Plane 1 10 2

Author: p | 2025-04-24

Plane 1: This is a large, angled wing plane with 1 seat for the red circle customer. br / Plane 2: This is a large, straight wing plane with 2 seats for the blue square Plane 1: This is a large, angled wing plane with 1 seat for the red circle customer. br / Plane 2: This is a large, straight wing plane with 2 seats for the blue square customer. br /

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17. Comparison of simulated and measured radiation patterns of the proposed antenna (a) H-plane at 1.8 GHz (b) E-plane at 1.8 GHz (c) H-plane at 2.45 GHz (d) E-plane at 2.45 GHz (e) H-plane at 5.8 GHz (f) E-plane at 5.8 GHz. Figure 18. 3D radiation patterns of the wearable textile antenna: (a) 1.8 GHz, (b) 2.45 GHz, (c) 5.8 GHz. Figure 18. 3D radiation patterns of the wearable textile antenna: (a) 1.8 GHz, (b) 2.45 GHz, (c) 5.8 GHz. Figure 19. Comparison of measured radiation patterns in different bending scenarios (at 25 mm, 35 mm, and 45 mm) (a) H-plane at 1.8 GHz (b) E-plane at 1.8 GHz (c) H-plane at 2.45 GHz (d) E-plane at 2.45 GHz (e) H-plane at 5.8 GHz (f) E-plane at 5.8 GHz. Figure 19. Comparison of measured radiation patterns in different bending scenarios (at 25 mm, 35 mm, and 45 mm) (a) H-plane at 1.8 GHz (b) E-plane at 1.8 GHz (c) H-plane at 2.45 GHz (d) E-plane at 2.45 GHz (e) H-plane at 5.8 GHz (f) E-plane at 5.8 GHz. Figure 20. Simulated vs. measured efficiency of the proposed wearable antenna. Figure 20. Simulated vs. measured efficiency of the proposed wearable antenna. Figure 21. Simulated average SAR distribution on the cuboid phantom: at (a) 1.8 GHz (b) 2.45 GHz (c) 5.8 GHz. Figure 21. Simulated average SAR distribution on the cuboid phantom: at (a) 1.8 GHz (b) 2.45 GHz (c) 5.8 GHz. Figure 22. Link margin between Tx (proposed ant.) and Rx (monopole ant.) antennas at 1.8/2.45/5.8 GHz frequency bands. Figure 22. Link margin between Tx (proposed ant.) and Rx (monopole ant.) antennas at 1.8/2.45/5.8 GHz frequency bands. Table 1. Proposed antenna and previously reported textile antennas. Table 1. Proposed antenna and previously reported textile antennas. Ref. (Year)[19](2020)[20](2021)[21](2022)[22](2022)[23](2023)This WorkArea(mm2)65 × 6060 × 6060 × 6055 × 4084 × 6960 × 60Area(λ02)0.18 × 0.17(0.03)0.49 × 0.49(0.24)0.64 × 0.64(0.41)1.46 × 1.06(1.55)0.55 × 0.67(0.37)0.36 × 0.36(0.13)Frequency (GHz)0.868/2.452.45/3.452.4/3.32/3.93/5.882.4/51.8/2.45/5.8B.W. (%)NG/3.54.9/6.73.7/5.7/5.85/9.813.15/7617.2/39.1/19.6Peak gain (dBi)NG/−1.46.7/8.9−0.81/−2.81/−1.16/2.85.27.23.7/5.3/9.6SAR (W/Kg)1 gm/10 gmNG0.1/0.04(at 0.5 W)0.11/0.33(at 1 W)0.7/---(at 1 W)NG0.0796/0.07590.0575/0.05520.0226/0.0204(at 1 W) Table 2. Antenna’s design parameters (in mm). Table 2. Antenna’s design parameters (in mm). SymbolValueSymbolValueSymbolValueSymbolValueL60L718WF3.6X118.5LP50L813W1–W66.0X234.5LF15L918W705X321.0L125L1012W811X417.0L209L113.0W903X57.5L303L1208W1011Y14.0L407L1338W1103Y2–Y45.0L503W60W1206 L63.5WP40W1331 Table 3. Boresight peak gain values (dBi). Table 3. Boresight peak gain values (dBi). Frequency(GHz)Simulation(Chest Phantom)Measured(on Human Chest)1.83.72.82.455.34.65.89.68.2 Table 4. Boresight peak gain values (dBi) at different bending radii: 45 mm, 35 mm, and 25 mm. Table 4. Boresight peak gain values (dBi) at different bending radii: 45 mm, 35 mm, and 25 mm. Frequency (GHz)At 45 mmAt 35 mmAt 25 mm1.82.12.00.2 2.454.54.63.25.88.28.17.8 Table 5. Maximum SAR of the proposed antenna (at 1 W input power). Table 5. Maximum SAR of the proposed antenna (at 1 W input power). Frequency(GHz)Maximum SAR (on Phantom)1 gm10 gm1.80.07960.07592.450.05750.05525.80.02260.0204 Table 6. Link budget parameters. Table 6. Link budget parameters. Transmitter Frequency (GHz)1.8/2.45/5.8GtAntenna gain (dBi)3.7/5.3/9.6PtTransmitted power (dBm)16 EIRP (dBm)19.7/21.3/25.6ReceiverGrReceiver antenna gain (dBi)(external antenna)2.15ToAmbient temperature (K)293 Boltzmann constant1.38 × 10−23NoNoise power density (dB/Hz)−203.9Signal qualityBrBit rate (Mbps)0.250, 1, 10 Eb/NoIdeal PSK (dB)9.6GcCoding gain (dB)0GdFixing deterioration (dB)2.5 Disclaimer/Publisher’s Note: The statements, opinions and data contained in

