Council of Legislation 2020/23 Election
Election of the District 1080 Representative
for the Council of Legislation 2020/23
Click here to cast your club’s vote Click here to download a pdf copy
Every district selects a representative to serve at the Councils on Resolutions and the Council on Legislation that take place during their Council term from 1 July 2020 through 30 June 2023.
The Council representatives will represent their district at the:
• 2020, 2021, and 2022 Councils on Resolutions
• 2022 Council on Legislation
Representatives and alternates must be selected by 30 June 2020 and reported to Rotary International. Selection requirements and duties for representatives can be found in Article 9 of the RI Bylaws.
To serve as representative, a candidate must:
• Be a member of a club in the district;
• Have served a full term as District Governor at the time of election;
• Be able to complete mandatory Council training;
• Be digitally literate for communication, reviewing documents, and voting; and
• Be able to attend the Council on Legislation for its entire duration.
Duties of a Council representative are as follows:
• Assist clubs in preparing proposed resolutions and enactments;
• Discuss proposed resolutions and enactments at district meetings;
• Be knowledgeable of the attitudes of Rotarians within the district;
• Give critical consideration to all proposed resolutions and enactments and effectively communicate their views on such proposals to the Councils;
• Act as an objective legislator;
• Participate in the Councils on Resolutions;
• Attend the Council on Legislation for its full duration; and
• Report on the deliberations of the Councils to the clubs in their district following the meetings of the Councils.
Representatives are also required to complete an online representative course and Council training at their 2021 Rotary Institute in preparation for the 2022 Council on Legislation. Their participation at the Institute is not funded by Rotary International.
Selecting a Representative
The selection of the representative and alternate must be conducted in the 2019-2020 Rotary year and reported to Rotary by 30 June 2020 using the online representative selection form. Procedures for the selection can be found in Article 9 of the RI Bylaws.
Details of the Past District Governors who have indicated an interest in standing as the District 1080 Council of Legislation Representative are shown in alphabetical order:
Ray Burman Swaffham
Barry Catchpole Norwich St Edmund
Tom Griffin Woodbridge Deben
Robert Lovick Great Yarmouth Haven
Nigel Neville Cambridge
David Simpson Woodbridge Deben
The representative and alternate will be elected by the Single Transferable Voting system through an online ballot.
RIBI VOTING RULES
By District Councils – “The RIBI Rules”
The members of the District Council entitled to a vote are:
The District Governor
Past Governors (of any district) holding membership (other than honorary membership) of a club in the district
The District Governor elect
The District Governor nominee
The Assistant Governor(s)
The District Secretary
The District Treasurer
The chairmen of the committees appointed by the District Council
The representatives of the clubs
For voting purposes, Clubs get the same number of votes as they do for a District Council Meeting and must cast all their votes for a single candidate.
District Officers and PDGs are to vote individually and not as part of their club vote.
Each club is entitled to one representative, and one further representative for every 25 or major fraction thereof of its active members.
1-12 members = 1 vote
13-37 members = 2 votes
38-62 members = 3 votes
63-87 members = 4 votes
88-112 members = 5 votes
Voting will close on Monday 18th May 2020. Votes received after that date will not be counted.
Notes on the Single Transferable Vote
As our election system in Rotary is largely based on what is known as the “single transferable vote” system, it is most important that you understand how this system works. It may also be used by districts if there are three or more nominees for District Office. Let us assume that you have to select one candidate from among five nominees for a Rotary office. The five nominees, let us say, are Rotarians Black, Brown, Green, Grey and White. Their names will be printed in that order on a voting slip and you will be required to vote by putting a “1” against the candidate of your choice, a “2” against your second choice, a “3” against your third choice, and so on. You must not vote with a cross. You will be wise to number every candidate but need not do so if there are some about whom you have no opinion.
Let us assume that 100 valid votes are cast. The scrutineers will then sort out these 100 into five piles, each containing all the votes cast as first choice for a particular candidate. After counting, let us say the result is as follows:
Green 33 (i.e., 33 papers with a “1″ against Green)
White – 27 (i.e., 27 papers with a “1” against White)
Grey 15 and so on
None has secured a majority over all the others combined, so some might urge that the leader, Green, should be elected and others that he should not on the grounds that, for instance, if Black had not stood all Black’s supporters would have voted for White, putting him ahead. To find out whether this is so or not, we ask Black’s supporters what they would have done, Black is declared defeated and each of his papers is transferred to the candidate that voter has marked “2”. The distributed votes are then counted and added to the first figures. We might then get this:
Green 33 + 2 = 35 (Two delegates voted Black 1, Green 2)
White 27 + 3 = 30 (Three delegates voted Black 1, White 2)
Brown 15 + 3 = 18 (Three delegates voted Black 1, Brown 2)
Grey – 15 + 2 = 17 (Two delegates voted Black 1, Grey 2)
Grey is now eliminated and each of his votes is transferred to whichever of the three remaining candidates is marked by that voter as his next choice. For the original 15 papers this will usually be the second choice, but if there is any paper marked 1 for Grey and 2 for Black it will go to the third choice; the additional votes picked up from Black will also go to the third choice. After counting again we would reach this result:
White – 30 + 8 = 38 (Eight delegates voted Grey 1, White 2)
Green – 35 + 2 = 37 (Two delegates voted Grey 1, Green 2)
Brown 18 + 6 = 24 (Four delegates voted Grey 1, Brown 2 and two voted Black 1, Grey 2, Brown 3)
Non transferable – 1 (One delegate voted Grey 1 and no more)
Nobody has yet secured an absolute majority over all the others combined. Had the result after the second elimination been:
Green 35 + 16 = 51;
White 30 + 0 = 30;
Brown 18 + 1 = 19
there would have been no need to eliminate Brown and redistribute his votes. Green would have had an absolute majority and if all Brown’s 19 voters had White as their next choice, White could not overtake Green’s 51 votes.
To complete our example Brown’s votes are reallocated each being transferred to Green or White according to which is marked by that voter as his earlier preference. After the last count the position might be:
White 38 + 13 = 51 (Thirteen delegates put White before Green)
Green 37 + 11 = 48 (Eleven delegates put Green before White)
Non transferable 1
The one non transferable paper could not have altered the result but had there been more they might have done; it is unwise to stop numbering so long as there are any candidates left about whom you have any opinion.
Suppose the final result had been a tie between Green and White, then whichever of those two had received the more votes on the first count would have been declared elected. If they were equal on the first count, the one leading on the second count would have been declared elected, and so on. Similarly, if there is a tie for bottom place, that one of the equal candidates is eliminated who was lowest on the first count (or on the earliest count in which they were unequal). In the unlikely event of their tying all the way through, the returning officer must draw lots.