数控技术与装备 第三版 韩建梅主编 第一章课后习题

第1章数控机床基础知识1-1数控机床具有哪些特点？1、具有柔性化和灵活性。

当改变加工工件时，只要改变数控程序即可，所以合适产品更新换代快的要求。

2、可以采用较高的切削速度和进给速度（或进给量）。

3、加工精度高，质量稳定。

数控机床本身精度高，此外还可以利用参数的修改进行。

精度校正和补偿。

1-2数控机床由哪几部分组成？1、程序及程序载体。

数控装置由数控机床自动加工工件的工作指令组成。

包括切靴过程中所需要的机械运动，工件，轮廓，尺寸。

工艺参数等加工信息。

2、输入装置。

输入装置的作用是将程序载体上的数控代码信息转化成相映的电脉冲信号。

并传送至数控装置的储存器。

3、数控装置。

数控装置是数控机床的核心。

包括微型计算机，各种接口电路，显示器。

和硬件及相应软件。

4、强电控制装置。

5、伺服控制装置。

6、机床的机械部件。

1-3伺服控制装置的主要作用是什么？伺服控制装置主要完成机床的运动，其运动控制。

包括进给运动主轴，运动位置控制等。

1-4先进制造技术包括哪些内容？1-5数控机床按伺服控制系统和加工运动轨迹方式分为哪几类，各有什么特点？一、按控制方式分最常用的数控机床可分为以下三类:1、开环数控机床，这类机床通常为经济型、中小型数控机床。

具有结构简单，价格低廉，调试方便等优点。

但通常输出的转距大小受到限制，而且当输入的频率较高时，容易湿不男女，实现运动部件的控制，因此已不能完全满足数控机床提高功率。

运动速度和加工精度的要求。

2、闭环数控机床，相比开环数控机床，闭环数控机床的精度更高。

速度更快，驱动功率更大，但是这类机床价格昂贵，对机床结构及传动链依然提出了严格的要求。

3、半闭环数控机床。

半闭环数控机床可以获得比开环系统更高的加工精度。

但由于机械传动链的误差无法得到消除或校正。

因此它的位移精度比闭环系统低，大多数数控机床采用半闭环控制系统。

二、按机械加工运动轨迹方式分类1、点位控制数控机床（孔加工）点位控制数控机床的要求点在空间的位置准确。

数控技术课后完整答案1.1数控机床的工作流程是什么？数控加工程序的编制，输入，译码，刀具补偿，插补，位置控制和机床加工1.2数控机床有哪几部分组成？各部分的基本功能是什么？输入输出设备：主要实现程序编制、程序和数据的输入以及显示、存储和打印。

数控装置：接收来自输入设备的程序和数据，并按输入信息的要求完成数值的计算、逻辑判断和输入输出控制等功能。

（多坐标控制，插补功能，程序输入、编辑和修改功能、故障自诊断功能、补偿功能、信息转换功能、多种加工方式选择、辅助功能、显示功能、通信和联网功能）1.3什么是点位控制、直线控制、轮廓控制数控机床？三者如何区别？点位控制数控机床点位控制是指道具从某一位置移到下一个位置的过程中，不考虑其运动轨迹，只要求道具能最终准确达到目标位置。

直线控制数控机床这类数控机床不仅要保证点与点之间的准确定位，而且要控制两相关点之间的位移速度和路线。

轮廓控制数控机床这类机床的数控装置能够同时控制两轴或两个以上的轴，对未知和速度进行严格的不间断控制。

区别：点位控制不考虑运动轨迹，直线控制要求保证两点之间的精确定位，轮廓控制对于位置和速度有严格的要求。

1.4数控机床有哪些特点？加工零件的适应性强，灵活性好；加工精度高，产品质量稳定；生产率高；减少工人劳动强度；生产管理水平高1.5按伺服系统的控制原理分类，分为哪几类数控机床？各有何特点？开环控制的数控机床受步进电动机的步距精度和工作频率以及传动机构的传动精度的影响，速度和精度都较低。

结构简单、成本较低、调试维修方便闭环控制的数控机床定位精度高、速度调节快，工作台惯量大所以系统设计和调整存在困难，系统稳定性受到不利影响半闭环控制的数控机床控制精度没有闭环高，但机床工作的稳定性却有毒大惯量工作台被排除在控制环外而提高，调试方便什么是控制编程？手工编程的内容有哪些？从零件图样到制成控制介质的全部过程。

分析零件图样，确定加工工艺过程，数值计算，编写零件加工程序，制作控制介质，程序校验，试切削数控编程有哪几种方法？各有何特点？手工编程对于几何形状较为简单的零件，数值计算较简单，程序段不错，采用手工编彻骨较容易完成，而且经济、及时。

数控技术第三版章节练习答案第一章绪论1.1数控机床的工作流程是什么？答：数控机床由输入装置、CNC装置、伺服系统和机床的机械部件构成。

数控加工程序的编制-输入-译码-刀具补偿-插补-位置控制和机床加工1.2 数控机床由哪几部分组成？各部分的基本功能是什么？答：组成：由输入输出设备、数控装置、伺服系统、测量反馈装置和机床本体组成输入输出设备：实现程序编制、程序和数据的输入以及显示、存储和打印数控装置：接受来自输入设备的程序和数据，并按输入信息的要求完成数值计算、逻辑判断和输入输出控制等功能。

伺服系统：接受数控装置的指令，驱动机床执行机构运动的驱动部件。

测量反馈装置：检测速度和位移，并将信息反馈给数控装置，构成闭环控制系统。

机床本体：用于完成各种切削加工的机械部分。

1.3．什么是点位控制、直线控制、轮廓控制数控机床？三者如何区别？答：（1）点位控制数控机床特点：只与运动速度有关，而与运动轨迹无关。

如：数控钻床、数控镗床和数控冲床等。

（2）直线控制数控机床特点：a.既要控制点与点之间的准确定位，又要控制两相关点之间的位移速度和路线。

b.通常具有刀具半径补偿和长度补偿功能，以及主轴转速控制功能。

如：简易数控车床和简易数控铣床等。

（3）连续控制数控机床（轮廓控制数控机床）：对刀具相对工件的位置，刀具的进给速度以及它的运动轨迹严加控制的系统。

具有点位控制系统的全部功能，适用于连续轮廓、曲面加工。

1.4．数控机床有哪些特点？答：a．加工零件的适用性强，灵活性好；b．加工精度高，产品质量稳定；c．柔性好；d．自动化程度高，生产率高；e．减少工人劳动强度；f．生产管理水平提高。

适用范围：零件复杂、产品变化频繁、批量小、加工复杂等1.5．按伺服系统的控制原理分类，分为哪几类数控机床？各有何特点？答：（1）开环控制的数控机床；其特点：a.驱动元件为步进电机；b.采用脉冲插补法：逐点比较法、数字积分法；c.通常采用降速齿轮；d. 价格低廉，精度及稳定性差。