How to combine 2 planes together into 1 object

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Word Problems: Algebra 1 and 2 - Math Plane

Nity. If the attempt fails, the opponent cannot react to trip the death dog. Poison Immunity (Ex): All demodands are immune to poison. Scent (Ex): A death dog can detect approaching enemies, sniff out hidden foes, and track by sense of smell. Outsider Traits: A demodand has darkvision (60-foot range, or 120-foot range for shators). It cannot be raised Skills: Death dogs have a +4 racial bonus on Listen and or resurrected. Spot checks. FARASTU DEMODAND Medium-Size Outsider (Evil, Extraplanar) The prison plane of Carceri is home to many sorts of crea- Hit Dice: 11d8+22 (71 hp) tures. Its petitioners plot endlessly to find an escape from Initiative: +5 their hellish existence, but watchful eyes see to it that Speed: 40 ft. such plots never succeed. The demodands are the self- AC: 25 (+1 Dex, +14 natural), touch 11, flat-footed 24 appointed keepers of the Carcerian population. They are the Base Attack/Grapple: +11/+23 jailers and wardens, although they themselves are mostly Attack: Claw +15 melee prisoners of the plane as surely as the petitioners they try Full Attack: 2 claws +15 melee and bite +10 melee to watch. Damage: Claw 1d4+4, bite 1d6+2 Face/Reach: 5 ft./5 ft. Of course, the number of petitioners on Carceri makes Special Attacks: Adhesive slime, improved grab, rage, the demodands’ task an impossible one, but this doesn’t trouble them. After their exile from some other plane spell-like abilities, summon demodand many millennia ago, for a reason long forgotten (although Special Qualities: Acid immunity, cold resistance 10, DR the kelubars and shators both agree it was the fault of the farastus), the demodands were exiled to the Tarterian 10/+1, outsider traits, poison immunity, fire resistance Depths of Carceri to serve a penance. Once there, they 10, scent, SR 23 took it upon themselves to set up. Plane 1: This is a large, angled wing plane with 1 seat for the red circle customer. br / Plane 2: This is a large, straight wing plane with 2 seats for the blue square

