Frida Winnerstrand

Frida Winnerstrand
Frida Winnerstrand
	Winnerstrand in 1908.
Born	Karolina Alfrida Kumlin; 23 January 1881; Lövsta, Järfälla, Sweden
Died	1 December 1943 (aged 62); Stockholm, Sweden
Nationality	Swedish
Occupation	Actress
Years active	1901–1943
Spouse(s)	Olof Winnerstrand (m. 1906⁠–⁠1943)

Karolina Alfrida ("Frida") Winnerstrand, née Kumlin (23 January 1881 – 1 December 1943) was a Swedish actress.

Biography

Born in Lövsta, Järfälla, Stockholm, Frida Winnerstrand made her debut 1896 in the Anna Lundbergs Theatre Company and later studied acting at the Royal Dramatic Theatre's acting school (Dramatens elevskola) in Stockholm 1900-1901. Later she joined the Hillberg Company (1901–02) and the Selander Company (1902–03) on tour. Performed at Albert Ranft's theatre, old Östermalmsteatern 1903, at the Swedish Theatre 1904 and from 1905-19 at Vasateatern. From 1919 to her death in 1943 she worked at the Royal Dramatic Theatre in Stockholm where some of her most notable parts were as Lady Chiltern in Oscar Wildes An Ideal Husband (1919), Adelaide i play Johan Ulfstjerna, Julia Swedenhielm in Hjalmar Bergmans Swedenhielms (1925), Essie Miller in Eugene O'Neills Ah, Wilderness! (1935–36), Mrs Heyst in Strindbergs Easter and as Mor i Falla in Selma Lagerlöfs Kejsarn av Portugallien (The Emperor of Portugal) (1939).

Frida Winnerstrand made her film debut in 1920 in Pauline Brunius silent short film De läckra skaldjuren, and played altogether 10 parts on film (only three talkies). She was since 1906 married to Swedish actor Olof Winnerstrand. They often appeared together on stage and played opposite each other in the popular long-running radio series Familjen Björck (The Bjorck Family), as Mr and Mrs Bjorck (that aired in Swedish Radio, SR, between 1936-1943), and later also in the popular film adaption of the series, shot in 1940.

Selected filmography

Familjen Björck (The Bjorck Family) (1940)
Thunder and Lightning (1938)
Äventyret (1936)
De läckra skaldjuren (1920)

gollark: Did you just randomly decide to calculate that?

gollark: Well, you can, or also "it would have about the same mass as the atmosphere".

gollark: Wikipedia says that spider silk has a diameter of "2.5–4 μm", which I approximated to 3μm for convenience, so a strand has a 1.5μm radius. That means that its cross-sectional area (if we assume this long thing of spider silk is a cylinder) is (1.5e-6)², or ~7e-12. Wikipedia also says its density is about 1.3g/cm³, which is 1300kg/m³, and that the observable universe has a diameter of 93 billion light-years (8.8e26 meters). So multiply the length of the strand (the observable universe's diameter) by the density of spider silk by the cross-sectional area of the strand and you get 8e18 kg, while the atmosphere's mass is about 5e18 kg, so close enough really.

gollark: Okay, so by mass it actually seems roughly correct.

gollark: So, spider silk comes in *very* thin strands and is somewhat denser than water, interesting.

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