第一章1.(Q1) What is the difference between a host and an end system? List the types of endsystems. Is a Web server an end system?Answer: There is no difference. Throughout this text, the words “host” and “end system” are used interchangeably. End systems include PCs, workstations, Web servers, mail servers, Internet-connected PDAs, WebTVs, etc.2.(Q2) The word protocol is often used to describe diplomatic relations. Give an example of adiplomatic protocol.Answer: Suppose Alice, an ambassador of country A wants to invite Bob, an ambassador of country B, over for dinner. Alice doesn’t simply just call Bob on the phone and say, come to our dinner table now”. Instead, she calls Bob and suggests a date and time. Bob may respond by saying he’s not available that particular date, but he is available another date. Alice and Bob continue to send “messages” back and forth until they agree on a date and time. Bob then shows up at the embassy on the agreed date, hopefully not more than 15 minutes before or after the agreed time. Diplomatic protocols also allow for either Alice or Bob to politely cancel the engagement if they have reasonable excuses.3.(Q3) What is a client program? What is a server program? Does a server program requestand receive services from a client program?Answer: A networking program usually has two programs, each running on a different host, communicating with each other. The program that initiates the communication is the client.Typically, the client program requests and receives services from the server program.4.(Q4) List six access technologies. Classify each one as residential access, company access, ormobile access.Answer:1. Dial-up modem over telephone line: residential; 2. DSL over telephone line: residential or small office; 3. Cable to HFC: residential; 4. 100 Mbps switched Etherent: company; 5. Wireless LAN: mobile; 6. Cellular mobile access (for example, 3G/4G): mobile5.(Q5) List the available residential access technologies in your city. For each type of access,provide the advertised downstream rate, upstream rate, and monthly price.Answer: Current possibilities include: dial-up (up to 56kbps); DSL (up to 1 Mbps upstream, up to 8 Mbps downstream); cable modem (up to 30Mbps downstream, 2 Mbps upstream.6.(Q7) What are some of the physical media that Ethernet can run over?Answer: Ethernet most commonly runs over twisted-pair copper wire and “thin” coaxial cable. It also can run over fibers optic links and thick coaxial cable.7.(Q8) Dial-up modems, HFC, and DSL are all used for residential access. For each of theseaccess technologies, provide a range of transmission rates and comment on whether the transmission rate is shared or dedicated.Answer:Dial up modems: up to 56 Kbps, bandwidth is dedicated; ISDN: up to 128 kbps, bandwidth is dedicated; ADSL: downstream channel is .5-8 Mbps, upstream channel is up to1 Mbps, bandwidth is dedicated; HFC, downstream channel is 10-30 Mbps and upstreamchannel is usually less than a few Mbps, bandwidth is shared.8.(Q13) Why is it said that packet switching employs statistical multiplexing? Contraststatistical multiplexing with the multiplexing that takes place in TDM.Answer: In a packet switched network, the packets from different sources flowing on a link do not follow any fixed, pre-defined pattern. In TDM circuit switching, each host gets the same slot in a revolving TDM frame.9.(Q14) Suppose users share a 2Mbps link. Also suppose each user requires 1Mbps whentransmitting, but each user transmits only 20 percent of the time. (See the discussion of statistical multiplexing in Section 1.3.)a.When circuit switching is used, how many users can be supported?b.For the remainder of this problem, suppose packet switching is used. Why will there beessentially no queuing delay before the link if two or fewer users transmit at the same time? Why will there be a queuing delay if three users transmit at the same time?c.Find the probability that a given user is transmitting.d.Suppose now there are three users. Find the probability that at any given time, allthree users are transmitting simultaneously. Find the fraction of time during which the queue grows.Answer:a. 2 users can be supported because each user requires half of the link bandwidth.b.Since each user requires 1Mbps when transmitting, if two or fewer users transmitsimultaneously, a maximum of 2Mbps will be required. Since the available bandwidth of the shared link is 2Mbps, there will be no queuing delay before the link. Whereas, if three users transmit simultaneously, the bandwidth required will be 3Mbps which is more than the available bandwidth of the shared link. In this case, there will be queuing delay before the link.c.Probability that a given user is transmitting = 0.2d.Probability that all three users are transmitting simultaneously=(33)p3(1−p)0=0.23=0.008. Since the queue grows when all the users are transmitting, the fraction oftime during which the queue grows (which is equal to the probability that all three users are transmitting simultaneously) is 0.008.10.