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Booster Performance License for 4430 Series Router for up to 3.4 Gbps CEF* FL-44-BOOST-K9 (=) Booster Performance License for 4450 Series Router for up to 3.8 Gbps CEF* FL-4460-BOOST-K9 (=) Booster Performance License for 4460 Series Router for up to 10 Gbps CEF* * Test results for IP Routing (CEF) @ IMIX Ordering information The Cisco ISR 4000 Family is orderable and shipping. To place an order, refer to Table 9 below and visit the Cisco Ordering Home Page. Table 9. Cisco ISR 4000 Series ordering information Product Name Product Description ISR4461/K9 Cisco ISR 4461 with 4 onboard GE, 3 NIM slots, 1 ISC slot, 3 SM slots, 8 GB Flash Memory default, 2 GB DRAM default (data plane), 4 GB DRAM default (control plane) ISR4451-X/K9 ISR 4451 with 4 onboard GE, 3 NIM slots, 1 ISC slot, 2 SM slots, 8 GB Flash Memory default, 2 GB DRAM default (data plane), 4 GB DRAM default (control plane) ISR4431/K9 ISR 4431 with 4 onboard GE, 3 NIM slots, 1 ISC slot, 8GB Flash Memory default, 2 GB DRAM default (data plane), 4 GB DRAM default (control plane) ISR4351/K9 ISR 4351 with 3 onboard GE, 3 NIM slots, 1 ISC slot, 2 SM slots, 4 GB Flash Memory default, 4 GB DRAM default ISR4331/K9 ISR 4331 with 3 onboard GE, 2 NIM slots, 1 ISC slot, 1 SM slots, 4 GB Flash Memory default, 4 GB DRAM default ISR4321/K9 ISR 4321 with 2 onboard GE, 2 NIM slots, 1 ISC slot, 4 GB Flash Memory default, 4 GB DRAM default ISR4221/K9 ISR 4221 with 2 onboard GE, 2 NIM slots, 1 ISC slot, 8 GB Flash Memory default, 4 GB DRAM default ISR4221X/K9 ISR 4221 with 2 onboard GE, 2 NIM slots, 1 ISC slot, 8 GB Flash Memory default, 8 GB DRAM default For additional product numbers, including the Cisco 4000 Family bundle offerings, please contact your local Cisco account representative. To place an order, visit the Cisco Ordering Home Page. To download software, visit the Cisco Software Center. Integrated Services Router Migration Options The Cisco ISR 4000 Family is included in the standard Cisco Technology Migration Program (TMP). Refer to and contact your local Cisco account representative for program details. Warranty information The Cisco ISR 4000 Series Integrated Services Routers have a 90-day limited liability warranty. Product sustainability Information about Cisco’s Environmental, Social and Governance (ESG) initiatives and performance

PROBLEMS IN PLANE AND SOLID GEOMETRY v.1 Plane

For the Support, choose the plane created in Step 2. Place this in any XYZ direction that is the least perpendicular to the support plane. The length Start and End values can be any values chosen by you.5) Next, we will define the contour of the emboss walls. Create a line using “Angle/Normal to Curve” as the Line type. Use the line created in the previous step for the Curve, the Plane created in Step 2 for the support, and the point created in Step 1 for the Point. The angle will define one side of your emboss wall and can be chosen by you. The length can be chosen by you too, however, it must be long enough to intersect the other walls that will be created shortly. These values can be adjusted at any point in the process and will be reviewed later on.6) Parallel this line using the plane created in Step 2 as the support. The offset value, or “Constant” will define where you want your wall to be with regards to the emboss center.7) Create a sweep with “Line as the Profile” Type using the newly created Parallel as well as the plane created in Step 2 the as input for the Reference surface and Guide curve 1, respectively.8) Copy the Side 1 geoset and paste in the Emboss 1 geoset. Rename this geoset “Side 2”. Renaming can be done by right clicking the geoset, selecting Properties, and then going to the tab named “Feature Properties”. Here, there will be a box to edit the Feature Name. Edit this box to change the name of the geoset.Note: The copy/paste portion of this process can be replicated using Powercopy. This process will be covered shortly using the finished emboss geometry.9) Edit the Parallel line within Side 2 so that it is at the desired offset for your next wall. Reverse parallel and sweep directions if necessary.10) You now have 2 of the 4 walls of your emboss.11) Copy Side 1 and Side 2, paste them in Emboss 1 and rename them Side 3 and Side 4,

