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A final task is to make an enclosure for the WiMicro and carrier board. Having an exposed circuit board looks cool but as the WiMicro is expensive it s a good idea to protect it. Any small enclosure of the right size can be used. Figure 12-13 shows one example using a small project box. Small cut-aways on the sides are made where the antenna and Roomba cable come out. Figure 12-14 shows the finished enclosure mounted on the Roomba.
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A long list of registry keys appears. Scroll down the list until you nd Directory, and then double-click it.You may have dif culty locating this key because it is buried between other keys that have long names. On most Windows XP computers, the Directory key is located between the DirectFrame key and the DirectPlay key. Underneath Directory, single-click the Shell registry key (see Figure 8-4).
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This period with constant currents J g lasts typically tens to hundreds of milliseconds, after which the vapour is depleted appreciably and the conditions de ning our problem are no longer constant. The condensation rate C g is given by the gas kinetic theory as 4 R 2 Pv /(2 mkT )1/2 , where R is the radius of the g -mer, m is the mass of the vapour molecule, k is Boltzmann s constant and T is temperature. The evaporation rate, E g +1 , is not dependent on the pressure of the vapour; it is solely determined by the temperature and the size of the cluster. We can therefore calculate it at thermodynamic equilibrium (S = 1), where J = 0 and the balanced steady state holds:
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PROBLEMS 16.1 Growth rate. Let (1, a) X= (1, 1/a)
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number of unsuccessful retransmissions. The retransmission of a datagram is accomplished using the Go-back-to-N routine. Any number of outstanding bytes may not be acknowledged. When the destination station does acknowledge the receipt of a series of bytes, the source will look at the ACK number. All sequence numbers up to but not including the ACK number are considered received in good condition. This means that a source station can start the sequence number with 3 and then send two datagrams containing 100 TCP segment bytes each. When it receives an ACK from the destination of 203 (3 to 102, and then 102 to 202, leaving the ACK at 203 as the next expected byte), it will know that the data in both datagrams previously sent are considered received in good condition. The number of outstanding packets allowed is the next topic of discussion.
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