Mary Staniforth b. 11 November 1621

From Rodovid EN
Person:93754
Lineage Staniforth
Sex Female
Full name (at birth) Mary Staniforth
Parents

♂ Thomas Staniforth b. 7 March 1593

♀ Anne

Events

11 November 1621 birth: Eckington (Derbyshire), England


From grandparents to grandchildren

Grandparents
♂ Ralph Staniforth
birth: 1557, Eckington (Derbyshire), England
♀ Dorothy Staniforth
birth: 1558, Eckington (Derbyshire), England
♀ Elizabeth Staniforth
birth: 5 December 1559, Eckington (Derbyshire), England
♂ John Staniforth
birth: 14 March 1560, Eckington (Derbyshire), England
♂ William Staniforth
birth: 27 March 1563, Eckington (Derbyshire), England
♂ Robert Staniforth
birth: 31 August 1567, Eckington (Derbyshire), England
♀ Agnes Staniforth
birth: 31 May 1570, Eckington (Derbyshire), England
♂ Thomas Staniforth
birth: 1556
marriage: ♀ Margaret Scotte , Eckington (Derbyshire), England
♀ Margaret Scotte
birth: 1564, Eckington (Derbyshire), England
marriage: ♂ Thomas Staniforth , Eckington (Derbyshire), England
Grandparents
Parents
♀ Grace Staniforth
birth: 8 October 1587, Eckington (Derbyshire), England
♀ Ann Staniforth
birth: 5 March 1589, Eckington (Derbyshire), England
♀ Jane Staniforth
birth: 5 March 1589, Eckington (Derbyshire), England
♂ Thomas Staniforth
birth: 7 March 1593, Eckington (Derbyshire), England
marriage: ♀ Anne
Parents
 
== 3 ==
♂ George Staniforth
birth: 27 November 1625, Eckington (Derbyshire), England
marriage: ♀ Dorothy
♀ Anne Staniforth
birth: 2 June 1618, Eckington (Derbyshire), England
♀ Elizabeth Staniforth
birth: 26 December 1619, Eckington (Derbyshire), England
♀ Troth Staniforth
birth: 1627, Eckington (Derbyshire), England
♀ Anne Staniforth
birth: 8 June 1628, Eckington (Derbyshire), England
♀ Jane Staniforth
birth: 27 July 1629, Eckington (Derbyshire), England
♀ Margaret Staniforth
birth: 29 August 1630, Eckington (Derbyshire), England
♀ Jane Staniforth
birth: 25 August 1633, Eckington (Derbyshire), England
♀ Mary Staniforth
birth: 11 November 1621, Eckington (Derbyshire), England
== 3 ==
Scotte
Staniforth