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DimensionalD. Perpendicular on OC (produced) from point object, e.g., a ring, a disc, any plane sheet, etc.D is DN. Moment of inertia of the object about N³the axis ACB is IC (DC)2 dm , and about Mthe axis MOP it is IO ³ DO 2 dm . ? IO ³ DO 2 dm ³ >DN@2 >NO@2 dm³ >DN@2 [NC]2 2.NC.CO >CO@2 dm Fig. 1.16: Theorem of perpendicular axes. 15Figure 1.16 shows a rigid laminar object able (II) Consider any two mutually perpendicularto rotate about three mutually perpendicularaxes x, y and z. Axes x and y are in the plane diameters x and y of the flywheel. If theof the object while the z axis is perpendicularto it, and all are concurrent at O. Consider a flywheel rotates about these diameters, thesemass element dm located at any point P. PM= y and PN = x are the perpendiculars drown three axes (own axis and two diameters) willfrom P respectively on the x and y axes. Therespective perpendicular distances of point be mutually perpendicular and concurrent.M from x, y and z axes will then be y, x and Thus, perpendicular axes theorem is y 2 + x2 . If Ix, Iy and Iz are the respectivemoment of inertias of the body about x, y and applicable. Let Id be the moment of inertiaz axes, we can write, of the flywheel, when rotating about its ³ ³? I x y 2dm, I y x2dm and diameter. ? Id I x I y Thus, according to the theorem of ³Iz y2 x2 dm perpendicular axes, 1 MR2 Iz 2 Ix Iy 2Id ? Id 1 MR2 4³ ³? I z y 2dm x2dm I x I yThis is the mathematical form of thetheorem of perpendicular axes. As the diameter passes through the centre It states that, “The moment of inertia of mass of the (uniform) disc, Id = IC(Iz) of a laminar object about an axis (z) Consider a tangent in the plane of the discperpendicular to its plane is the sum of itsmoment of inertias about two mutuallyperpendicular axes (x and y) in its plane, allthe three axes being concurrent”.Example 1.7: A flywheel is a mechanical and parallel to this diameter. It is at thedevice specifically designed to efficiently distance h = R from the diameter. Thus,store rotational energy. For a particular parallel axes theorem is applicable aboutmachine it is in the form of a uniform 20 kg these two axes.disc of diameter 50 cm, able to rotate about ∴ IT, parallel = Io = Ic + Mh2 = Id + MR2its own axis. Calculate its kinetic energy = 1 MR2 + MR2 = 5 MR2when rotating at 1200 rpm. Use S 2 10. 4 4Calculate its moment of inertia, in case it is 5 5 4 4rotated about a tangent in its plane. ∴ IT, parallel = MR2 = 20 × 0.252Solution: (I) As the flywheel is in the form = 1.5625 kg m2of a uniform disc rotating about its own 1.8