(Q16) Consider sending a packet from a source host to a destination host over a fixed route.List the delay components in the end-to-end delay. Which of these delays are constant and which are variable?Answer:The delay components are processing delays, transmission delays, propagation delays, and queuing delays. All of these delays are fixed, except for the queuing delays, which are variable.11.(Q19) Suppose Host A wants to send a large file to Host B. The path from Host A to Host Bhas three links, of rates R1 = 250 kbps, R2 = 500 kbps, and R3 = 1 Mbps.a.Assuming no other traffic in the network, what is the throughput for the file transfer.b.Suppose the file is 2 million bytes. Roughly, how long will it take to transfer the file toHost B?c.Repeat (a) and (b), but now with R2 reduced to 200 kbps.Answer:a.250 kbpsb.64 secondsc.200 kbps; 80 seconds12.(P2) Consider the circuit-switched network in Figure 1.8. Recall that there are n circuits oneach link.a.What is the maximum number of simultaneous connections that can be in progress atany one time in this network?b.Suppose that all connections are between the switch in the upper-left-hand cornerand the switch in the lower-right-hand corner. What is the maximum number ofsimultaneous connections that can be in progress?Answer:a.We can n connections between each of the four pairs of adjacent switches. This gives amaximum of 4n connections.b.We can n connections passing through the switch in the upper-right-hand corner andanother n connections passing through the switch in the lower-left-hand corner, giving a total of 2n connections.13.(P4) Review the car-caravan analogy in Section 1.4. Assume a propagation speed of 50km/hour.a.Suppose the caravan travels 150 km, beginning in front of one tollbooth, passingthrough a second tollbooth, and finishing just before a third tollbooth. What is theend-to-end delay?b.Repeat (a), now assuming that there are five cars in the caravan instead of ten.Answer: Tollbooths are 150 km apart, and the cars propagate at 50 km/hr, A tollbooth services a car at a rate of one car every 12 seconds.a.There are ten cars. It takes 120 seconds, or two minutes, for the first tollbooth to servicethe 10 cars. Each of these cars has a propagation delay of 180 minutes before arriving at the second tollbooth. Thus, all the cars are lined up before the second tollbooth after 182 minutes. The whole process repeats itself for traveling between the second and third tollbooths. Thus the total delay is 364 minutes.b.Delay between tollbooths is 5*12 seconds plus 180 minutes, i.e., 181minutes. The totaldelay is twice this amount, i.e., 362 minutes.14.(P5) This elementary problem begins to explore propagation delay and transmission delay,two central concepts in data networking. Consider two hosts, A and B, connected by a single link of rate R bps. Suppose that the two hosts are separated by m meters, and suppose the propagation speed along the link is s meters/sec. Host A is to send a packet of size L bits to Host B.a.Express the propagation delay, d prop , in terms of m and s.b.Determine the transmission time of the packet, d trans , in terms of L and R.c.Ignoring processing and queuing delays, obtain an expression for the end-to-enddelay.d.Suppose Host A begins to transmit the packet at time t = 0. At time t = d trans , whereis the last bit of the packet?e.Suppose d prop is greater than d trans . At time t = d trans , where is the first bit of thepacket?f.Suppose d prop is less than d trans . At time t = d trans , where is the first bit of thepacket?g.Suppose s = 2.5*108, L = 100bits, and R = 28kbps. Find the distance m so that d propequals d trans .Answer:a. d prop = m/s seconds.b. d trans = L/R seconds.c. d end-to-end = (m/s + L/R) seconds.d.The bit is just leaving Host A.e.The first bit is in the link and has not reached Host B.f.The first bit has reached Host B.g.Wantm=LRS=10028∗103(2.5∗108)=893 km.15.(P6) In this problem we consider sending real-time voice from Host A to Host B over apacket-switched network (VoIP). Host A converts analog voice to a digital 64 kbps bit stream on the fly. Host A then groups the bits into 56-Byte packets. There is one linkbetween Host A and B; its transmission rate is 500 kbps and its propagation delay is 2 msec.As soon as Host A gathers a packet, it sends it to Host B. As soon as Host B receives an entire packet, it converts the packet’s bits to an analog signal. How much time elapses from the time a bit is created (from the original analog signal at Host A) until the bit is decoded (as part of the analog signal at Host B)?Answer: Consider the first bit in a packet. Before this bit can be transmitted, all of the bits in the packet must be generated. This requires56∗8sec=7 msec64∗103The time required to transmit the packet is56∗8sec=896 μsec500∗103Propagation delay = 2 msec.The delay until decoding is7msec + 896μsec + 2msec = 9.896 msecA similar analysis shows that all bits experience a delay of 9.896 msec.16.