rest.
No problem! This is a good build.
This looks awesome - please tag me when it’s out! The soonest that I can help with code will be this weekend, as I have a very tight exam schedule rn
A vertical disk of radius r rolls around a horizontal track of radius R. This track rotates at a constant rate of Ω0 about the vertical axis. The center of the disk moves at a constant angular rate of φ ̇ about the same vertical axis. Using the set of unit vectors, eˆr,eˆφ,eˆz, obtain the acceleration of point P, located on the edge of the disk when the point is at its highest position. (from Greenwood, Principles of Dynamics, 1988)
If you don't get any help soon enough, let me know. I can make some quick code for the thruster tomorrow.
The code for the engine shouldn't be affected by the weight of the aircraft. If it is, you can adjust the powermultiplier accordingly. Looks like you deleted the comment that I needed to make the code though, what conditions were there for this engine to activate again? Some combination of trim and an activation group?
.
As for that code, I'm not too familiar with PID controllers, so I wont be very useful deciphering it. I think that the VTOL bar changes the sign (negative or positive) of the correcting action made from the thruster, or makes the correcting action evaluate to zero. Not sure if that is exactly what is going on though.
sure! Does the engine go to 100% throttle when we meet these conditions? Or should it still be throttlable by the user?
I have some pretty heavy school stuff at the moment, so I wont be able to actually play SP. I can give some advice and code though!
The community works in mysterious ways lmao
I see MC Ride, I upvote.
An air bearing is constructed from a circular disk 2R=1 m in diameter that issues
air from many small holes in its lower surface. The total cross-sectional area of the holes is