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Plane) And 1 If bitVal = 1 Then Dim byteIndex As Integer, bitIndex As Integer byteIndex = x \ 8 bitIndex = 7 - (x Mod 8) Dim currByte As Integer currByte = Asc(Mid$(planeData(plane), byteIndex + 1, 1)) currByte = currByte Or _ShL(1, bitIndex) Mid$(planeData(plane), byteIndex + 1, 1) = Chr$(currByte) End If Next plane Next x ' RLE kódování pro obě roviny daného řádku RLE encode each plane for the current line Dim p As Integer For p = 0 To 1 Dim rawLine As String, encoded As String rawLine = planeData(p) encoded = "" Dim iPos As Integer iPos = 1 Do While iPos Dim currentByte As Integer, count As Integer currentByte = Asc(Mid$(rawLine, iPos, 1)) count = 1 Do While (iPos + count If Asc(Mid$(rawLine, iPos + count, 1)) = currentByte Then count = count + 1 Else Exit Do End If Loop If (count = 1) And (currentByte encoded = encoded + Chr$(currentByte) Else encoded = encoded + Chr$(192 + count) + Chr$(currentByte) End If iPos = iPos + count Loop ' Zápis RLE kódovaných dat pro danou rovinu Write encoded data for this plane Put #fileNum, , encoded Next p Next y _Source s Close #fileNumEnd Sub' -------------------------------------------------------------------------------------------------------' SUB SavePCX16Clr – uloží obrázek jako 16barevný (4bitový) PCX soubor.' SUB SavePCX16Clr – saves the image as a 16-color (4-bit) PCX file.' Vstupní parametry: image (ukazatel na obrázek s indexovanými hodnotami 0–15), fileName (název souboru)' Input parameters: image (image pointer with indexed values 0–15), fileName (output file name)' -------------------------------------------------------------------------------------------------------Sub SavePCX16Clr (image As Long, fileName As String) ' Získání rozměrů obrázku / Get image dimensions Dim width As Integer, height As Integer width = _Width(image) height = _Height(image) ' Výpočet bajtů na rovinu: (width+7)\8 a zarovnání na sudé číslo Calculate bytes per line and align to even number Dim bytesPerLine As Integer bytesPerLine = (width + 7) \ 8 If (bytesPerLine Mod 2) 0 Then bytesPerLine = bytesPerLine + 1 status = GetUsedColors(image) myMask$ = TransformMask ' ----------------------------------------------------------- ' Sestavíme paletu 16 EGA barev Build a 16-color EGA palette ' EGA barvy: ' 0: černá (0,0,0) ' 1: modrá (0,0,170) ' 2: zelená (0,170,0) ' 3: cyan (0,170,170) ' 4: červená (170,0,0) ' 5: magenta (170,0,170) ' 6: hnědá (170,85,0) ' 7: světle šedá (170,170,170) ' 8: tmavě šedá (85,85,85) ' 9: jasně modrá (85,85,255) ' 10: jasně zelená (85,255,85) ' 11: jasně cyan (85,255,255) ' 12: jasně červená (255,85,85) ' 13: jasně magenta (255,85,255) ' 14: žlutá (255,255,85) ' 15: bílá (255,255,255) ' ------------------------------ Dim paletteData As String paletteData = "" ' V PCX 16barevném formátu se standardně očekává EGA paleta, ale zde může být nahrazena barvami z obrázku. ' In PCX 16-color format, the standard EGA. Plane 1: This is a large, angled wing plane with 1 seat for the red circle customer. br / Plane 2: This is a large, straight wing plane with 2 seats for the blue square

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Intersect view planeOne point, two point, three point perspective Three point perspective: Three principal axes intersect view planeOne point, two point, three point perspective View Plane Three point Two point One point3D Projections Rays converge on eye position Rays parallel Perspective Parallel Rays at angle to view plane Rays perpendicular to view plane Orthographic Oblique View plane aligned with principal axes View plane not aligned with principal axes Cabinet Cavalier Elevations Axonometric Isometric Trimetric DimetricFront Elevation • Parallel Orthogonal Elevation Front elevation of tallest buildings in the world From members.iinet.net.au/ ~paulkoh ElevationsIsometric View • In isometric view, the three principal axes of the object intersect the view plane at equal distance. Therefore, when projected, they are 120o apart. Projections Rays converge on eye position Rays parallel Perspective Parallel Rays at angle to view plane Rays perpendicular to view plane Orthographic Oblique Cabinet Cavalier Elevations Axonometric IsometricOblique projections • Projection lines are at an angle to the view plane. • Let the angle be a be the angle the projection line makes with the view plane. • tan a = 1 (or, a = 45o) called cavalier projection • tan a = 2 (or, a = 63.4o) called cabinet projection 1 2 a a 1 1 cavalier cabinet 1 1/2 1 1Orthographic Parallel Projection Matrix • Transform each vertex from Viewing Coordinates into Normalized Coordinates using orthographic projection • Suppose that a point is (x,y,z) in Viewing Coordinates, what’s the transformation necessary to transform it to (x’,y’,z’) in Normalized Coordinates? • Given: the dimensions of the view window: xwmin, xwmax, ywmin, ywmax • Orthogonal Projection Matrix on p. 362. • Basically Translate center of view to origin and then Scale to (-1,1) cube • Translate by -(min+max)/2, then scale by 2/(max-min). xwmax+ xwmin xwmax - xwmin 2 xwmax - xwmin 0