(P9) Consider a packet of length L which begins at end system A, travels over one link to apacket switch, and travels from the packet switch over a second link to a destination end system. Let d i, s i, and R i denote the length, propagation speed, and the transmission rate of link i, for i= 1, 2. The packet switch delays each packet by d proc. Assuming no queuing delays, in terms of d i, s i, R i, (i= 1, 2), and L, what is the total end-to-end delay for the packet? Suppose now the packet Length is 1,000 bytes, the propagation speed on both links is 2.5 * 108m/s, the transmission rates of both links is 1 Mbps, the packet switch processing delay is 2 msec, the length of the first link is 6,000 km, and the length of the last link is 3,000 km. For these values, what is the end-to-end delay?Answer: The first end system requires L/R1to transmit the packet onto the first link; the packet propagates over the first link in d1/s1; the packet switch adds a processing delay ofd proc; after receiving the entire packet, the packet switch requires L/R2to transmit the packetonto the second link; the packet propagates over the second link in d2/s2. Adding these five delays givesd end-end = L/R1 + L/R2 + d1/s1 + d2/s2 + d procTo answer the second question, we simply plug the values into the equation to get 8 + 8 +24 + 12 + 2 = 54 msec.17.(P10) In the above problem, suppose R1 = R2 = R and d proc= 0. Further suppose the packetswitch does not store-and-forward packets but instead immediately transmits each bit it receivers before waiting for the packet to arrive. What is the end-to-end delay?Answer: Because bits are immediately transmitted, the packet switch does not introduce any delay; in particular, it does not introduce a transmission delay. Thus,d end-end = L/R + d1/s1 + d2/s2For the values in Problem 9, we get 8 + 24 + 12 = 44 msec.18.(P11) Suppose N packets arrive simultaneously to a link at which no packets are currentlybeing transmitted or queued. Each packet is of length L and the link has transmission rate R.What is the average queuing delay for the N packets?Answer:The queuing delay is 0 for the first transmitted packet, L/R for the second transmitted packet, and generally, (n-1)L/R for the nth transmitted packet. Thus, the average delay for the N packets is(L/R + 2L/R + ....... + (N-1)L/R)/N = L/RN(1 + 2 + ..... + (N-1)) = LN(N-1)/(2RN) = (N-1)L/(2R)Note that here we used the well-known fact that1 +2 + ....... + N = N(N+1)/219.(P14) Consider the queuing delay in a router buffer. Let I denote traffic intensity; that is, I =La/R. Suppose that the queuing delay takes the form IL/R (1-I) for I<1.a.Provide a formula for the total delay, that is, the queuing delay plus the transmissiondelay.b.Plot the total delay as a function of L/R.Answer:a.The transmission delay is L / R . The total delay isILR(1−I)+LR=L/R1−Ib.Let x = L / R.Total delay=x 1−αx20.(P16) Perform a Traceroute between source and destination on the same continent at threedifferent hours of the day.a.Find the average and standard deviation of the round-trip delays at each of the threehours.b.Find the number of routers in the path at each of the three hours. Did the pathschange during any of the hours?c.Try to identify the number of ISP networks that the Traceroute packets pass throughfrom source to destination. Routers with similar names and/or similar IP addresses should be considered as part of the same ISP. In your experiments, do the largest delays occur at the peering interfaces between adjacent ISPs?d.Repeat the above for a source and destination on different continents. Compare theintra-continent and inter-continent results.Answer: Experiments.21.(P18) Suppose two hosts, A and B, are separated by 10,000 kilometers and are connectedby a direct link of R = 2 Mbps. Suppose the propagation speed over the link is 2.5•108 meters/sec.a.Calculate the bandwidth-delay product, R •d prop.b.Consider sending a file of 400,000 bits from Host A to Host B. Suppose the file is sentcontinuously as one large message. What is the maximum number of bits that will be in the link at any given time?c.Provide an interpretation of the bandwidth-delay product.d.What is the width (in meters) of a bit in the link? Is it longer than a football field?e.Derive a general expression for the width of a bit in terms of the propagation speed s,the transmission rate R, and the length of the link m.Answer:a.d prop = 107 / 2.5•108 = 0.04 sec; so R •d prop = 80,000bitsb.80,000bitsc.The bandwidth-delay product of a link is the maximum number of bits that can be in thelink.d. 1 bit is 125 meters long, which is longer than a football fielde.m / (R •d prop ) = m / (R * m / s) = s/R22.(P20) Consider problem P18 but now with a link of R = 1 Gbps.a.Calculate the bandwidth-delay product, R·d prop .b.Consider sending a file of 400,000 bits from Host A to Host B. Suppose the file is sentcontinuously as one big message. What is the maximum number of bits that will be inthe link at any given time?c.What is the width (in meters) of a bit in the link?Answer:a.40,000,000 bits.b.400,000 bits.c.0.25 meters.23.(P21) Refer again to problem P18.a.How long does it take to send the file, assuming it is sent continuously?b.Suppose now the file is broken up into 10 packet is acknowledged by the receiver andthe transmission time of an acknowledgment packet is negligible. Finally, assumethat the sender cannot send a packet until the preceding one is acknowledged. Howlong does it take to send the file?pare the results from (a) and (b).Answer:a. d trans + d prop = 200 msec + 40 msec = 240 msecb.10 * (t trans + 2 t prop ) = 10 * (20 msec + 80 msec) = 1.0 sec。