almost equal to the entire lower surface area of the disk. The bearing floats h=2.0 mm above the
table and the air flows through the bearing with an average velocity of =3 m/s.
(a) Using control volume analysis, find an algebraic expression for the radial velocity under
the bearing as a function of the radial coordinate r, assuming that the flow is uniform, steady and
incompressible. Your answer may also contain v, h, and R.
(b) Find the magnitude and location of the maximum radial acceleration experienced by a fluid
particle in the gap. You might want to use the expression for the del operator in cylindrical coordinates.
Hot
You would essentially need to do a geometry problem to find an angle offset for your turret. Besides that, I can't really help further, since I have not actually done the math myself.
@SavageMan I have some hidden thrusters that prevent autoroll. Think your system, but on all 3 axes, and more spaghetti. My code is under section 3.2 (Facilitating Rotational Stability) in the technical documentation.
@FuriousChicken Not sure yet. Feel free to make some recommendations!
Lmao. Not sure what their target demo is here
People like to put some lore behind their builds. Often, fictional companies are part of that lore.
ͬͬͤͬͦͬͬͤ ͤͬͦͬͬͤ ͬͦͬͬͤ ͦͬͬͤ ͬͬͤ ͬͤ ͤ Ỏ̷͖͈̞̩͎̻̫̫̜͉̠̫͕̭̭̫̫̹̗̹͈̼̠̖͍͚̥͈̮̼͕̠̤̯̻̥̬̗̼̳̤̳̬̪̹͚̞̼̠͕̼̠̦͚̫͔̯̹͉͉̘͎͕̼̣̝͙̱̟̹̩̟̳̦̭͉̮̖̭̣̣̞̙̗̜̺̭̻̥͚͙̝̦̲̱͉͖͉̰̦͎̫̣̼͎͍̠̮͓̹̹͉̤̰̗̙͕͇͔̱͕̭͈̳̗̭͔̘̖̺̮̜̠͖̘͓̳͕̟̠̱̫̤͓͔̘̰̲͙͍͇̙͎̣̼̗̖͙̯͉̠̟͈͍͕̪͓̝̩̦̖̹̼̠̘̮͚̟͉̺̜͍͓̯̳̱̻͕̣̳͉̻̭̭̱͍̪̩̭̺͕̺̼̥̪͖̦̟͎̻̰Ỏ̷͖͈̞̩͎̻̫̫̜͉̠̫͕̭̭̫̫̹̗̹͈̼̠̖͍͚̥͈̮̼͕̠̤̯̻̥̬̗̼̳̤̳̬̪̹͚̞̼̠͕̼̠̦͚̫͔̯̹͉͉̘͎͕̼̣̝͙̱̟̹̩̟̳̦̭͉̮̖̭̣̣̞̙̗̜̺̭̻̥͚͙̝̦̲̱͉͖͉̰̦͎̫̣̼͎͍̠̮͓̹̹͉̤̰̗̙͕͇͔̱͕̭͈̳̗̭͔̘̖̺̮̜̠͖̘͓̳͕̟̠̱̫̤͓͔̘̰̲͙͍͇̙͎̣̼̗̖͙̯͉̠̟͈͍͕̪͓̝̩̦̖̹̼̠̘̮͚̟͉̺̜͍͓̯̳̱̻͕̣̳͉̻̭̭̱͍̪̩̭̺͕̺̼̥̪͖̦̟͎̻̰ ͬͤ ͬͬͤ ͦͬͬͤ ͬͦͬͬͤ ͤͬͦͬͬͤ ͬͤͬͦͬͬͤ ͬͬͤͬͦͬͬͤ ͦͬͬͤͬͦͬͬͤ ͬͦͬͬͤͬͦͬͬͤ ͤͬͦͬͬͤͬͦͬͬͤ ͬͦͬͬͤͬͦͬͬͤ ͦͬͬͤͬͦͬͬͤ ͬͬͤͬͦͬͬͤ ͬͤͬͦͬͬͤ ͤͬͦͬͬͤ ͬͦͬͬͤ ͦͬͬͤ ͬͬͤ ͬͤ ͤ Ỏ̷͖͈̞̩͎̻̫̫̜͉̠̫͕̭̭̫̫̹̗̹͈̼̠̖͍͚̥͈̮̼͕̠̤̯̻̥̬̗̼̳̤̳̬̪̹͚̞̼̠͕̼̠̦͚̫͔̯̹͉͉̘͎͕̼̣̝͙̱̟̹̩̟̳̦̭͉̮̖̭̣̣̞̙̗̜̺̭̻̥͚͙̝̦̲̱͉͖͉̰̦͎̫̣̼͎͍̠̮͓̹̹͉̤̰̗̙͕͇͔̱͕̭͈̳̗̭͔̘̖̺̮̜̠͖̘͓̳͕̟̠̱̫̤͓͔̘̰̲͙͍͇̙͎̣̼̗̖͙̯͉̠̟͈͍͕̪͓̝̩̦̖̹̼̠̘̮͚̟͉̺̜͍͓̯̳̱̻͕̣̳͉̻̭̭̱͍̪̩̭̺͕̺̼̥̪͖̦̟͎̻̰Ỏ̷͖͈̞̩͎̻̫̫̜͉̠̫͕̭̭̫̫̹̗̹͈̼̠̖͍͚̥͈̮̼͕̠̤̯̻̥̬̗̼̳̤̳̬̪̹͚̞̼̠͕̼̠̦͚̫͔̯̹͉͉̘͎͕̼̣̝͙̱̟̹̩̟̳̦̭͉̮̖̭̣̣̞̙̗̜̺̭̻̥͚͙̝̦̲̱͉͖͉̰̦͎̫̣̼͎͍̠̮͓̹̹͉̤̰̗̙͕͇͔̱͕̭͈̳̗̭͔̘̖̺̮̜̠͖̘͓̳͕̟̠̱̫̤͓͔̘̰̲͙͍͇̙͎̣̼̗̖͙̯͉̠̟͈͍͕̪͓̝̩̦̖̹̼̠̘̮͚̟͉̺̜͍͓̯̳̱̻͕̣̳͉̻̭̭̱͍̪̩̭̺͕̺̼̥̪͖̦̟͎̻̰ ͬͤ ͬͬͤ ͦͬͬͤ ͬͦͬͬͤ ͤͬͦͬͬͤ ͬͤͬͦͬͬͤ ͬͬͤͬͦͬͬͤ ͦͬͬͤͬͦͬͬͤ ͬͦͬͬͤͬͦͬͬͤ ͤͬͦͬͬͤͬͦͬͬͤ ͬͦͬͬͤͬͦͬͬͤ ͦͬͬͤͬͦͬͬͤ ͬͬͤͬͦͬͬͤ ͬͤͬͦͬͬͤ ͤͬͦͬͬͤ ͬͦͬͬͤ ͦͬͬͤ ͬͬͤ ͬͤ ͤ Ỏ̷͖͈̞̩͎̻̫̫̜͉̠̫͕̭̭̫̫̹̗̹͈̼̠̖͍͚̥͈̮̼͕̠̤̯̻̥̬̗̼̳̤̳̬̪̹͚̞̼̠͕̼̠̦͚̫͔̯̹͉͉̘͎͕̼̣̝͙̱̟̹̩̟̳̦̭͉̮̖̭̣̣̞̙̗̜̺̭̻̥͚͙̝̦̲̱͉͖͉̰̦͎̫̣̼͎͍̠̮͓̹̹͉̤̰̗̙͕͇͔̱͕̭͈̳̗̭͔̘̖̺̮̜̠͖̘͓̳͕̟̠̱̫̤͓͔̘̰̲͙͍͇̙͎̣̼̗̖͙̯͉̠̟͈͍͕̪͓̝̩̦̖̹̼̠̘̮͚̟͉̺̜͍͓̯̳̱̻͕̣̳͉̻̭̭̱͍̪̩̭̺͕̺̼̥̪͖̦̟͎̻̰Ỏ̷͖͈̞̩͎̻̫̫̜͉̠̫͕̭̭̫̫̹̗̹͈̼̠̖͍͚̥͈̮̼͕̠̤̯̻̥̬̗̼̳̤̳̬̪̹͚̞̼̠͕̼̠̦͚̫͔̯̹͉͉̘͎͕̼̣̝͙̱̟̹̩̟̳̦̭͉̮̖̭̣̣̞̙̗̜̺̭̻̥͚͙̝̦̲̱͉͖͉̰̦͎̫̣̼͎͍̠̮͓̹̹͉̤̰̗̙͕͇͔̱͕̭͈̳̗̭͔̘̖̺̮̜̠͖̘͓̳͕̟̠̱̫̤͓͔̘̰̲͙͍͇̙͎̣̼̗̖͙̯͉̠̟͈͍͕̪͓̝̩̦̖̹̼̠̘̮͚̟͉̺̜͍͓̯̳̱̻͕̣̳͉̻̭̭̱͍̪̩̭̺͕̺̼̥̪͖̦̟͎̻̰ ͬͤ ͬͬͤ ͦͬͬͤ ͬͦͬͬͤ ͤͬͦͬͬͤ ͬͤͬͦͬͬͤ ͬͬͤͬͦͬͬͤ ͦͬͬͤͬͦͬͬͤ ͬͦͬͬͤͬͦͬͬͤ ͤͬͦͬͬͤͬͦͬͬͤ ͬͦͬͬͤͬͦͬͬͤ ͦͬͬͤͬͦͬͬͤ ͬͬͤͬͦͬͬͤ ͬͤͬͦͬͬͤ ͤͬͦͬͬͤ ͬͦͬͬͤ ͦͬͬͤ ͬͬͤ ͬͤ ͤ Ỏ̷͖͈̞̩͎̻̫̫̜͉̠̫͕̭̭̫̫̹̗̹͈̼̠̖͍͚̥͈̮̼͕̠̤̯̻̥̬̗̼̳̤̳̬̪̹͚̞̼̠͕̼̠̦͚̫͔̯̹͉͉̘͎͕̼̣̝͙̱̟̹̩̟̳̦̭͉̮̖̭̣̣̞̙̗̜̺̭̻̥͚͙̝̦̲̱͉͖͉̰̦͎̫̣̼͎͍̠̮͓̹̹͉̤̰̗̙͕͇͔̱͕̭͈̳̗̭͔̘̖̺̮̜̠͖̘͓̳͕̟̠̱̫̤͓͔̘̰̲͙͍͇̙͎̣̼̗̖͙̯͉̠̟͈͍͕̪͓̝̩̦̖̹̼̠̘̮͚̟͉̺̜͍͓̯̳̱̻͕̣̳͉̻̭̭̱͍̪̩̭̺͕̺̼̥̪͖̦̟͎̻̰Ỏ̷͖͈̞̩͎̻̫̫̜͉̠̫͕̭̭̫̫̹̗̹͈̼̠̖͍͚̥͈̮̼͕̠̤̯̻̥̬̗̼̳̤̳̬̪̹͚̞̼̠͕̼̠̦͚̫͔̯̹͉͉̘͎͕̼̣̝͙̱̟̹̩̟̳̦̭͉̮̖̭̣̣̞̙̗̜̺̭̻̥͚͙̝̦̲̱͉͖͉̰̦͎̫̣̼͎͍̠̮͓̹̹͉̤̰̗̙͕͇͔̱͕̭͈̳̗̭͔̘̖̺̮̜̠͖̘͓̳͕̟̠̱̫̤͓͔̘̰̲͙͍͇̙͎̣̼̗̖͙̯͉̠̟͈͍͕̪͓̝̩̦̖̹̼̠̘̮͚̟͉̺̜͍͓̯̳̱̻͕̣̳͉̻̭̭̱͍̪̩̭̺͕̺̼̥̪͖̦̟͎̻̰ ͬͤ ͬͬͤ ͦͬͬͤ ͬͦͬͬͤ ͤͬͦͬͬͤ ͬͤͬͦͬͬͤ ͬͬͤͬͦͬͬͤ ͦͬͬͤͬͦͬͬͤ ͬͦͬͬͤͬͦͬͬͤ ͤͬͦͬͬͤͬͦͬͬͤ ͬͦͬͬͤͬͦͬͬͤ ͦͬͬͤͬͦͬͬͤ ͬͬͤͬͦͬͬͤ ͬͤͬͦͬͬͤ ͤͬͦͬͬͤ ͬͦͬͬͤ ͦͬͬͤ ͬͬͤ ͬͤ ͤ Ỏ̷͖͈̞̩͎̻̫̫̜͉̠̫͕̭̭̫̫̹̗̹͈̼̠̖͍͚̥͈̮̼͕̠̤̯̻̥̬̗̼̳̤̳̬̪̹͚̞̼̠͕̼̠̦͚̫͔̯̹͉͉̘͎͕̼̣̝͙̱̟̹̩̟̳̦̭͉̮̖̭̣̣̞̙̗̜̺̭̻̥͚͙̝̦̲̱͉͖͉̰̦͎̫̣̼͎͍̠̮͓̹̹͉̤̰̗̙͕͇͔̱͕̭͈̳̗̭͔̘̖̺̮̜̠͖̘͓̳͕̟̠̱̫̤͓͔̘̰̲͙͍͇̙͎̣̼̗̖͙̯͉̠̟͈͍͕̪͓̝̩̦̖̹̼̠̘̮͚̟͉̺̜͍͓̯̳̱̻͕̣̳͉̻̭