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Attacks: Improved grab, shadow strands, spell-like Treasure: Standard Alignment: Usually chaotic neutral abilities, Strength damage Advancement: By character class Special Qualities: All-around vision, cold resistance 10, Level Adjustment: 5 darkvision 60 ft., DR 5/+1, fast healing 3, shadow jump, Dark stalkers are the enigmatic leaders of the dark ones, shadowstuff armor, SR 16, sunlight vulnerability, tenta- although it is difficult to tell whether they actually belong to cle regeneration the same race as dark creepers. Saves: Fort +4, Ref +7, Will +9 Abilities: Str 17, Dex 18, Con 12, Int 15, Wis 16, Cha 17 Dark stalkers appear human in almost every way, and they Skills: Climb +18, Concentration +7, Hide +21, Listen +10, stand head and shoulders above their dark creeper kin. Move Silently +11, Search +9, Spot +10 They have dusky skin and lack the hooves of dark creepers. Feats: Alertness, Combat Reflexes, Power Attack, Weapon Dark stalkers cover themselves head to toe in black, somber Finesse clothing and never willingly reveal their faces. Climate/Terrain: Any land (Plane of Shadow) Organization: Solitary or coven (2–4)Combat Challenge Rating: 10 Dark stalkers lead dark creepers into battle, and they place Treasure: Standard poison on their short swords to deal terrible wounds. They Alignment: Usually neutral evil use fog cloud to escape from powerful opponents. Advancement: 10–15 HD (Medium-size); 16–27 HD Fog Cloud (Sp): Twice per day, a dark stalker can use fog (Large) cloud as the spell cast by a 5th-level sorcerer. Poison Use (Ex): Dark stalkers never risk accidentally Darkweavers are sinister and alien beings from the Plane poisoning themselves when applying poison to a blade. of Shadow that have found that the Material Plane offers They typically use shadow essence poison (Fort DC 17, ini- far more plentiful hunting grounds than their home. Rela- tial damage 1 point. Plane 1: This is a large, angled wing plane with 1 seat for the red circle customer. br / Plane 2: This is a large, straight wing plane with 2 seats for the blue square Plane 1: This is a large, angled wing plane with 1 seat for the red circle customer. br / Plane 2: This is a large, straight wing plane with 2 seats for the blue square customer. br /

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Section formula,We know that the coordinates of the point R, which divide the line segment joining two points, P (x1, y1, z1) and Q (x2, y2, z2), externally in the ratio m: n, are given byUpon comparing, we havex1 = -2, y1 = 3, z1 = 5;x2 = 1, y2 = -4, z2 = 6 andm = 2, n = 3So, the coordinates of the point which divide the line segment joining the points P (– 2, 3, 5) and Q (1, – 4, 6) in the ratio 2: 3 externally are given by∴ The coordinates of the point which divides the line segment joining the points (-2, 3, 5) and (1, -4, 6) are (-8, 17, 3).2. Given that P (3, 2, – 4), Q (5, 4, – 6) and R (9, 8, –10) are collinear. Find the ratio in which Q divides PR.Solution:Let us consider Q divides PR in the ratio k: 1.By using the section formula,We know that the coordinates of the point R, which divides the line segment joining two points, P (x1, y1, z1) and Q (x2, y2, z2), internally in the ratio m : n, are given byUpon comparing, we havex1 = 3, y1 = 2, z1 = -4;x2 = 9, y2 = 8, z2 = -10 andm = k, n = 1So, we have9k + 3 = 5 (k+1)9k + 3 = 5k + 59k – 5k = 5 – 34k = 2k = 2/4= ½Hence, the ratio in which Q divides PR is 1: 2.3. Find the ratio in which the YZ-plane divides the line segment formed by joining the points (–2, 4, 7) and (3, –5, 8).Solution:Let the line segment formed by joining the points P (-2, 4, 7) and Q (3, -5, 8) be PQ.We know that any point on the YZ-plane is of the form (0, y, z).So now, let R (0, y, z) divide the line segment PQ in the ratio k: 1.Then,Upon comparing, we havex1 = -2, y1 = 4, z1 = 7x2 = 3, y2 = -5, z2 = 8 andm = k, n = 1By