数控技术与数控机床学习通课后章节答案期末考试题库2023年1.数控机床总的发展趋势是（）参考答案:工序集中###高速、高效、高精度###提高可靠性2.半闭环进给伺服系统的数控机床，其定位精度主要取决于（）参考答案:检测装置的精度###机床传动机构的精度3.目前广泛采用的程序段格式是地址符可变长度程序段格式。

参考答案:对4.零件程序是按程序段号的顺序执行的，而与程序段的输入顺序无关参考答案:错5.规定空间直角坐标系X、Y、Z三者的关系及其方向用（）判定参考答案:右手法则6.辅助功能由地址字M和其后两位数字组成，主要用于（）参考答案:控制机床各种辅助功能的开关动作7.G41和G42是半径补偿指令。

刀具加工工件时，若沿着刀具前进方向观察，刀具偏移在工件轮廓右边，则选用G41；刀具偏移在工件轮廓的左边，则选G42。

参考答案:错8.辅助功能M03代码表示（）参考答案:主轴顺时转动9.在加工过程中工件坐标系是静态的，即工件坐标系一旦建立或选定便一直有效，不得更改。

参考答案:错10.设H01=6mm，G91 G43 G01 Z-15；执行后的实际移动量为（）参考答案:9mmC系统是指（）参考答案:计算机数控系统12.机床零点与机床参考点是数控机床上同一点的不同说法。

参考答案:错13.零件程序是按（）的顺序执行的。

参考答案:程序段的输入先后14.一个完整的数控加工程序由程序编号、程序内容和程序结束三个部分组成。

参考答案:对15.属于点位控制系统的数控机床是（）。

参考答案:数控钻床16.圆弧插补程序中，若采用圆弧半径R编程时，从起始点到终点存在两条圆弧线段，当（）时，用-R表示圆弧半径。

参考答案:圆弧大于180°17.在数控机床编程中，进给速度指令是以字母开头的。

主轴转速指令是以字母开头的。

刀具号指令是以字母开头的参考答案:F, S, T18.在直线进给时，F指令所指定的速度是（）参考答案:各进给轴合成的进给速度19.( )坐标是用来确定工件坐标系的基本坐标系，其原点称为( )。

数控技术答案第三版数控技术答案第三版【篇一：数控技术试题及答案3】/p> 在每小题列出的四个备选项中只有一个是符合题目要求的，请将其代码填写在题后的括号内。

错选、多选或未选均无分。

1．脉冲编码器输出的信号类型是【】a．脉冲串 b.电压 c．电流d．相位2．步进电机每接收一个电脉冲信号，相应转过的一个固定角度称为【】a．齿距角 b．步距角 c.转角 d．夹角3．加工工件同一轮廓时采用不同直径的刀具，或同一把刀具因磨损后使直径发生变化的情况下，为了不改变已经编制好的程序，这时可以调用数控系统的【】a．刀具长度补偿功能 b．刀具半径补偿功能c．刀具位置补偿功能d．刀尖圆弧半径补偿功能4．插补圆弧时，逼近给定的圆弧曲线采用【】a微小圆弧 b．微小曲线 c．微小角度d．微小直线5．步进电机输人的脉冲频率越高，输出的电磁转矩【】a．越小 b．越大 c．相等 d．无关6.多微处理器的结构特点是【】a．多任务集中控制、分时处理 b.多任务并行处理c.单任务分时处理d.结构简单7.数控系统的管理软件不包括【】a输入 b．i／0处理 c.显示 d．插补8水磁式直流伺服电机常用的调述装置是【】a.晶体管脉冲宽度调速系统 b .正弦波脉冲宽度调速系统c.交流变频凋速系统d.步进电机凋速系统9.大量粗车削外园车刀之主偏角一般宜选用：【】(a)0； (b)20；（c）30；（d）45 度。

10.过定位是指定位时工件的同一【】被两个定位元件重复限制的定位状态。

(a)平面； (b)自由度； (c)圆柱面；(d)方向。

二、填空题(本大题共15空，每空2分，共30分)请在每小题的空格中填上正确答案。

错填、不填均无分。

1.数控系统按功能水平进行分类．有__________、__________和__________三种形式。

2．主轴速度功能由字母__________加__________表示，单位是r／min。

3.数控系统的控制轴数指的是__________。

《数控技术》思考与练习《数控技术》习题集数控技术各章节习题第一章思考题：1、数控机床、数控技术的基本概念2、数控机床的特点、组成、各部分作用、分类3、数控机床的加工原理、使用范围4、数控技术的发展趋势5、什么是点位控制、直线控制、轮廓控制数控机床？6、按伺服系统的控制原理数控机床分哪几类？7、数控机床的发展趋势主要有哪些？8、如何提高数控机床的精度、速度和可靠性C装置的主要技术指标有哪些？第二章思考题一、判断题1．（）立铣刀的刀位点是刀具中心线与刀具底面的交点。

2．（）球头铣刀的刀位点是刀具中心线与球头球面交点。

3．（）由于数控机床的先进性，因此任何零件均适合在数控机床上加工。

4．（）换刀点应设置在被加工零件的轮廓之外，并要求有一定的余量。

5．（）为保证工件轮廓表面粗糙度，最终轮廓应在一次走刀中连续加工出来。

二、简答题1、何谓对刀点？2、何谓刀位点？3、何谓换刀点？4、数控工艺与传统工艺相比有哪些特点？5、数控编程开始前，进行工艺分析的目的是什么？6、如何从经济观点出发来分析何种零件在数控机床上加工合适？7、确定对刀点时应考虑哪些因素？8、指出立铣刀、球头铣刀和钻头、车刀和镗刀的刀位点。

9、什么是数控编程？简要说明数控编程的内容与步骤10、准备功能（G）与辅助功能（M）在数控编程中的作用如何？11、M00、M01、M02、M30的区别在哪里？12、编程中如何选择对刀点、换刀点？13、简述数控机床的坐标系及编程特点。

14、什么是“字地址程序段各式”有何特点？15、数控机床的X、Y、Z坐标轴及其方向如何确定？16、数控机床坐标系和工件坐标系之间的关系如何？17、F、S、T功能指令各自的作用？18、G00、G01、G02、G03的不同点在哪里？19、什么是零件的加工路线？确定加工路线应遵循的原则？20、什么是基点和节点？21、G40、G41、G42、G43、G44的含义如何22、何谓模态代码和非模态代码？举列说明23、编写该零件的精加工程序第三章思考题1、数控机床对位置检测装置的要求有哪些？2、位置检测装置可按哪些方式分类？3、分析感应同步器与旋转变压器的结构特点。